Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 13 Section And Mid Point Formula

Exercise 13A

Question 1. Calculate the co-ordinates of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1: 2.
(ii) A (-4, 6) and B (3, -5) in the ratio 3: 2.
Answer:
(i) Let the co-ordinates of the point P be (x, y).
Using section formula:
\( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{1 \times 5 + 2 \times 1}{1 + 2} = \frac{7}{3} \)

\( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{1 \times 9 + 2 \times 3}{1 + 2} = \frac{15}{3} = 5 \)

Thus, the co-ordinates of point P are \( \left(\frac{7}{3}, 5\right) \).

(ii) Let the co-ordinates of the point P be (x, y).
\( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{3 \times 3 + 2 \times (-4)}{3 + 2} = \frac{1}{5} \)

\( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{3 \times (-5) + 2 \times 6}{3 + 2} = \frac{-3}{5} \)

Thus, the co-ordinates of point P are \( \left(\frac{1}{5}, -\frac{3}{5}\right) \).
In simple words: We use the section formula to find where a point divides a line. We put the given ratios and coordinates into the formula to get the answer.

📝 Teacher's Note: Show students the section formula on the board first. Make them write it down. Then substitute values step by step. This helps them remember the pattern.

🎯 Exam Tip: Always write "Let the coordinates be (x, y)" first. Then write the section formula clearly. Show each step of calculation to get full marks.

Question 2. In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis.
Answer: Let the line joining points A (2, -3) and B (5, 6) be divided by point P (x, 0) in the ratio k: 1.
Using section formula for y-coordinate:
\( y = \frac{ky_2 + y_1}{k + 1} \)

\( 0 = \frac{k \times 6 + 1 \times (-3)}{k + 1} \)

\( 0 = 6k - 3 \)

\( k = \frac{1}{2} \)

Thus, the required ratio is 1: 2.
In simple words: When a line crosses the x-axis, the y-coordinate becomes 0. We use this fact in the section formula to find the ratio.

📝 Teacher's Note: Remind students that x-axis means y = 0, and y-axis means x = 0. This is the key to solving these problems easily.

🎯 Exam Tip: Write "Point on x-axis has coordinates (x, 0)" first. This shows the examiner you understand the concept. Then use section formula.

Question 3. In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis.
Answer: Let the line joining points A (2, -4) and B (-3, 6) be divided by point P (0, y) in the ratio k: 1.
Using section formula for x-coordinate:
\( x = \frac{kx_2 + x_1}{k + 1} \)

\( 0 = \frac{k \times (-3) + 1 \times 2}{k + 1} \)

\( 0 = -3k + 2 \)

\( k = \frac{2}{3} \)

Thus, the required ratio is 2: 3.
In simple words: When a line crosses the y-axis, the x-coordinate becomes 0. We put this in the section formula to find the ratio.

📝 Teacher's Note: Draw a simple diagram showing the line crossing y-axis. This helps students see why x = 0 at the y-axis.

🎯 Exam Tip: Write "Point on y-axis has coordinates (0, y)" clearly. Then substitute in the x-coordinate formula only.

Question 4. In what ratio does the point (1, a) divided the join of (-1, 4) and (4, -1)? Also, find the value of a.
Answer: Let the point P (1, a) divides the line segment AB in the ratio k: 1.
Using section formula for x-coordinate:
\( 1 = \frac{4k + (-1)}{k + 1} \)

\( k + 1 = 4k - 1 \)

\( 2 = 3k \)

\( k = \frac{2}{3} \) ...(1)

Using section formula for y-coordinate:
\( a = \frac{-k + 4}{k + 1} \)

From equation (1), \( k = \frac{2}{3} \):
\( a = \frac{-\frac{2}{3} + 4}{\frac{2}{3} + 1} = \frac{\frac{10}{3}}{\frac{5}{3}} = 2 \)

Hence, the required ratio is 2: 3 and the value of a is 2.
In simple words: We know one coordinate of the dividing point. We use this to find the ratio first, then use the ratio to find the other coordinate.

📝 Teacher's Note: Show students that we use the known coordinate (x = 1) first to find k. Then we use this k value to find the unknown coordinate a.

🎯 Exam Tip: Solve for the ratio using the known coordinate first. Write "From equation (1)" when substituting k value. This shows clear working.

Question 5. In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)? Also, find the value of a.
Answer: Let the point P (a, 6) divides the line segment joining A (-4, 3) and B (2, 8) in the ratio k: 1.
Using section formula for y-coordinate:
\( 6 = \frac{8k + 3}{k + 1} \)

\( 6k + 6 = 8k + 3 \)

\( 3 = 2k \)

\( k = \frac{3}{2} \) ...(1)

Using section formula for x-coordinate:
\( a = \frac{2k + (-4)}{k + 1} \)

From equation (1):
\( a = \frac{2 \times \frac{3}{2} - 4}{\frac{3}{2} + 1} = \frac{3 - 4}{\frac{5}{2}} = \frac{-1}{\frac{5}{2}} = -\frac{2}{5} \)

Hence, the required ratio is 3: 2 and the value of a is \( -\frac{2}{5} \).
In simple words: We use the known y-coordinate (y = 6) to find the ratio. Then we use this ratio to find the unknown x-coordinate a.

📝 Teacher's Note: Emphasize that we can use either known coordinate to find the ratio. Here we used y = 6 because it was easier to work with.

🎯 Exam Tip: Choose the coordinate that gives simpler numbers to work with. Show clear substitution steps to avoid calculation errors.

Question 6. In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Answer: Let the point P (x, 0) on x-axis divides the line segment joining A (4, 3) and B (2, -6) in the ratio k: 1.
Using section formula for y-coordinate:
\( 0 = \frac{-6k + 3}{k + 1} \)

\( 0 = -6k + 3 \)

\( k = \frac{1}{2} \)

Thus, the required ratio is 1: 2.
Also, using section formula for x-coordinate:
\( x = \frac{2k + 4}{k + 1} = \frac{2 \times \frac{1}{2} + 4}{\frac{1}{2} + 1} = \frac{1 + 4}{\frac{3}{2}} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \)

Thus, the required co-ordinates of the point of intersection are \( \left(\frac{10}{3}, 0\right) \).
In simple words: First we find the ratio using y = 0. Then we use this ratio to find the exact x-coordinate where the line crosses the x-axis.

📝 Teacher's Note: Remind students that intersection with x-axis always has y = 0. This makes the y-coordinate calculation very simple.

🎯 Exam Tip: Always find the ratio first, then find the intersection point. Write both answers clearly - the ratio and the coordinates.

Question 7. Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.
Answer: Let S (0, y) be the point on y-axis which divides the line segment PQ in the ratio k: 1.
Using section formula for x-coordinate:
\( 0 = \frac{3k + (-4)}{k + 1} \)

\( 0 = 3k - 4 \)

\( 3k = 4 \)

\( k = \frac{4}{3} \) ...(1)

Using section formula for y-coordinate:
\( y = \frac{0 + 7}{k + 1} = \frac{7}{\frac{4}{3} + 1} = \frac{7}{\frac{7}{3}} = 3 \)

Hence, the required ratio is 4: 3 and the required point is S(0, 3).
In simple words: We use x = 0 to find the ratio because the line crosses the y-axis. Then we use this ratio to find where exactly it crosses.

📝 Teacher's Note: Draw the y-axis on the board and mark the point (0, 3). Show students how this point divides the line segment in the given ratio.

🎯 Exam Tip: For y-axis intersection, use x = 0 in the section formula. Always write both the ratio and the intersection coordinates as final answers.

Question 8. Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Answer: The line segment from P(5, -10) to O(0, 0) is divided into five equal parts by points A, B, C, and D.

Point A divides PO in the ratio 1: 4.
Co-ordinates of point A are:
\( \left(\frac{1 \times 0 + 4 \times 5}{1 + 4}, \frac{1 \times 0 + 4 \times (-10)}{1 + 4}\right) = \left(\frac{20}{5}, \frac{-40}{5}\right) = (4, -8) \)

Point B divides PO in the ratio 2: 3.
Co-ordinates of point B are:
\( \left(\frac{2 \times 0 + 3 \times 5}{2 + 3}, \frac{2 \times 0 + 3 \times (-10)}{2 + 3}\right) = \left(\frac{15}{5}, \frac{-30}{5}\right) = (3, -6) \)

Point C divides PO in the ratio 3: 2.
Co-ordinates of point C are:
\( \left(\frac{3 \times 0 + 2 \times 5}{3 + 2}, \frac{3 \times 0 + 2 \times (-10)}{3 + 2}\right) = \left(\frac{10}{5}, \frac{-20}{5}\right) = (2, -4) \)

Point D divides PO in the ratio 4: 1.
Co-ordinates of point D are:
\( \left(\frac{4 \times 0 + 1 \times 5}{4 + 1}, \frac{4 \times 0 + 1 \times (-10)}{4 + 1}\right) = \left(\frac{5}{5}, \frac{-10}{5}\right) = (1, -2) \)

In simple words: When we divide a line into 5 equal parts, we get 4 dividing points. Each point divides the line in a different ratio - 1:4, 2:3, 3:2, and 4:1.

📝 Teacher's Note: Draw a line segment and mark all 5 equal parts. Show students how point A is 1 part from P and 4 parts from O, so ratio is 1:4.

🎯 Exam Tip: For n equal parts, you get (n-1) dividing points. Write the ratios clearly: 1:(n-1), 2:(n-2), etc. Use section formula for each point.

Question 9. The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that \( \frac{PB}{AB} = \frac{1}{5} \). Find the co-ordinates of P.
Answer:
Let the co-ordinates of point P are (x, y).

Given: PB : AB = 1 : 5
∴ PB : PA = 1 : 4
Coordinates of P are
\( (x, y) = \left(\frac{4 \times (-2) + 1 \times (-3)}{5}, \frac{4 \times 6 + 1 \times (-10)}{5}\right) = \left(-\frac{11}{5}, \frac{14}{5}\right) \)

In simple words: We use the section formula to find where point P divides the line. We know the ratio and the end points, so we can find P's coordinates.

📝 Teacher's Note: Draw a line segment on the board. Show how P divides AB in the ratio 4:1 internally. Students often mix up the ratios - remind them PB:AB = 1:5 means PB:PA = 1:4.

🎯 Exam Tip: Always write the given ratio clearly first. Then use the section formula. Show all substitution steps. Don't forget to simplify fractions in your final answer.

Question 10. P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the coordinates of P.
Answer:
5AP = 2BP
\( \frac{AP}{BP} = \frac{2}{5} \)
The co-ordinates of the point P are
\( \left(\frac{2 \times (-2) + 5 \times 4}{2 + 5}, \frac{2 \times 6 + 5 \times 3}{2 + 5}\right) \)
\( = \left(\frac{16}{7}, \frac{27}{7}\right) \)

In simple words: We find the ratio AP:BP first, then use the section formula. Point P divides AB in the ratio 2:5 internally.

📝 Teacher's Note: Show students how to convert 5AP = 2BP into the ratio AP:BP = 2:5. This step confuses many students. Practice with more examples.

🎯 Exam Tip: First convert the given condition into a simple ratio. Then apply section formula carefully. Write fractions clearly - don't convert to decimals.

Question 11. Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Answer:
The co-ordinates of every point on the line x = 2 will be of the type (2, y).
Using section formula, we have:

\( x = \frac{m_1 \times 5 + m_2 \times (-3)}{m_1 + m_2} \)

\( 2 = \frac{5m_1 - 3m_2}{m_1 + m_2} \)

2m₁ + 2m₂ = 5m₁ - 3m₂
5m₂ = 3m₁
\( \frac{m_1}{m_2} = \frac{5}{3} \)
Thus, the required ratio is 5: 3.

\( y = \frac{m_1 \times 7 + m_2 \times (-1)}{m_1 + m_2} \)

\( y = \frac{5 \times 7 + 3 \times (-1)}{5 + 3} \)

\( y = \frac{35 - 3}{8} = \frac{32}{8} = 4 \)

Thus, the required co-ordinates of the point of intersection are (2, 4).

In simple words: The vertical line x = 2 cuts the line segment. We find where it cuts using the section formula. The ratio is 5:3 and the intersection point is (2, 4).

📝 Teacher's Note: Draw the coordinate plane. Show the two points and the vertical line x = 2. Mark where they intersect. This visual helps students understand the problem.

🎯 Exam Tip: Remember that points on line x = 2 have coordinates (2, y). Use this fact in the section formula. Always find both the ratio and intersection point.

Question 12. Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Answer:
The co-ordinates of every point on the line y = 2 will be of the type (x, 2).
Using section formula, we have:

\( y = \frac{m_1 \times (-3) + m_2 \times 5}{m_1 + m_2} \)

\( 2 = \frac{-3m_1 + 5m_2}{m_1 + m_2} \)

2m₁ + 2m₂ = -3m₁ + 5m₂
5m₁ = 3m₂
\( \frac{m_1}{m_2} = \frac{3}{5} \)
Thus, the required ratio is 3: 5.

In simple words: The horizontal line y = 2 cuts the line segment AB. Using the section formula, we find it divides AB in the ratio 3:5.

📝 Teacher's Note: This is similar to Question 11, but now we have a horizontal line y = 2 instead of vertical line x = 2. Show the difference on the graph.

🎯 Exam Tip: For line y = 2, all points have form (x, 2). Use the y-coordinate in section formula. The method is same as previous question but with y instead of x.

Question 13. The point P (5, -4) divides the line segment AB, as shown in the figure, in the ratio 2: 5. Find the co-ordinates of points A and B.
Answer:
Point A lies on x-axis. So, let the co-ordinates of A be (x, 0).
Point B lies on y-axis. So, let the co-ordinates of B be (0, y).
P divides AB in the ratio 2: 5.
We have:
\( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \)

\( 5 = \frac{2 \times 0 + 5 \times x}{2 + 5} \)

\( 5 = \frac{5x}{7} \)

x = 7
Thus, the co-ordinates of point A are (7, 0).

\( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \)

\( -4 = \frac{2 \times y + 5 \times 0}{2 + 5} \)

\( -4 = \frac{2y}{7} \)

\( -2 = \frac{y}{7} \)

y = -14
Thus, the co-ordinates of point B are (0, -14).

In simple words: From the diagram, A is on the x-axis and B is on the y-axis. We use the section formula backwards to find A and B when we know P and the ratio.

📝 Teacher's Note: This is a reverse problem. We know the dividing point and ratio, but need to find the endpoints. Show students how A(x,0) means it's on x-axis and B(0,y) means it's on y-axis.

🎯 Exam Tip: When a point is on x-axis, its y-coordinate is 0. When on y-axis, its x-coordinate is 0. Use this fact to set up the section formula equations.

Question 14. Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Answer:
Let P and Q be the point of trisection of the line segment joining the points A (-3, 0) and B (6, 6).

So, AP = PQ = QB

We have AP: PB = 1: 2
Co-ordinates of the point P are
\( \left(\frac{1 \times 6 + 2 \times (-3)}{1 + 2}, \frac{1 \times 6 + 2 \times 0}{1 + 2}\right) \)

\( = \left(\frac{6 - 6}{3}, \frac{6}{3}\right) = (0, 2) \)

We have AQ: QB = 2: 1
Co-ordinates of the point Q are
\( \left(\frac{2 \times 6 + 1 \times (-3)}{2 + 1}, \frac{2 \times 6 + 1 \times 0}{2 + 1}\right) \)

\( = \left(\frac{9}{3}, \frac{12}{3}\right) = (3, 4) \)

In simple words: Trisection means dividing into three equal parts. We find two points P and Q that divide the line segment into three equal pieces.

📝 Teacher's Note: Draw a line segment and show how trisection creates three equal parts. Point P divides AB in ratio 1:2, and point Q divides AB in ratio 2:1.

🎯 Exam Tip: For trisection, always find both points. First point divides in ratio 1:2, second point divides in ratio 2:1. Don't confuse the ratios.

Question 15. Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Answer:
Let P and Q be the point of trisection of the line segment joining the points A (-5, 8) and B (10, -4).

So, AP = PQ = QB

We have AP: PB = 1: 2
Co-ordinates of the point P are
\( \left(\frac{1 \times 10 + 2 \times (-5)}{1 + 2}, \frac{1 \times (-4) + 2 \times 8}{1 + 2}\right) \)

\( = \left(\frac{10 - 10}{3}, \frac{12}{3}\right) = (0, 4) \)

So, point P lies on the y-axis.
We have AQ: QB = 2: 1
Co-ordinates of the point Q are
\( \left(\frac{2 \times 10 + 1 \times (-5)}{2 + 1}, \frac{2 \times (-4) + 1 \times 8}{2 + 1}\right) \)

\( = \left(\frac{20 - 5}{3}, \frac{-8 + 8}{3}\right) = (5, 0) \)

So, point Q lies on the x-axis.

Hence, the line segment joining the given points A and B is trisected by the co-ordinate axes.

In simple words: We find the two trisection points. One point is on the y-axis and the other is on the x-axis. This proves that the coordinate axes trisect the line segment.

📝 Teacher's Note: This is a special case where the trisection points happen to lie on the coordinate axes. Show this on a graph to make it clear to students.

🎯 Exam Tip: Calculate both trisection points. Show that one has x-coordinate 0 (on y-axis) and other has y-coordinate 0 (on x-axis). This proves the statement.

Question 16. Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Answer:
Let A and B be the point of trisection of the line segment joining the points P (2, 1) and Q (5, -8).

So, PA = AB = BQ

We have AP: PQ = 1: 2
Co-ordinates of the point P are
\( \left(\frac{1 \times 10 + 2 \times (-5)}{1 + 2}, \frac{1 \times (-4) + 2 \times 8}{1 + 2}\right) \)

\( = \left(\frac{10 - 10}{3}, \frac{12}{3}\right) = (0, 4) \)

So, point P lies on the y-axis.
We have AQ: QB = 2: 1
Co-ordinates of the point Q are
\( \left(\frac{2 \times 10 + 1 \times (-5)}{2 + 1}, \frac{2 \times (-4) + 1 \times 8}{2 + 1}\right) \)

\( = \left(\frac{20 - 5}{3}, \frac{-8 + 8}{3}\right) = (5, 0) \)

So, point Q lies on the x-axis.

In simple words: We need to check if A (3, -2) is indeed a trisection point, and then find the other trisection point using the section formula.

📝 Teacher's Note: First verify that A (3, -2) is actually a trisection point by checking if it divides PQ in ratio 1:2. Then find the other point that divides in ratio 2:1.

🎯 Exam Tip: To prove A is a trisection point, show it divides the line segment in ratio 1:2 or 2:1. Then find the other trisection point using the opposite ratio.

Question 17. If A = (-4, 3) and B = (8, -6)
(i) Find the length of AB.
(ii) In what ratio is the line joining A and B, divided by the x-axis?
Answer:
(i) A (-4, 3) and B (8, -6)
AB = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
= \( \sqrt{(8 + 4)^2 + (-6 - 3)^2} \)
= \( \sqrt{144 + 81} \)
= \( \sqrt{225} \)
= 15 units

(ii) Let P be the point which divides AB on the x-axis in the ratio k : 1.
Therefore, y-coordinate of P = 0.
\( \frac{-6k + 3}{k + 1} = 0 \)
\( \implies -6k + 3 = 0 \)
\( \implies k = \frac{1}{2} \)
∴ Required ratio is 1:2.

In simple words: First we use distance formula to find how far apart the two points are. Then we find where the line crosses the x-axis by setting y = 0 and solving for the ratio.

📝 Teacher's Note: Show students that distance formula is like using a ruler on graph paper. When line crosses x-axis, the y-value becomes zero - this is the key to finding ratios.

🎯 Exam Tip: Always write the distance formula first. For ratio problems, remember that on x-axis y = 0, and on y-axis x = 0. Show all working steps clearly.

Question 18. The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Answer:
Since, point L lies on y-axis, its abscissa is 0.
Let the co-ordinates of point L be (0, y). Let L divides MN in the ratio k: 1.

Using section formula, we have:
\( x = \frac{k \times (-3) + 1 \times 5}{k + 1} \)
\( 0 = \frac{-3k + 5}{k + 1} \)
-3k + 5 = 0
\( k = \frac{5}{3} \)

Thus, the required ratio is 5: 3.
Now, \( y = \frac{k \times 2 + 1 \times 7}{k + 1} \)
\( = \frac{\frac{5}{3} \times 2 + 7}{\frac{5}{3} + 1} \)
\( = \frac{10 + 21}{5 + 3} \)
\( = \frac{31}{8} \)

Therefore, coordinates of L are \( (0, \frac{31}{8}) \).

In simple words: When a line crosses the y-axis, the x-value is always 0. We use this fact to find where exactly it crosses and in what ratio it divides the line.

📝 Teacher's Note: Emphasize that abscissa means x-coordinate and ordinate means y-coordinate. On y-axis, x is always 0. Students often confuse these terms.

🎯 Exam Tip: Write "abscissa = 0" first, then apply section formula. Always state the final coordinates clearly with brackets. Show fraction calculations step by step.

Question 19. A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that AP: PB = AQ: QC = 1: 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = 1/3 BC.
Answer:
(i) Co-ordinates of P are
\( \left(\frac{1 \times (-1) + 2 \times 2}{1 + 2}, \frac{1 \times 2 + 2 \times 5}{1 + 2}\right) \)
\( = \left(\frac{3}{3}, \frac{12}{3}\right) \)
= (1, 4)

Co-ordinates of Q are
\( \left(\frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times 8 + 2 \times 5}{1 + 2}\right) \)
\( = \left(\frac{9}{3}, \frac{18}{3}\right) \)
= (3, 6)

(ii) Using distance formula, we have:
BC = \( \sqrt{(5 + 1)^2 + (8 - 2)^2} = \sqrt{36 + 36} = 6\sqrt{2} \)
PQ = \( \sqrt{(3 - 1)^2 + (6 - 4)^2} = \sqrt{4 + 4} = 2\sqrt{2} \)

Hence, PQ = \( \frac{1}{3} \) BC.

In simple words: We find points P and Q using section formula. Then we calculate distances using distance formula. PQ is exactly one-third of BC length.

📝 Teacher's Note: This shows the midpoint theorem in triangles. When points divide sides in same ratio, the line joining them is parallel to third side and has length equal to 1/3 of it.

🎯 Exam Tip: Use section formula carefully - numerator has cross multiplication. Always verify your answer by checking if PQ = (1/3)BC numerically. Write "Hence proved" at the end.

Question 20. A (-3, 4), B (3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2: 3.
Answer:
BP: PC = 2: 3
Co-ordinates of P are
\( \left(\frac{2 \times (-2) + 3 \times 3}{2 + 3}, \frac{2 \times 4 + 3 \times (-1)}{2 + 3}\right) \)
\( = \left(\frac{-4 + 9}{5}, \frac{8 - 3}{5}\right) \)
= (1, 1)

Using distance formula, we have:
AP = \( \sqrt{(1 + 3)^2 + (1 - 4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units

In simple words: First we find where point P is located on line BC using the given ratio. Then we find the distance from A to P using distance formula.

📝 Teacher's Note: Remind students that when finding coordinates using ratios, we use section formula. The ratio BP:PC means P divides BC in ratio 2:3.

🎯 Exam Tip: First find coordinates of P using section formula, then find AP using distance formula. Always write "units" after the final answer. Show all calculation steps clearly.

Question 21. The line segment joining A (2, 3) and B (6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Answer:
Since, point K lies on x-axis, its ordinate is 0.
Let the point K (x, 0) divides AB in the ratio k: 1.
We have,
\( y = \frac{k \times (-5) + 1 \times 3}{k + 1} \)
\( 0 = \frac{-5k + 3}{k + 1} \)
\( k = \frac{3}{5} \)

Thus, K divides AB in the ratio 3: 5.
Also, we have:
\( x = \frac{k \times 6 + 1 \times 2}{k + 1} \)
\( x = \frac{\frac{3}{5} \times 6 + 2}{\frac{3}{5} + 1} \)
\( x = \frac{18 + 10}{3 + 5} \)
\( x = \frac{28}{8} = \frac{7}{2} = 3\frac{1}{2} \)

Thus, the co-ordinates of the point K are \( \left(3\frac{1}{2}, 0\right) \).

In simple words: When line crosses x-axis, y-coordinate is 0. We use this to find the ratio and then find the x-coordinate of the crossing point.

📝 Teacher's Note: Ordinate means y-coordinate. On x-axis, y is always 0. Students should remember this basic fact about coordinate axes.

🎯 Exam Tip: Write "ordinate = 0" first. Apply section formula for y-coordinate, set equal to 0, solve for k. Then find x-coordinate. Write final answer as coordinates with brackets.

Question 22. The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Answer:
Since, point K lies on y-axis, its abscissa is 0.
Let the point K (0, y) divides AB in the ratio k: 1.
We have,
\( x = \frac{k \times (-6) + 1 \times 4}{k + 1} \)
\( 0 = \frac{-6k + 4}{k + 1} \)
\( k = \frac{4}{6} = \frac{2}{3} \)

Thus, K divides AB in the ratio 2: 3.
Also, we have:
\( y = \frac{k \times (-2) + 1 \times 7}{k + 1} \)
\( y = \frac{-2k + 7}{k + 1} \)
\( y = \frac{-2 \times \frac{2}{3} + 7}{\frac{2}{3} + 1} \)
\( y = \frac{-\frac{4}{3} + 21}{\frac{2}{3} + 1} \)
\( y = \frac{-4 + 21}{2 + 3} \)
\( y = \frac{17}{5} \)

Thus, the co-ordinates of the point K are \( \left(0, \frac{17}{5}\right) \).

In simple words: When line crosses y-axis, x-coordinate is 0. We use this to find the ratio and then find the y-coordinate of the crossing point.

📝 Teacher's Note: Abscissa means x-coordinate. On y-axis, x is always 0. This is opposite to the previous question where ordinate was 0 on x-axis.

🎯 Exam Tip: Write "abscissa = 0" first. Apply section formula for x-coordinate, set equal to 0, solve for k. Then find y-coordinate. Show fraction calculations clearly.

Question 23. The line joining P (-4, 5) and Q (3, 2) intersects the y-axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) the ratio PR: RQ.
(ii) the co-ordinates of R.
(iii) the area of the quadrilateral PMNQ.
Answer:
(i) Since, point R lies on y-axis, its abscissa is 0.
Let the point R (0, y) divides PQ in the ratio k: 1.
We have,
\( x = \frac{k \times 3 + 1 \times (-4)}{k + 1} \)
\( 0 = \frac{3k - 4}{k + 1} \)
\( k = \frac{4}{3} \)

Thus, the ratio PR: RQ is 4: 3.

(ii) Also, we have:
\( y = \frac{k \times 2 + 1 \times 5}{k + 1} \)
\( y = \frac{\frac{4}{3} \times 2 + 5}{\frac{4}{3} + 1} \)
\( y = \frac{\frac{8}{3} + 5}{\frac{4}{3} + 1} \)
\( y = \frac{8 + 15}{4 + 3} \)
\( y = \frac{23}{7} \)

Thus, the co-ordinates of R are \( \left(0, \frac{23}{7}\right) \).

(iii) M is foot of perpendicular from P to x-axis, so M(-4, 0)
N is foot of perpendicular from Q to x-axis, so N(3, 0)
Area of quadrilateral PMNQ = \( \frac{1}{2} \times |PM + QN| \times MN \)
= \( \frac{1}{2} \times (5 + 2) \times (3 - (-4)) \)
= \( \frac{1}{2} \times 7 \times 7 = \frac{49}{2} \) square units

In simple words: We find where line PQ crosses y-axis, then calculate the area of the shape formed by dropping perpendiculars to x-axis.

📝 Teacher's Note: For perpendiculars to x-axis, y-coordinate becomes 0 while x stays same. Area of trapezoid uses formula: (1/2) × (sum of parallel sides) × height.

🎯 Exam Tip: Draw a rough diagram to visualize. M and N are directly below P and Q on x-axis. Use trapezoid area formula. Write "square units" for area.

[Diagram: This shows a coordinate plane with points P(-4, 5), Q(3, 2), M on the negative x-axis, and N on the positive x-axis. There are line segments connecting these points forming what appears to be a geometric figure.]

Question 24. In the given figure, line APB meets the x-axis at point A and y-axis at point B. P is the point (-4, 2) and AP: PB = 1: 2. Find the co-ordinates of A and B.
Answer:
Given, A lies on x-axis and B lies on y-axis.
Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.
Given, P is the point (-4, 2) and AP: PB = 1: 2.

Using section formula:
\( -4 = \frac{1 \times 0 + 2 \times x}{1 + 2} \)
\( -4 = \frac{2x}{3} \)
\( x = \frac{-4 \times 3}{2} = -6 \)

Also,
\( 2 = \frac{1 \times y + 2 \times 0}{1 + 2} \)
\( 2 = \frac{y}{3} \)
\( y = 6 \)

Thus, the co-ordinates of points A and B are (-6, 0) and (0, 6) respectively.
In simple words: We use the section formula to find where A and B are. Since P divides AB in ratio 1:2, we can find the exact positions of A and B on the axes.

📝 Teacher's Note: Draw the axes and mark point P first. Show students that A is on x-axis (so y = 0) and B is on y-axis (so x = 0). This makes the section formula easier to use.

🎯 Exam Tip: Always write "Let A = (x, 0)" and "Let B = (0, y)" at the start. Apply section formula for both x and y coordinates separately. Check your answer by verifying the ratio.

Question 25. Given a line segment AB joining the points A(-4, 6) and B(8, -3). Find:
(i) the ratio in which AB is divided by the y-axis
(ii) find the coordinates of the point of intersection
(iii) the length of AB
Answer:
(i)
Let the required ratio be \( m_1 : m_2 \).
Consider A(-4, 6) = \( (x_1, y_1) \); B(8, -3) = \( (x_2, y_2) \) and let
P(x, y) be the point of intersection of the line segment and the y-axis.

By section formula, we have,
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \), \( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)

\( x = \frac{8m_1 - 4m_2}{m_1 + m_2} \), \( y = \frac{-3m_1 + 6m_2}{m_1 + m_2} \)

The equation of the y-axis is x = 0
\( x = \frac{8m_1 - 4m_2}{m_1 + m_2} = 0 \)
\( 8m_1 - 4m_2 = 0 \)
\( 8m_1 = 4m_2 \)
\( \frac{m_1}{m_2} = \frac{4}{8} = \frac{1}{2} \)

So the ratio is 1:2.

(ii)
From the previous subpart, we have, \( \frac{m_1}{m_2} = \frac{1}{2} \)
Let \( m_1 = k \) and \( m_2 = 2k \), where k is any constant.
Also, we have,
\( x = \frac{8m_1 - 4m_2}{m_1 + m_2} \), \( y = \frac{-3m_1 + 6m_2}{m_1 + m_2} \)

\( x = \frac{8 \times k - 4 \times 2k}{k + 2k} \), \( y = \frac{-3 \times k + 6 \times 2k}{k + 2k} \)

\( x = \frac{8k - 8k}{3k} \), \( y = \frac{-3k + 12k}{3k} \)

\( x = \frac{0}{3k} \), \( y = \frac{9k}{3k} \)

\( x = 0 \), \( y = 3 \)

Thus, the point of intersection is P(0, 3)

(iii)
The length of AB = Distance between two points A and B.
The distance between two given points A\( (x_1, y_1) \) and B\( (x_2, y_2) \) is given by,
Distance AB = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{(8 + 4)^2 + (-3 - 6)^2} \)
\( = \sqrt{(12)^2 + (9)^2} \)
\( = \sqrt{144 + 81} \)
\( = \sqrt{225} \)
\( = 15 \text{ units} \)

In simple words: The y-axis cuts the line AB at point (0, 3) in the ratio 1:2. We use the section formula and distance formula to find all the answers.

📝 Teacher's Note: Show students that any point on y-axis has x-coordinate = 0. This is the key to solving part (i). For distance, remind them that we square both differences to make them positive.

🎯 Exam Tip: For ratio problems, always set x = 0 for y-axis intersection. Show all steps clearly. For distance, write the formula first, then substitute values. Don't forget to take the square root at the end.

Question 26. If P(-b, 9a - 2) divides the line segment joining the points A(-3, 3a + 1) and B(5, 8a) in the ratio 3: 1, find the values of a and b.
Answer:
Take \( (x_1, y_1) = (-3, 3a + 1) \); \( (x_2, y_2) = B(5, 8a) \) and
\( (x, y) = (-b, 9a - 2) \)
Here \( m_1 = 3 \) and \( m_2 = 1 \)

Coordinate of P(x, y) = \( \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right) \)

\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \) and \( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)

\( -b = \frac{3 \times 5 + 1 \times (-3)}{3 + 1} \) and \( 9a - 2 = \frac{3 \times 8a + 1(3a + 1)}{3 + 1} \)

\( -b = \frac{15 - 3}{4} \) and \( 9a - 2 = \frac{24a + 3a + 1}{4} \)

\( -4b = 12 \) and \( 36a - 8 = 27a + 1 \)

\( b = -3 \) and \( 9a = 9 \)

Therefore, \( a = 1 \) and \( b = -3 \)
In simple words: We use the section formula to find the coordinates of P. Then we match the given coordinates with our calculated values to find a and b.

📝 Teacher's Note: Remind students to be careful with signs when substituting negative coordinates. Show them to solve for x and y coordinates separately to avoid confusion.

🎯 Exam Tip: Write the section formula clearly at the start. Match the x-coordinates and y-coordinates separately. Solve the resulting equations step by step. Always check your answer by substituting back.

Exercise 13B

Question 1. Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3) and (-1, 7)
Answer:
(i) A(-6, 7) and B(3, 5)
Mid-point of AB = \( \left(\frac{-6 + 3}{2}, \frac{7 + 5}{2}\right) = \left(\frac{-3}{2}, 6\right) \)

(ii) A(5, -3) and B(-1, 7)
Mid-point of AB = \( \left(\frac{5 - 1}{2}, \frac{-3 + 7}{2}\right) = (2, 2) \)

In simple words: To find the mid-point, we add both x-coordinates and divide by 2, then add both y-coordinates and divide by 2. This gives us the exact middle point.

📝 Teacher's Note: Show students that mid-point is like finding the average of x-coordinates and average of y-coordinates. Use the midpoint formula: \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).

🎯 Exam Tip: Always write the midpoint formula first. Be careful with negative signs when adding coordinates. Write your final answer as an ordered pair (x, y).

Question 2. Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Answer:
Mid-point of AB = (2, 3)
\( \left(\frac{3 + x}{2}, \frac{5 + y}{2}\right) = (2, 3) \)

\( \frac{3 + x}{2} = 2 \) and \( \frac{5 + y}{2} = 3 \)
\( \implies 3 + x = 4 \) and \( 5 + y = 6 \)
\( \implies x = 1 \) and \( y = 1 \)

In simple words: We use the mid-point formula. The middle point between two points is the average of their x-values and y-values. We solve simple equations to find x = 1 and y = 1.

📝 Teacher's Note: Show students that mid-point formula is like finding the average. If you want to find the middle between 3 and x, you add them and divide by 2. Students remember averages easily.

🎯 Exam Tip: Always write the mid-point formula first. Then substitute the given values and solve step by step. Show all working clearly.

Question 3. A (5, 3), B (-1, 1) and C (7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = 1/2 BC.
Answer:
Given, L is the mid-point of AB and M is the mid-point of AC.

Co-ordinates of L are:
\( L = \left(\frac{5 - 1}{2}, \frac{3 + 1}{2}\right) = (2, 2) \)

Co-ordinates of M are:
\( M = \left(\frac{5 + 7}{2}, \frac{3 - 3}{2}\right) = (6, 0) \)

Using distance formula:
\( BC = \sqrt{(7 + 1)^2 + (-3 - 1)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \)

\( LM = \sqrt{(6 - 2)^2 + (0 - 2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)

Hence, \( LM = \frac{1}{2}BC \)

In simple words: We find the middle points L and M first. Then we calculate the distances LM and BC using the distance formula. We see that LM is exactly half of BC.

📝 Teacher's Note: This proves the mid-point theorem. The line joining mid-points of two sides of a triangle is half the third side. Draw a triangle on the board to show this clearly.

🎯 Exam Tip: First find coordinates of mid-points L and M. Then use distance formula for both LM and BC. Show that LM = (1/2)BC by simplifying the square roots.

Question 4. Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10)
(ii) B; if A = (3, -1) and M = (-1, 3).
Answer:
(i) Let the co-ordinates of A be (x, y).
\( (1, 7) = \left(\frac{x - 5}{2}, \frac{y + 10}{2}\right) \)

\( 1 = \frac{x - 5}{2} \) and \( 7 = \frac{y + 10}{2} \)
\( \implies 2 = x - 5 \) and \( 14 = y + 10 \)
\( \implies x = 7 \) and \( y = 4 \)

Hence, the co-ordinates of A are (7, 4).

(ii) Let the co-ordinates of B be (x, y).
\( (-1, 3) = \left(\frac{3 + x}{2}, \frac{-1 + y}{2}\right) \)

\( -1 = \frac{3 + x}{2} \) and \( 3 = \frac{-1 + y}{2} \)
\( \implies -2 = 3 + x \) and \( 6 = -1 + y \)
\( \implies x = -5 \) and \( y = 7 \)

Hence, the co-ordinates of B are (-5, 7).

In simple words: When we know the mid-point and one end point, we can find the other end point. We use the mid-point formula and solve for the unknown coordinates.

📝 Teacher's Note: Explain that if M is the middle of A and B, then A + B = 2M. This helps students remember the relationship between the three points.

🎯 Exam Tip: Set up the mid-point formula correctly. Remember that mid-point = (sum of x-coordinates)/2, (sum of y-coordinates)/2. Solve each coordinate separately.

Question 5. P (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the coordinates of points A and B.
Answer:
Point A lies on y-axis, so let its co-ordinates be (0, y).
Point B lies on x-axis, so let its co-ordinates be (x, 0).
P (-3, 2) is the mid-point of line segment AB.

\( (-3, 2) = \left(\frac{0 + x}{2}, \frac{y + 0}{2}\right) \)

\( (-3, 2) = \left(\frac{x}{2}, \frac{y}{2}\right) \)

\( -3 = \frac{x}{2} \) and \( 2 = \frac{y}{2} \)
\( \implies -6 = x \) and \( 4 = y \)

Thus, the co-ordinates of points A and B are (0, 4) and (-6, 0) respectively.

In simple words: Looking at the diagram, point A is on the y-axis (so x = 0) and point B is on the x-axis (so y = 0). We use these facts with the mid-point formula to find the exact coordinates.

[Diagram: A coordinate plane showing point P(-3, 2) as the mid-point of line segment AB, where A is on the y-axis and B is on the x-axis.]

📝 Teacher's Note: Show students that points on the y-axis have x-coordinate = 0, and points on the x-axis have y-coordinate = 0. Use the diagram to make this clear.

🎯 Exam Tip: From the diagram, identify which axis each point lies on. This gives you one coordinate for free. Then use the mid-point formula to find the other coordinate.

Question 6. In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.
Answer:
Point A lies on x-axis, so let its co-ordinates be (x, 0).
Point B lies on y-axis, so let its co-ordinates be (0, y).
P (4, 2) is mid-point of line segment AB.

\( (4, 2) = \left(\frac{x + 0}{2}, \frac{0 + y}{2}\right) \)

\( 4 = \frac{x}{2} \) and \( 2 = \frac{y}{2} \)
\( \implies 8 = x \) and \( 4 = y \)

Hence, the co-ordinates of points A and B are (8, 0) and (0, 4) respectively.

In simple words: From the diagram, A is on the x-axis and B is on the y-axis. We use this information with the mid-point formula to find where exactly A and B are located.

[Diagram: A coordinate plane showing point P(4, 2) as the mid-point of line segment AB, where A is on the x-axis and B is on the y-axis.]

📝 Teacher's Note: This is similar to the previous question but with A and B switched. Emphasize that students must look at the diagram carefully to see which axis each point is on.

🎯 Exam Tip: Always look at the diagram first. Identify which axis each unknown point lies on. This tells you one coordinate is zero, making the problem much easier.

Question 7. (-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the lengths of its median through the vertex (3, -6)
Answer:
Let A (-5, 2), B (3, -6) and C (7, 4) be the vertices of the given triangle.
The median through vertex B (3, -6) goes to the mid-point of the opposite side AC.

Co-ordinates of mid-point E of AC are:
\( E = \left(\frac{-5 + 7}{2}, \frac{2 + 4}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3) \)

The median through vertex B(3, -6) is BE.

Using distance formula:
\( BE = \sqrt{(1 - 3)^2 + (3 - (-6))^2} = \sqrt{4 + 81} = \sqrt{85} = 9.22 \)

In simple words: A median connects a vertex to the middle of the opposite side. First we find the mid-point of the side opposite to vertex B. Then we find the distance from B to this mid-point.

[Diagram: A triangle ABC with vertices at A(-5,2), B(3,-6), and C(7,4), showing the medians from each vertex meeting at the centroid.]

📝 Teacher's Note: Explain that a median is like drawing a line from a corner to the middle of the opposite side. Every triangle has three medians, and they all meet at one point called the centroid.

🎯 Exam Tip: First find the mid-point of the side opposite to the given vertex. Then use the distance formula between the vertex and this mid-point. Always approximate the final answer if needed.

Question 8. Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
Answer:
Given, AB = BC = CD
So, B is the mid-point of AC. Let the co-ordinates of point A be (x, y).

\( (0, 3) = \left(\frac{x + 1}{2}, \frac{y + 8}{2}\right) \)

\( 0 = \frac{x + 1}{2} \) and \( 3 = \frac{y + 8}{2} \)
\( \implies 0 = x + 1 \) and \( 6 = y + 8 \)
\( \implies -1 = x \) and \( -2 = y \)

Thus, the co-ordinates of point A are (-1, -2).

Also, C is the mid-point of BD. Let the co-ordinates of point D be (p, q).

\( (1, 8) = \left(\frac{0 + p}{2}, \frac{3 + q}{2}\right) \)

\( 1 = \frac{0 + p}{2} \) and \( 8 = \frac{3 + q}{2} \)
\( \implies 2 = 0 + p \) and \( 16 = 3 + q \)
\( \implies 2 = p \) and \( 13 = q \)

Thus, the co-ordinates of point D are (2, 13).

In simple words: Since all segments are equal, B is the middle of A and C, and C is the middle of B and D. We use the mid-point formula twice to find A and D.

[Diagram: A straight line showing points A, B, C, and D in order, with B at (0,3) and C at (1,8), and equal distances marked between consecutive points.]

📝 Teacher's Note: When segments are equal on a straight line, each point becomes the mid-point of its neighbors. This creates a pattern that students can use to solve similar problems.

🎯 Exam Tip: If AB = BC = CD, then B is mid-point of AC and C is mid-point of BD. Apply the mid-point formula twice. Keep track of which point is the mid-point of which pair.

Question 9. One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).
Answer:
Let A (-2, 5) and B (x, y) be the two ends of the diameter.
The centre of the circle is the mid-point of the diameter.

Centre = \( \left(\frac{-2 + x}{2}, \frac{5 + y}{2}\right) = (2, -1) \)

\( \frac{-2 + x}{2} = 2 \) and \( \frac{5 + y}{2} = -1 \)
\( \implies -2 + x = 4 \) and \( 5 + y = -2 \)
\( \implies x = 6 \) and \( y = -7 \)

Therefore, the co-ordinates of the other end of the diameter are (6, -7).

In simple words: The center of a circle is always at the middle of any diameter. So if we know the center and one end of the diameter, we can find the other end using the mid-point formula.

📝 Teacher's Note: Draw a circle on the board and show that the center is always at the middle of any diameter. This helps students understand why we use the mid-point formula here.

🎯 Exam Tip: Remember that center of circle = mid-point of diameter. Set up the mid-point formula with center as the result, and solve for the unknown endpoint. This is a common exam question.

Question 10. A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of a quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Answer:
Step 1: Find the mid-point of AC.
Co-ordinates of the mid-point of AC are:
\( \left(\frac{2 + (-4)}{2}, \frac{5 + 3}{2}\right) = \left(\frac{-2}{2}, \frac{8}{2}\right) = (-1, 4) \)

Step 2: Find the mid-point of BD.
Co-ordinates of the mid-point of BD are:
\( \left(\frac{1 + (-3)}{2}, \frac{0 + 8}{2}\right) = \left(\frac{-2}{2}, \frac{8}{2}\right) = (-1, 4) \)

Step 3: Compare the mid-points.
Since, mid-point of AC = mid-point of BD
Hence, ABCD is a parallelogram.

Final Answer: Mid-point of AC = (-1, 4), Mid-point of BD = (-1, 4). The quadrilateral is a parallelogram.
In simple words: We found the middle point of both diagonals AC and BD. Both middle points are the same. When diagonals cut each other at the middle, the shape is a parallelogram.

📝 Teacher's Note: Show students that in a parallelogram, diagonals always cut each other in half. Draw a parallelogram and mark the diagonals. This helps students remember the property.

🎯 Exam Tip: Always find both mid-points using the formula. If they are same, write "parallelogram". The examiner looks for this word and the mid-point formula.

Question 11. P (4, 2) and Q (-1, 5) are the vertices of a parallelogram PQRS and (-3, 2) are the coordinates of the points of intersection of its diagonals. Find the coordinates of R and S.
Answer:
Step 1: Let the coordinates of R and S be (x,y) and (a,b) respectively.
Mid-point of PR is O.

Step 2: Use the property that diagonals bisect each other.
\( O(-3,2) = \left(\frac{4 + x}{2}, \frac{2 + y}{2}\right) \)

\( -3 = \frac{4 + x}{2}, 2 = \frac{2 + y}{2} \)

\( -6 = 4 + x, 4 = 2 + y \)
\( x = -10, y = 2 \)

Hence, R = (-10, 2)

Step 3: Similarly, find S using mid-point of SQ.
\( O(-3,2) = \left(\frac{a - 1}{2}, \frac{b + 5}{2}\right) \)

\( -3 = \frac{a - 1}{2}, 2 = \frac{b + 5}{2} \)

\( -6 = a - 1, 4 = b + 5 \)
\( a = -5, b = -1 \)

Hence, S = (-5, -1)

Final Answer: The coordinates of R and S are (-10, 2) and (-5, -1).
In simple words: In a parallelogram, the diagonals cut each other exactly in the middle. We used this fact to find the missing corners by working backwards from the center point.

📝 Teacher's Note: Draw a parallelogram PQRS with diagonals PR and QS crossing at O. Show students that O is exactly halfway between P and R, and also halfway between Q and S.

🎯 Exam Tip: Write the mid-point formula clearly. Set it equal to the given center point. Solve the equations step by step. Show all working for full marks.

Question 12. A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the coordinates of vertex C.
Answer:
Step 1: Let the coordinates of vertex C be (x, y).
ABCD is a parallelogram.

Step 2: Use the property that diagonals bisect each other.
Mid-point of AC = Mid-point of BD

\( \left(\frac{-1 + x}{2}, \frac{0 + y}{2}\right) = \left(\frac{1 + 3}{2}, \frac{3 + 5}{2}\right) \)

\( \left(\frac{-1 + x}{2}, \frac{y}{2}\right) = (2, 4) \)

Step 3: Solve for x and y.
\( \frac{-1 + x}{2} = 2 \) and \( \frac{y}{2} = 4 \)

\( x = 5 \) and \( y = 8 \)

Final Answer: The coordinates of vertex C are (5, 8).
In simple words: We know three corners of a parallelogram. To find the fourth corner, we use the fact that diagonals cut each other in the middle. This gives us two equations to solve.

📝 Teacher's Note: Explain that in a parallelogram, opposite corners are connected by diagonals. These diagonals always meet at their middle points. Use this property to find missing vertices.

🎯 Exam Tip: Always write "Mid-point of AC = Mid-point of BD" first. Then use the mid-point formula. Solve the equations clearly to get the missing coordinates.

Question 13. The points (2, -1), (-1, 4) and (-2, 2) are mid-points of the sides of a triangle. Find its vertices.
Answer:
Step 1: Let A(\( x_1, y_1 \)), B(\( x_2, y_2 \)) and C(\( x_3, y_3 \)) be the coordinates of the vertices of triangle ABC.

Step 2: Set up equations using mid-point conditions.
Midpoint of AB, i.e. D
\( D(2, -1) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

\( 2 = \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} = -1 \)

\( x_1 + x_2 = 4 \) ... (1) \( y_1 + y_2 = -2 \) ... (2)

Similarly,
\( x_1 + x_3 = -2 \) ... (3) \( y_1 + y_3 = 8 \) ... (4)
\( x_2 + x_3 = -4 \) ... (5) \( y_2 + y_3 = 4 \) ... (6)

Step 3: Solve the system of equations.
Adding (1), (3) and (5), we get:
\( 2(x_1 + x_2 + x_3) = -2 \)
\( x_1 + x_2 + x_3 = -1 \)
\( 4 + x_3 = -1 \) [from (1)]
\( x_3 = -5 \)

From (3): \( x_1 - 5 = -2 \) \( \implies x_1 = 3 \)
From (5): \( x_2 - 5 = -4 \) \( \implies x_2 = 1 \)

Adding (2), (4) and (6), we get:
\( 2(y_1 + y_2 + y_3) = 10 \)
\( y_1 + y_2 + y_3 = 5 \)
\( -2 + y_3 = 5 \) [from (2)]
\( y_3 = 7 \)

From (4): \( y_1 + 7 = 8 \) \( \implies y_1 = 1 \)
From (6): \( y_2 + 7 = 4 \) \( \implies y_2 = -3 \)

Final Answer: The coordinates of the vertices of triangle ABC are (3, 1), (1, -3) and (-5, 7).
In simple words: We have the middle points of all three sides. We make equations using the mid-point formula. Then we solve these equations to find the three corners of the triangle.

📝 Teacher's Note: Draw a triangle and mark the mid-points of each side. Explain that each mid-point is the average of two vertices. This creates a system of six equations to solve.

🎯 Exam Tip: Write all six equations first. Add them in groups to eliminate variables. Show each step clearly. Check your answer by verifying the mid-points.

Question 14. Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e., lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Answer:
Step 1: Since AB = BC, point B is the mid-point of AC.

\( (y, 7) = \left(\frac{-5 + 1}{2}, \frac{x + (-3)}{2}\right) \)

\( (y, 7) = \left(-2, \frac{x - 3}{2}\right) \)

Step 2: Equate coordinates.
\( y = -2 \) and \( 7 = \frac{x - 3}{2} \)

\( y = -2 \) and \( x = 17 \)

Final Answer: The values are x = 17 and y = -2.
In simple words: When three points lie on a straight line and the middle one is exactly halfway between the other two, we use the mid-point formula to find the unknown coordinates.

📝 Teacher's Note: Explain that "AB = BC" means B is exactly in the middle of A and C. Draw three points on a line to show this clearly.

🎯 Exam Tip: When you see "AB = BC", immediately think "B is the mid-point of AC". Use the mid-point formula and solve for the unknowns.

Question 15. Points P (a, -4), Q (-2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.
Answer:
Step 1: Understand the given condition.
Given, PR = 2QR
Now, Q lies between P and R, so PR = PQ + QR
\( \therefore \) PQ + QR = 2QR
\( \Rightarrow \) PQ = QR
\( \Rightarrow \) Q is the mid-point of PR.

Step 2: Use mid-point formula.
\( (-2, b) = \left(\frac{a + 0}{2}, \frac{-4 + 2}{2}\right) \)

\( (-2, b) = \left(\frac{a}{2}, -1\right) \)

Step 3: Solve for a and b.
\( \Rightarrow a = -4, \quad b = -1 \)

Final Answer: The values are a = -4 and b = -1.
In simple words: When PR = 2QR and Q is between P and R, it means Q is exactly in the middle of P and R. So Q is the mid-point of PR.

📝 Teacher's Note: Draw a line segment PR with Q in between. Show that if PR = 2QR, then PQ must also equal QR. This makes Q the middle point.

🎯 Exam Tip: When you see "PR = 2QR" with Q between P and R, always conclude "Q is mid-point of PR". Then use the mid-point formula directly.

Question 16. Calculate the co-ordinates of the centroid of a triangle ABC, if A = (7, -2), B = (0, 1) and C = (-1, 4).
Answer:
Step 1: Use the centroid formula.
The centroid of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \) is:

\( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)

Step 2: Substitute the given coordinates.
Centroid = \( \left(\frac{7 + 0 + (-1)}{3}, \frac{-2 + 1 + 4}{3}\right) \)

Centroid = \( \left(\frac{6}{3}, \frac{3}{3}\right) \)

Centroid = \( (2, 1) \)

Final Answer: The coordinates of the centroid are (2, 1).
In simple words: The centroid is the balance point of a triangle. To find it, we add all x-coordinates and divide by 3, then add all y-coordinates and divide by 3.

📝 Teacher's Note: Explain that centroid is where the triangle would balance if made of cardboard. It's the average of all three vertices. Show students how to add coordinates and divide by 3.

🎯 Exam Tip: Remember the centroid formula: add all x-coordinates and divide by 3, add all y-coordinates and divide by 3. Always check your arithmetic twice.

Exercise 13C

Question 1. Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP: PC = 3: 2. Find the length of line segment AP.
Answer:
Given: BP: PC = 3: 2
Using section formula, the co-ordinates of point P are:
\( P = \left(\frac{3 \times 5 + 2 \times 0}{3 + 2}, \frac{3 \times 10 + 2 \times 5}{3 + 2}\right) \)
\( = \left(\frac{15}{5}, \frac{40}{5}\right) = (3, 8) \)

Using distance formula:
\( AP = \sqrt{(3 - 4)^2 + (8 + 4)^2} = \sqrt{1 + 144} = \sqrt{145} = 12.04 \)

Length of line segment AP = \( \sqrt{145} \) units
In simple words: We found point P using the section formula. Then we used the distance formula to find how far P is from A. Like measuring the straight line distance between two dots on paper.

📝 Teacher's Note: Draw the triangle on graph paper first. Show students how to mark the point P that divides BC in ratio 3:2. This makes the calculation clear.

🎯 Exam Tip: Always write "Given" first, then use section formula to find P, then distance formula for AP. Show all steps clearly to get full marks.

Question 2. A (20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of a point P in AB such that: 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Answer:
Part 1: Finding point P where 3PB = AB
Given: 3PB = AB
\( \Rightarrow \frac{AB}{PB} = \frac{3}{1} \)
\( \Rightarrow \frac{AB - PB}{PB} = \frac{3 - 1}{1} = \frac{2}{1} \)
\( \Rightarrow \frac{AP}{PB} = \frac{2}{1} \)

Using section formula:
\( P = \left(\frac{2 \times 10 + 1 \times 20}{2 + 1}, \frac{2 \times (-20) + 1 \times 0}{2 + 1}\right) \)
\( = \left(\frac{40}{3}, \frac{-40}{3}\right) \)

Part 2: Finding point Q where AB = 6AQ
Given: AB = 6AQ
\( \Rightarrow \frac{AQ}{AB} = \frac{1}{6} \)
\( \Rightarrow \frac{AQ}{AB - AQ} = \frac{1}{6 - 1} = \frac{1}{5} \)
\( \Rightarrow \frac{AQ}{QB} = \frac{1}{5} \)

Using section formula:
\( Q = \left(\frac{1 \times 10 + 5 \times 20}{1 + 5}, \frac{1 \times (-20) + 5 \times 0}{1 + 5}\right) \)
\( = Q\left(\frac{110}{6}, \frac{-20}{6}\right) = Q\left(\frac{55}{3}, \frac{-10}{3}\right) \)

Coordinates: P = \( \left(\frac{40}{3}, \frac{-40}{3}\right) \) and Q = \( \left(\frac{55}{3}, \frac{-10}{3}\right) \)
In simple words: We found two special points on line AB. Point P divides AB in ratio 2:1, and point Q divides AB in ratio 1:5. Like finding two specific marks on a ruler.

📝 Teacher's Note: Explain that 3PB = AB means P is closer to B. Draw a line and mark these points to show the ratios visually.

🎯 Exam Tip: Convert the given conditions to ratios first (like AP:PB). Then apply section formula. Write coordinates as fractions if they don't simplify to whole numbers.

Question 3. A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP: PB = 3: 5 and AQ: QC = 3: 5. Show that: PQ = 3/8 BC.
Answer:
Finding coordinates of point P:
Given that point P lies on AB such that AP: PB = 3: 5.
Using section formula:
\( P = \left(\frac{3 \times 0 + 5 \times (-8)}{3 + 5}, \frac{3 \times 16 + 5 \times 0}{3 + 5}\right) \)
\( = \left(\frac{-40}{8}, \frac{48}{8}\right) = (-5, 6) \)

Finding coordinates of point Q:
Given that point Q lies on AC such that AQ: QC = 3: 5.
Using section formula:
\( Q = \left(\frac{3 \times 0 + 5 \times (-8)}{3 + 5}, \frac{3 \times 0 + 5 \times 0}{3 + 5}\right) \)
\( = \left(\frac{-40}{8}, \frac{0}{8}\right) = (-5, 0) \)

Calculating distances:
Using distance formula:
\( PQ = \sqrt{(-5 + 5)^2 + (0 - 6)^2} = \sqrt{0 + 36} = 6 \)
\( BC = \sqrt{(0 - 0)^2 + (0 - 16)^2} = \sqrt{0 + (16)^2} = 16 \)

Verification:
\( \frac{3}{8} BC = \frac{3}{8} \times 16 = 6 = PQ \)

Hence, proved: PQ = \( \frac{3}{8} \)BC
In simple words: We found the coordinates of points P and Q using the section formula. Then we calculated the lengths PQ and BC. Finally, we showed that PQ equals exactly 3/8 of BC.

📝 Teacher's Note: Draw the triangle first and mark points P and Q. Show how P and Q divide the sides in the same ratio 3:5. This helps students see why PQ is parallel to BC.

🎯 Exam Tip: Always find coordinates first, then calculate distances, then verify the relationship. Write "Hence, proved" at the end for proof questions.

Question 4. Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.
Answer:
Let P and Q be the points of trisection of the line segment joining A (6, -9) and B (0, 0).
P divides AB in the ratio 1: 2. Therefore, the co-ordinates of point P are:
\( P = \left(\frac{1 \times 0 + 2 \times 6}{1 + 2}, \frac{1 \times 0 + 2 \times (-9)}{1 + 2}\right) \)
\( = \left(\frac{12}{3}, \frac{-18}{3}\right) = (4, -6) \)

Q divides AB in the ratio 2: 1. Therefore, the co-ordinates of point Q are:
\( Q = \left(\frac{2 \times 0 + 1 \times 6}{2 + 1}, \frac{2 \times 0 + 1 \times (-9)}{2 + 1}\right) \)
\( = \left(\frac{6}{3}, \frac{-9}{3}\right) = (2, -3) \)

The points of trisection are (4, -6) and (2, -3)
In simple words: Trisection means dividing into three equal parts. We found the two points that divide the line into three equal pieces. Like cutting a rope into three equal parts.

📝 Teacher's Note: Draw a line from origin to (6, -9) and show the three equal parts. Explain that P is 1/3 of the way and Q is 2/3 of the way from A to B.

🎯 Exam Tip: For trisection, one point divides in ratio 1:2 and the other in ratio 2:1. Apply section formula for both points separately.

Question 5. A line segment joining A(-1, 5/3) and B (a, 5) is divided in the ratio 1: 3 at P, point where the line segment AB intersects the y-axis.
(i) Calculate the value of 'a'.
(ii) Calculate the co-ordinates of 'P'.
Answer:
Since the line segment AB intersects the y-axis at point P, let the co-ordinates of point P be (0, y).
P divides AB in the ratio 1: 3.

Using section formula:
\( (0, y) = \left(\frac{1 \times a + 3 \times (-1)}{1 + 3}, \frac{1 \times 5 + 3 \times \frac{5}{3}}{1 + 3}\right) \)
\( (0, y) = \left(\frac{a - 3}{4}, \frac{10}{4}\right) \)

\( 0 = \frac{a - 3}{4} \) and \( y = \frac{10}{4} \)
\( a = 3 \) and \( y = \frac{5}{2} = 2\frac{1}{2} \)

(i) The value of a is 3
(ii) The co-ordinates of point P are \( \left(0, 2\frac{1}{2}\right) \)
In simple words: Since P is on the y-axis, its x-coordinate is 0. We used this fact to find the value of a. Then we found the y-coordinate of P.

📝 Teacher's Note: Explain that any point on y-axis has x-coordinate = 0. Use this to set up the equation and solve for a.

🎯 Exam Tip: Remember that y-axis points have x = 0 and x-axis points have y = 0. Use this to solve for unknown values.

Question 6. In what ratio is the line joining A (0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates of the point where AB intersects the x-axis.
Answer:
Let the line segment AB intersects the x-axis at point P (x, 0) in the ratio k: 1.
Using section formula:
\( (x, 0) = \left(\frac{k \times 4 + 1 \times 0}{k + 1}, \frac{k \times (-1) + 1 \times 3}{k + 1}\right) \)
\( (x, 0) = \left(\frac{4k}{k + 1}, \frac{-k + 3}{k + 1}\right) \)

\( \Rightarrow 0 = \frac{-k + 3}{k + 1} \)
\( \Rightarrow k = 3 \)

Thus, the required ratio in which P divides AB is 3: 1.
Also, we have:
\( x = \frac{4k}{k + 1} = \frac{4 \times 3}{3 + 1} = \frac{12}{4} = 3 \)

The line AB is divided by x-axis in the ratio 3: 1
The co-ordinates of point P are (3, 0)
In simple words: We found where the line crosses the x-axis. Since it's on x-axis, y-coordinate is 0. We used this to find the ratio and the exact point.

📝 Teacher's Note: Draw the line from A(0, 3) to B(4, -1) and show where it crosses the x-axis. Students can see that it divides the line in a specific ratio.

🎯 Exam Tip: For x-axis intersection, y = 0. For y-axis intersection, x = 0. Use this condition to find the ratio and coordinates.

Question 17. The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Answer:
Let G be the centroid of triangle PQR whose coordinates are (2, -5) and let (x, y) be the coordinates of vertex P.

Coordinates of G are:
\( G(2, -5) = G\left(\frac{x + (-6) + 11}{3}, \frac{y + 5 + 8}{3}\right) \)

\( 2 = \frac{x + 5}{3} \), \( -5 = \frac{y + 13}{3} \)
\( 6 = x + 5 \), \( -15 = y + 13 \)
\( x = 1 \), \( y = -28 \)

Coordinates of vertex P are (1, -28)
In simple words: The centroid is the center point of a triangle. We used the centroid formula to work backwards and find the missing vertex P.

📝 Teacher's Note: Explain that centroid divides each side in ratio 2:1. It's like the balance point of the triangle. Show the centroid formula clearly.

🎯 Exam Tip: Centroid formula: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). Substitute known values and solve for the unknown coordinates.

Question 18. A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Answer:
Given, centroid of triangle ABC is the origin.
\( \therefore (0, 0) = \left(\frac{5 + (-4) + y}{3}, \frac{x + 3 + (-2)}{3}\right) \)
\( (0, 0) = \left(\frac{1 + y}{3}, \frac{x + 1}{3}\right) \)

\( 0 = \frac{1 + y}{3} \) and \( 0 = \frac{x + 1}{3} \)
\( y = -1 \) and \( x = -1 \)

The values are x = -1 and y = -1
In simple words: Since the centroid is at origin (0, 0), we used the centroid formula to find the missing coordinates x and y.

📝 Teacher's Note: Show that when centroid is at origin, the sum of all x-coordinates equals 0 and sum of all y-coordinates equals 0.

🎯 Exam Tip: When centroid is origin, set up two equations: sum of x-coordinates = 0 and sum of y-coordinates = 0. Solve for unknowns.

Question 7. The mid-point of the segment AB, as shown in diagram, is C (4, -3). Write down the coordinates of A and B.
Answer: Since point A lies on x-axis, let the coordinates of point A be (x, 0). Since point B lies on y-axis, let the coordinates of point B be (0, y). Given, mid-point of AB is C (4, -3).

Using mid-point formula:
\( (4, -3) = \left(\frac{x + 0}{2}, \frac{0 + y}{2}\right) \)

\( \Rightarrow (4, -3) = \left(\frac{x}{2}, \frac{y}{2}\right) \)

\( \Rightarrow 4 = \frac{x}{2} \) and \( -3 = \frac{y}{2} \)

\( \Rightarrow x = 8 \) and \( y = -6 \)

Thus, the coordinates of point A are (8, 0) and the coordinates of point B are (0, -6).
In simple words: We used the mid-point formula. Since A is on x-axis and B is on y-axis, we found their exact positions using the given mid-point.

[Diagram: A coordinate system showing a line segment AB where A is on the positive x-axis, B is on the negative y-axis, and C(4,-3) is marked as the midpoint of the segment]

📝 Teacher's Note: Show students how points on axes have one coordinate as zero. Draw the diagram to make it clear. This makes the problem much easier.

🎯 Exam Tip: Always write the mid-point formula first. Then substitute the known values. Check your answer by putting it back in the formula.

Question 8. AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find (i) the length of radius AC (ii) the coordinates of B.
Answer:
(i) Radius AC = \( \sqrt{(3 + 2)^2 + (-7 - 5)^2} \)
= \( \sqrt{5^2 + (-12)^2} \)
= \( \sqrt{25 + 144} \)
= \( \sqrt{169} \)
= 13 units

(ii) Let the coordinates of B be (x, y). Using mid-point formula, we have:
\( -2 = \frac{3 + x}{2} \) and \( 5 = \frac{-7 + y}{2} \)

\( \Rightarrow -4 = 3 + x \) and \( 10 = -7 + y \)

\( \Rightarrow x = -7 \) and \( y = 17 \)

Thus, the coordinates of B are (-7, 17).
In simple words: The radius is the distance from center to any point on circle. Since AB is diameter, center C is exactly in the middle of A and B.

📝 Teacher's Note: Remind students that diameter means the center is the mid-point of the two end points. Use distance formula for radius calculation.

🎯 Exam Tip: For radius, use distance formula. For finding other end of diameter, use mid-point formula with center as the mid-point.

Question 9. Find the co-ordinates of the centroid of a triangle ABC whose vertices are: A (-1, 3), B (1, -1) and C (5, 1)
Answer: Co-ordinates of the centroid of triangle ABC are:
\( \left(\frac{-1 + 1 + 5}{3}, \frac{3 - 1 + 1}{3}\right) \)
= \( \left(\frac{5}{3}, 1\right) \)
In simple words: Centroid is the point where all three medians of triangle meet. We find it by taking average of all x-coordinates and average of all y-coordinates.

📝 Teacher's Note: Tell students centroid is like the balance point of triangle. It divides each median in ratio 2:1. The formula is just adding coordinates and dividing by 3.

🎯 Exam Tip: Remember the centroid formula: add all x-coordinates and divide by 3, add all y-coordinates and divide by 3. Don't forget the brackets.

Question 10. The mid-point of the line-segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Answer: Using mid-point formula:
\( (2, -2a) = \left(\frac{4a - 4}{2}, \frac{2b - 3 + 3b}{2}\right) \)

\( \Rightarrow 2 = \frac{4a - 4}{2} \)
\( \Rightarrow 4a - 4 = 4 \)
\( \Rightarrow 4a = 8 \)
\( \Rightarrow a = 2 \)

Also,
\( -2a = \frac{2b - 3 + 3b}{2} \)
\( \Rightarrow -2 \times 2 = \frac{5b - 3}{2} \)
\( \Rightarrow 5b - 3 = -8 \)
\( \Rightarrow 5b = -5 \)
\( \Rightarrow b = -1 \)
In simple words: We used the mid-point formula and made two equations. Then we solved both equations to find a and b.

📝 Teacher's Note: Show students how to set up two separate equations from the x and y coordinates. Solve one equation at a time to avoid confusion.

🎯 Exam Tip: Write the mid-point formula first. Then equate x-coordinates separately and y-coordinates separately to get two equations. Solve step by step.

Question 11. The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
Answer: Mid-point of (2a, 4) and (-2, 2b) is (1, 2a + 1), therefore using mid-point formula:

\( x = \frac{x_1 + x_2}{2} \), \( y = \frac{y_1 + y_2}{2} \)

\( 1 = \frac{2a - 2}{2} \), \( 2a + 1 = \frac{4 + 2b}{2} \)

\( 1 = a - 1 \)
\( \Rightarrow a = 2 \)

\( 2a + 1 = 2 + b \)
Putting a = 2 in 2a + 1 = 2 + b, we get:
5 - 2 = b \( \Rightarrow \) b = 3

Therefore, a = 2, b = 3.
In simple words: We made two equations from x and y coordinates. First we found a = 2, then used this value to find b = 3.

📝 Teacher's Note: Show students how to substitute the value of a back into the second equation. This helps them check their work and avoid mistakes.

🎯 Exam Tip: After finding one variable, always substitute it back to find the other. Write your final answer clearly: a = 2, b = 3.

Question 12. (i) Write down the co-ordinates of the point P that divides the line joining A (-4, 1) and B (17, 10) in the ratio 1: 2. (ii) Calculate the distance OP, where O is the origin. (iii) In what ratio does the y-axis divide the line AB?
Answer:
(i) Co-ordinates of point P are:
\( \left(\frac{1 \times 17 + 2 \times (-4)}{1 + 2}, \frac{1 \times 10 + 2 \times 1}{1 + 2}\right) \)
= \( \left(\frac{17 - 8}{3}, \frac{10 + 2}{3}\right) \)
= \( \left(\frac{9}{3}, \frac{12}{3}\right) \)
= (3, 4)

(ii) OP = \( \sqrt{(0 - 3)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units

(iii) Let AB be divided by the point P (0, y) lying on y-axis in the ratio k: 1.
\( (0, y) = \left(\frac{k \times 17 + 1 \times (-4)}{k + 1}, \frac{k \times 10 + 1 \times 1}{k + 1}\right) \)

\( \Rightarrow (0, y) = \left(\frac{17k - 4}{k + 1}, \frac{10k + 1}{k + 1}\right) \)

\( \Rightarrow 0 = \frac{17k - 4}{k + 1} \)
\( \Rightarrow 17k - 4 = 0 \)
\( \Rightarrow k = \frac{4}{17} \)

Thus, the ratio in which the y-axis divide the line AB is 4: 17.
In simple words: We used section formula to find point P. Distance formula gave us OP. For y-axis division, we found where line AB crosses y-axis.

📝 Teacher's Note: Explain that section formula works when we know the ratio. For y-axis intersection, x-coordinate becomes 0, which helps find the ratio.

🎯 Exam Tip: For section formula, write m₁ and m₂ clearly above the fraction. For distance, use \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). Show all steps.

Question 13. Prove that the points A (-5, 4), B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D so that ABCD is a square.
Answer: We have:

AB = \( \sqrt{(-1 + 5)^2 + (-2 - 4)^2} = \sqrt{16 + 36} = \sqrt{52} \)

BC = \( \sqrt{(-1 - 5)^2 + (-2 - 2)^2} = \sqrt{36 + 16} = \sqrt{52} \)

AC = \( \sqrt{(5 + 5)^2 + (2 - 4)^2} = \sqrt{100 + 4} = \sqrt{104} \)

AB² + BC² = 52 + 52 = 104
AC² = 104

\( \because \) AB = BC and AB² + BC² = AC²
\( \therefore \) ABC is an isosceles right-angled triangle.

Let the coordinates of D be (x, y).
If ABCD is a square, then,
Mid-point of AC = Mid-point of BD
\( \left(\frac{-5 + 5}{2}, \frac{4 + 2}{2}\right) = \left(\frac{x - 1}{2}, \frac{y - 2}{2}\right) \)

\( 0 = \frac{x - 1}{2}, 3 = \frac{y - 2}{2} \)

x = 1, y = 8
Thus, the co-ordinates of point D are (1, 8).
In simple words: We found all three side lengths. Two sides are equal and they follow Pythagoras theorem, so it is isosceles right triangle. For square, diagonals bisect each other.

📝 Teacher's Note: Show students how to check for right angle using Pythagoras theorem: a² + b² = c². For square, diagonals have same mid-point.

🎯 Exam Tip: Calculate all three distances first. Check AB = BC for isosceles, and AB² + BC² = AC² for right angle. Write both conditions clearly.

Question 14. M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1). Find the co-ordinates of point M. Further, if R (2, 2) divides the line segment joining M and the origin in the ratio p: q, find the ratio p: q.
Answer: Given, M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1).
The co-ordinates of point M are:
\( \left(\frac{-3 + 9}{2}, \frac{7 - 1}{2}\right) \)
= \( \left(\frac{6}{2}, \frac{6}{2}\right) \)
= (3, 3)

Now, R (2, 2) divides the line segment joining M (3, 3) and origin O (0, 0) in ratio p: q.
Using section formula:
\( (2, 2) = \left(\frac{p \times 0 + q \times 3}{p + q}, \frac{p \times 0 + q \times 3}{p + q}\right) \)
\( (2, 2) = \left(\frac{3q}{p + q}, \frac{3q}{p + q}\right) \)

\( 2 = \frac{3q}{p + q} \)
\( 2(p + q) = 3q \)
\( 2p + 2q = 3q \)
\( 2p = q \)
\( \frac{p}{q} = \frac{1}{2} \)

Therefore, p: q = 1: 2.
In simple words: First we found M using mid-point formula. Then we used section formula to find how R divides the line MO.

📝 Teacher's Note: Show students how both x and y coordinates give the same equation when the point divides in same ratio. This confirms our answer.

🎯 Exam Tip: Find M first using mid-point formula. Then use section formula with M and O as endpoints and R as dividing point. Both coordinates will give same ratio.

Question 15. Calculate the ratio in which the line joining A(-4, 2) and B(3, 6) is divided by point P(x, 3). Also, find (a) x (b) length of AP.
Answer:
Given:
Points A(-4, 2) and B(3, 6)
Point P(x, 3) divides AB

Step 1: Let P divide AB in ratio k : 1
Using section formula: \( P = \left(\frac{3k - 4}{k + 1}, \frac{6k + 2}{k + 1}\right) \)

Step 2: Since y-coordinate of P is 3
\( \frac{6k + 2}{k + 1} = 3 \)
\( \implies 6k + 2 = 3(k + 1) \)
\( \implies 6k + 2 = 3k + 3 \)
\( \implies 3k = 1 \)
\( \implies k = \frac{1}{3} \)

Step 3: Required ratio is 1 : 3

Step 4: Find x-coordinate
\( x = \frac{3k - 4}{k + 1} = \frac{3 \times \frac{1}{3} - 4}{\frac{1}{3} + 1} = \frac{1 - 4}{\frac{4}{3}} = \frac{-3}{\frac{4}{3}} = -\frac{9}{4} \)

Step 5: Find length of AP
\( AP = \sqrt{(-\frac{9}{4} + 4)^2 + (3 - 2)^2} = \sqrt{(\frac{7}{4})^2 + 1^2} = \sqrt{\frac{49}{16} + 1} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4} \) units

Final answers:
(a) \( x = -\frac{9}{4} \)
(b) Length of AP = \( \frac{\sqrt{65}}{4} \) units
Ratio = 1 : 3

In simple words: We found where point P cuts the line AB. The ratio is 1:3. We used the section formula to find the exact position and distance.

📝 Teacher's Note: Show students how to use section formula step by step. The y-coordinate gives us the ratio first. Then we can find x-coordinate easily.

🎯 Exam Tip: Always write the ratio in simplest form. Use distance formula carefully for length calculations. Show all steps clearly.

Question 16. Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7).
Answer:
Given:
Points P(2, -2) and Q(3, 7)
Line: 2x + y = 4

Step 1: Let the line divide PQ in ratio k : 1
Point of division = \( \left(\frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1}\right) \)

Step 2: This point lies on line 2x + y = 4
\( 2\left(\frac{3k + 2}{k + 1}\right) + \frac{7k - 2}{k + 1} = 4 \)

Step 3: Solve for k
\( \frac{6k + 4 + 7k - 2}{k + 1} = 4 \)
\( \implies 6k + 4 + 7k - 2 = 4k + 4 \)
\( \implies 13k + 2 = 4k + 4 \)
\( \implies 9k = 2 \)
\( \implies k = \frac{2}{9} \)

Required ratio is 2 : 9

In simple words: We found where the given line cuts the line segment PQ. We used section formula and substituted in the line equation to find the ratio.

📝 Teacher's Note: Explain that when a line cuts a line segment, we find the cutting point using section formula. Then use the line equation to find the ratio.

🎯 Exam Tip: Write the section formula correctly. Substitute carefully in the line equation. Simplify the ratio to lowest terms.

Question 17. If the abscissa of a point P is 2. Find the ratio in which this point divides the line segment joining the point (-4, 3) and (6, 3). Also, find the co-ordinates of point P.
Answer:
Given:
Points A(-4, 3) and B(6, 3)
Point P has x-coordinate = 2

Step 1: Let P divide AB in ratio k : 1
Using section formula: \( P = \left(\frac{6k - 4}{k + 1}, \frac{3k + 3}{k + 1}\right) \)

Step 2: Since x-coordinate of P is 2
\( \frac{6k - 4}{k + 1} = 2 \)
\( \implies 6k - 4 = 2k + 2 \)
\( \implies 4k = 6 \)
\( \implies k = \frac{3}{2} \)

Step 3: Required ratio is 3 : 2

Step 4: Find y-coordinate
\( y = \frac{3k + 3}{k + 1} = \frac{3 \times \frac{3}{2} + 3}{\frac{3}{2} + 1} = \frac{\frac{9}{2} + 3}{\frac{5}{2}} = \frac{\frac{15}{2}}{\frac{5}{2}} = 3 \)

Final answers:
Ratio = 3 : 2
Coordinates of P = (2, 3)

In simple words: Since both given points have the same y-coordinate (3), point P will also have y-coordinate 3. We used the x-coordinate to find the ratio.

📝 Teacher's Note: Point out that both A and B have the same y-coordinate, so they lie on a horizontal line. Any point dividing this segment will have the same y-coordinate.

🎯 Exam Tip: When points have same y-coordinate, the dividing point will also have same y-coordinate. Use the given x-coordinate to find the ratio.

Question 18. The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q, point P lies on the line 2x - y + k = 0, find the value of k. Also, find the co-ordinates of point Q.
Answer:
Given:
Points A(2, 1) and B(5, -8)
Line AB is trisected by points P and Q
P lies on line 2x - y + k = 0

Step 1: Find coordinates of P
P divides AB in ratio 1 : 2
\( P = \left(\frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2}\right) = \left(\frac{9}{3}, \frac{-6}{3}\right) = (3, -2) \)

Step 2: Find k
Since P(3, -2) lies on 2x - y + k = 0
\( 2(3) - (-2) + k = 0 \)
\( \implies 6 + 2 + k = 0 \)
\( \implies k = -8 \)

Step 3: Find coordinates of Q
Q divides AB in ratio 2 : 1
\( Q = \left(\frac{2 \times 5 + 1 \times 2}{2 + 1}, \frac{2 \times (-8) + 1 \times 1}{2 + 1}\right) = \left(\frac{12}{3}, \frac{-15}{3}\right) = (4, -5) \)

Final answers:
k = -8
Coordinates of Q = (4, -5)

In simple words: Trisection means dividing into three equal parts. P is 1/3 of the way from A to B, and Q is 2/3 of the way from A to B.

📝 Teacher's Note: Explain that trisection creates two points - one at 1:2 ratio and other at 2:1 ratio. Draw a line segment and show where these points fall.

🎯 Exam Tip: For trisection, always use ratios 1:2 and 2:1. Substitute the point coordinates correctly in the given line equation to find k.

Question 19. M is the mid-point of the line segment joining the points A(0, 4) and B(6, 0). M also divides the line segment OP in the ratio 1 : 3. Find: (a) co-ordinates of M (b) co-ordinates of P (c) length of BP
Answer:
[Diagram: This diagram shows points O, A, B, M, and P plotted on a coordinate system with O at origin, A at (0,4), B at (6,0), and lines connecting these points.]

Step 1: Find coordinates of M (midpoint of AB)
\( M = \left(\frac{0 + 6}{2}, \frac{4 + 0}{2}\right) = (3, 2) \)

Step 2: Find coordinates of P
M divides OP in ratio 1 : 3
Let P = (x, y), O = (0, 0)
\( M = \left(\frac{x + 0}{1 + 3}, \frac{y + 0}{1 + 3}\right) = \left(\frac{x}{4}, \frac{y}{4}\right) \)

But M = (3, 2), so:
\( \frac{x}{4} = 3 \text{ and } \frac{y}{4} = 2 \)
\( \implies x = 12 \text{ and } y = 8 \)

Step 3: Find length of BP
\( BP = \sqrt{(12 - 6)^2 + (8 - 0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units

Final answers:
(a) Coordinates of M = (3, 2)
(b) Coordinates of P = (12, 8)
(c) Length of BP = 10 units

In simple words: M is exactly halfway between A and B. M is also 1/4 of the way from O to P. We used these facts to find all coordinates.

📝 Teacher's Note: Draw the diagram clearly. Show students that M has two roles - midpoint of AB and also divides OP in 1:3. This helps find P.

🎯 Exam Tip: Use midpoint formula for part (a). For part (b), use section formula with ratio 1:3. Apply distance formula carefully for part (c).

Question 20. Find the image of the point A(5, -3) under reflection in the point P(-1, 3).
Answer:
Given:
Point A(5, -3)
Reflection point P(-1, 3)

Step 1: Let A' be the image of A under reflection in P
P is the midpoint of AA'

Step 2: Let A' = (x, y)
Using midpoint formula:
\( P = \left(\frac{5 + x}{2}, \frac{-3 + y}{2}\right) = (-1, 3) \)

Step 3: Solve for x and y
\( \frac{5 + x}{2} = -1 \)
\( \implies 5 + x = -2 \)
\( \implies x = -7 \)

\( \frac{-3 + y}{2} = 3 \)
\( \implies -3 + y = 6 \)
\( \implies y = 9 \)

Image of A is A'(-7, 9)

In simple words: When we reflect a point in another point, the reflection point becomes the midpoint of the original point and its image. Like folding paper at point P.

📝 Teacher's Note: Show students how reflection works using a folded paper. The fold line passes through P, and A and A' are equidistant from P on opposite sides.

🎯 Exam Tip: Remember that in point reflection, the center of reflection is the midpoint of the original point and its image. Use midpoint formula to find the image.

Question 21. A(-4, 2), B(0, 2) and C(-2, -4) are the vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of ∆ PQR is the same as the centroid of ∆ ABC.
Answer:
A(-4, 2), B(0, 2) and C(-2, -4) are the vertices of ∆ABC.

Centroid of ∆ABC = \( \left(\frac{-4 + 0 - 2}{3}, \frac{2 + 2 - 4}{3}\right) = \left(\frac{-6}{3}, \frac{0}{3}\right) = (-2, 0) \)

P, Q and R are the mid-points of sides BC, CA and AB respectively.

Coordinates of P = \( \left(\frac{0 - 2}{2}, \frac{2 - 4}{2}\right) = \left(\frac{-2}{2}, \frac{-2}{2}\right) = (-1, -1) \)

Coordinates of Q = \( \left(\frac{-2 - 4}{2}, \frac{-4 + 2}{2}\right) = \left(\frac{-6}{2}, \frac{-2}{2}\right) = (-3, -1) \)

Coordinates of R = \( \left(\frac{-4 + 0}{2}, \frac{2 + 2}{2}\right) = \left(\frac{-4}{2}, \frac{4}{2}\right) = (-2, 2) \)

Centroid of ∆PQR = \( \left(\frac{-1 - 3 - 2}{3}, \frac{-1 + 1 + 2}{3}\right) = \left(\frac{-6}{3}, \frac{0}{3}\right) = (-2, 0) \)

Therefore, Centroid of ∆ABC = Centroid of ∆PQR.
In simple words: We found the center point of both triangles using the centroid formula. Both triangles have the same center point at (-2, 0).

📝 Teacher's Note: Show students that the centroid is always at the same point. Draw the original triangle and the triangle formed by joining midpoints. Both have the same center.

🎯 Exam Tip: Write the centroid formula clearly: \( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \). Show all midpoint calculations step by step. You get marks for each step.

Question 22. Centroid of ∆ABC = \( \left(\frac{3 - y - 1}{3}, \frac{1 - 4 + x}{3}\right) = \left(\frac{4 - y}{3}, \frac{5 + x}{3}\right) \) ... (i)
Answer:
P, Q and R are the mid points of the sides BC, CA and AB.

By mid-point formula, we get

P = \( \left(\frac{y - 1}{2}, \frac{4 + x}{2}\right) \), Q = \( \left(\frac{4}{2}, \frac{1 - x}{2}\right) \) and R = \( \left(\frac{3 - y}{2}, \frac{5}{2}\right) \)

Centroid of ∆PQR = \( \left(\frac{\frac{y - 1}{2} + \frac{4}{2} + \frac{3 - y}{2}}{3}, \frac{\frac{4 + x}{2} + \frac{1 - x}{2} + \frac{5}{2}}{3}\right) \)

= \( \left(\frac{y - 1 + 4 + 3 - y}{6}, \frac{4 + x + 1 - x + 5}{6}\right) \)

= \( \left(\frac{8 - 2y}{6}, \frac{10 - 2x}{6}\right) \)

= \( \left(\frac{4 + y}{3}, \frac{5 + x}{3}\right) \) ...... (ii)

From (i) and (ii), we get

Centroid of ∆ABC = Centroid of ∆PQR
In simple words: This is a general proof showing that any triangle and the triangle made by joining its midpoints will always have the same centroid. The algebra proves this works for any triangle.

📝 Teacher's Note: This is the general case using variables. Students should understand that this property is true for every triangle, not just specific examples with numbers.

🎯 Exam Tip: When doing general proofs, keep all algebra steps clear. Write (i) and (ii) to show both expressions are equal. This proves the statement for all triangles.