GSEB Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1. Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2: 3.
Answer: Let the given points be \( A(-1, 7) \) and \( B(4, -3) \).
Here, we have \( x_1 = -1 \), \( y_1 = 7 \), \( x_2 = 4 \), \( y_2 = -3 \).
Also, the ratio of division is \( m_1 = 2 \) and \( m_2 = 3 \).
Let the required point be \( P(x, y) \).
Then, the x-coordinate is given by \( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
And the y-coordinate is given by \( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
Now, we calculate \( x \):
\( x = \frac{2(4)+3(-1)}{2+3} \)
\( x = \frac{8-3}{5} \)
\( x = \frac{5}{5} \)
\( \implies x = 1 \)
Next, we calculate \( y \):
\( y = \frac{2(-3)+3(7)}{2+3} \)
\( y = \frac{-6+21}{5} \)
\( y = \frac{15}{5} \)
\( \implies y = 3 \)
Therefore, the point needed is \( (1, 3) \).
In simple words: We used the section formula to find the coordinates. The formula helps us determine a point that divides a line segment into a specific ratio. By putting in the numbers, we found the x-coordinate to be 1 and the y-coordinate to be 3.

Exam Tip: Remember the section formula \( (x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \) for internal division and clearly label \( m_1, m_2, x_1, y_1, x_2, y_2 \) before substituting values.

Question 2. Find the coordinates of the points of tri-section of the line segment joining (4, -1) and (-2, -3).
Answer: Let the line segment be AB, with \( A(4, -1) \) and \( B(-2, -3) \).
Let P and Q be the points that divide AB into three equal parts (tri-section).
This means \( AP = PQ = QB = 1 \).

Case I: For point P
Point P divides AB in the ratio \( 1:2 \). So, \( m_1 = 1 \) and \( m_2 = 2 \).
We have \( x_1 = 4 \), \( y_1 = -1 \), \( x_2 = -2 \), \( y_2 = -3 \).
The coordinates of P are given by the section formula:
\( P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \)
\( P(x, y) = \left( \frac{1(-2)+2(4)}{1+2}, \frac{1(-3)+2(-1)}{1+2} \right) \)
\( P(x, y) = \left( \frac{-2+8}{3}, \frac{-3-2}{3} \right) \)
\( P(x, y) = \left( \frac{6}{3}, \frac{-5}{3} \right) \)
\( \implies P(x, y) = \left( 2, \frac{-5}{3} \right) \)

Case II: For point Q
Point Q divides AB in the ratio \( 2:1 \). So, \( m_1 = 2 \) and \( m_2 = 1 \).
We have \( x_1 = 4 \), \( y_1 = -1 \), \( x_2 = -2 \), \( y_2 = -3 \).
The coordinates of Q are given by the section formula:
\( Q(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \)
\( Q(x, y) = \left( \frac{2(-2)+1(4)}{2+1}, \frac{2(-3)+1(-1)}{2+1} \right) \)
\( Q(x, y) = \left( \frac{-4+4}{3}, \frac{-6-1}{3} \right) \)
\( Q(x, y) = \left( \frac{0}{3}, \frac{-7}{3} \right) \)
\( \implies Q(x, y) = \left( 0, \frac{-7}{3} \right) \)
Thus, the coordinates of the required points are \( \left( 2, \frac{-5}{3} \right) \) and \( \left( 0, \frac{-7}{3} \right) \).
In simple words: To find the points that divide a line into three equal parts, we use the section formula twice. First, for the point P that splits the line in a 1:2 ratio. Second, for the point Q that splits it in a 2:1 ratio. We calculated the x and y values for both points using these ratios.

Exam Tip: For tri-section, remember there are two points, P and Q. P divides the segment in a 1:2 ratio, and Q divides it in a 2:1 ratio. Alternatively, Q can be found as the midpoint of PB after finding P, or P as the midpoint of AQ after finding Q.

Question 3. To conduct Sports Day activities in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Figure. Niharika runs \( \frac {1}{4} \)th the distance AD on the 2 line and posts a green flag. Preet runs \( \frac {1}{5} \)th distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Answer: Let's determine the coordinates of Niharika's (green flag) and Preet's (red flag) positions.
The distance AD has 100 flower pots, each 1m apart, so AD is 100m long.

Niharika's position:
She runs on the 2nd line, so her x-coordinate is 2.
She runs \( \frac {1}{4} \)th of the distance AD. Distance AD = 100m.
So, her y-coordinate is \( \frac {1}{4} \times 100 = 25 \).
Niharika's flag (M) is at \( (2, 25) \).

Preet's position:
She runs on the 8th line, so her x-coordinate is 8.
She runs \( \frac {1}{5} \)th of the distance AD. Distance AD = 100m.
So, her y-coordinate is \( \frac {1}{5} \times 100 = 20 \).
Preet's flag (N) is at \( (8, 20) \).

Distance between the flags (MN):
Using the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( MN = \sqrt{(8-2)^2 + (20-25)^2} \)
\( MN = \sqrt{6^2 + (-5)^2} \)
\( MN = \sqrt{36 + 25} \)
\( MN = \sqrt{61} \)
So, the distance between the flags is \( \sqrt{61} \) meters.

Rashmi's position (blue flag):
Rashmi posts her flag exactly halfway between M(2, 25) and N(8, 20).
This means we need to find the midpoint (P) of the line segment MN.
Using the midpoint formula \( P(x, y) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \):
\( P(x, y) = \left( \frac{2+8}{2}, \frac{25+20}{2} \right) \)
\( P(x, y) = \left( \frac{10}{2}, \frac{45}{2} \right) \)
\( P(x, y) = (5, 22.5) \)
Hence, Rashmi should post the blue flag on the fifth line at a distance of 22.5 m above it.
In simple words: First, we find where Niharika and Preet put their flags using the given fractions of the total distance. Then, we use the distance formula to calculate the separation between their flags. Finally, to find where Rashmi should put her flag, we calculate the middle point between Niharika's and Preet's flags.

Exam Tip: Clearly define the coordinates for each person's flag. Remember to use the distance formula for separation and the midpoint formula for the halfway point. Show all steps for full marks.

Question 4. Find the ratio in which the segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer: Let the segment joining \( A(-3, 10) \) and \( B(6, -8) \) be divided by point \( P(-1, 6) \) in the ratio \( k:1 \).
Here, \( x_1 = -3 \), \( y_1 = 10 \), \( x_2 = 6 \), \( y_2 = -8 \).
The point of division is \( P(-1, 6) \), and the ratio is \( m_1 = k \), \( m_2 = 1 \).
Using the section formula for the x-coordinate:
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
\( -1 = \frac{k(6)+1(-3)}{k+1} \)
\( -1 = \frac{6k-3}{k+1} \)
\( -1(k+1) = 6k-3 \)
\( -k-1 = 6k-3 \)
\( -1+3 = 6k+k \)
\( 2 = 7k \)
\( \implies k = \frac{2}{7} \)
We can also verify this using the y-coordinate:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( 6 = \frac{k(-8)+1(10)}{k+1} \)
\( 6 = \frac{-8k+10}{k+1} \)
\( 6(k+1) = -8k+10 \)
\( 6k+6 = -8k+10 \)
\( 6k+8k = 10-6 \)
\( 14k = 4 \)
\( k = \frac{4}{14} \)
\( \implies k = \frac{2}{7} \)
Since both coordinates give the same value for k, the ratio is \( k:1 = \frac{2}{7}:1 \).
To express this as whole numbers, multiply both sides by 7:
\( 2:7 \).
Hence, the required ratio is \( 2:7 \).
In simple words: We used the section formula and assumed the line segment is divided in a k:1 ratio. By plugging in the coordinates of the given points and the point of division, we solved for k using both the x and y coordinates. Both calculations gave k as 2/7, so the division ratio is 2:7.

Exam Tip: When finding the ratio, it's generally best to assume the ratio is \( k:1 \). This simplifies calculations by having only one unknown (k). Verify your answer using both x and y coordinates if possible to ensure consistency.

Question 5. Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis, and find the coordinates of the point of divisions.
Answer: Let the given points be \( A(1, -5) \) and \( B(-4, 5) \).
Let the x-axis cut the line segment AB at a point P in the ratio \( k:1 \).
The coordinates of any point on the x-axis are \( (x, 0) \). So, the y-coordinate of P is 0.
Here, \( x_1 = 1 \), \( y_1 = -5 \), \( x_2 = -4 \), \( y_2 = 5 \).
The ratio is \( m_1 = k \), \( m_2 = 1 \).
Using the section formula for the y-coordinate, which we know is 0:
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( 0 = \frac{k(5) + 1(-5)}{k+1} \)
\( 0 = \frac{5k-5}{k+1} \)
Since the numerator must be zero for the fraction to be zero:
\( 5k-5 = 0 \)
\( 5k = 5 \)
\( \implies k = 1 \)
Hence, the required ratio is \( 1:1 \).

Now, we need to find the coordinates of the point of division P. Since \( k=1 \), the point P is the midpoint of AB.
Using the section formula with \( k=1 \), or simply the midpoint formula:
\( P(x, y) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\( P(x, y) = \left( \frac{1+(-4)}{2}, \frac{-5+5}{2} \right) \)
\( P(x, y) = \left( \frac{-3}{2}, \frac{0}{2} \right) \)
\( \implies P(x, y) = \left( -\frac{3}{2}, 0 \right) \)
Therefore, the x-axis divides the line segment in the ratio \( 1:1 \), and the point of division is \( \left( -\frac{3}{2}, 0 \right) \).
In simple words: We assume the x-axis divides the line in a k:1 ratio. Since any point on the x-axis has a y-coordinate of 0, we use the y-part of the section formula and set it to 0. This helps us find k, which turns out to be 1. A ratio of 1:1 means the point is the midpoint. Then we use the midpoint formula to find the exact coordinates of that point.

Exam Tip: Remember that any point on the x-axis has y-coordinate 0, and any point on the y-axis has x-coordinate 0. This is a crucial piece of information for solving such problems.

Question 6. If (1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer: Let the vertices of the parallelogram be \( A(1, 2) \), \( B(4, y) \), \( C(x, 6) \), and \( D(3, 5) \).
We know that the diagonals of a parallelogram bisect each other. This means their midpoints are the same.
Let's find the midpoint of diagonal AC:
Midpoint of AC \( = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) = \left( \frac{1+x}{2}, \frac{2+6}{2} \right) = \left( \frac{1+x}{2}, \frac{8}{2} \right) = \left( \frac{1+x}{2}, 4 \right) \)

Now, let's find the midpoint of diagonal BD:
Midpoint of BD \( = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) = \left( \frac{4+3}{2}, \frac{y+5}{2} \right) = \left( \frac{7}{2}, \frac{y+5}{2} \right) \)

Since the midpoints are the same, we can equate their coordinates:
\( \left( \frac{1+x}{2}, 4 \right) = \left( \frac{7}{2}, \frac{y+5}{2} \right) \)

Equating the x-coordinates:
\( \frac{1+x}{2} = \frac{7}{2} \)
\( 1+x = 7 \)
\( x = 7-1 \)
\( \implies x = 6 \)

Equating the y-coordinates:
\( 4 = \frac{y+5}{2} \)
\( 4 \times 2 = y+5 \)
\( 8 = y+5 \)
\( y = 8-5 \)
\( \implies y = 3 \)
Hence, the required values are \( x = 6 \) and \( y = 3 \).
In simple words: In a parallelogram, the middle points of the diagonals are always the same. So, we found the midpoint for diagonal AC and the midpoint for diagonal BD. We then set their x-coordinates equal to each other and their y-coordinates equal to each other. Solving these two simple equations gave us the values for x and y.

Exam Tip: The key property for solving this type of problem is that the diagonals of a parallelogram bisect each other. This means their midpoints are identical. Always remember the midpoint formula.

Question 7. Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, -3) and B is (1, 4).
Answer: Let the given point be \( A(x, y) \). The diameter is AB, and the center of the circle C is \( (2, -3) \). Point B is \( (1, 4) \).
Since AB is the diameter, the center C is the midpoint of AB.
Using the midpoint formula for the x-coordinate:
\( \text{Midpoint x} = \frac{x_1+x_2}{2} \)
\( 2 = \frac{x+1}{2} \)
\( 2 \times 2 = x+1 \)
\( 4 = x+1 \)
\( x = 4-1 \)
\( \implies x = 3 \)

Using the midpoint formula for the y-coordinate:
\( \text{Midpoint y} = \frac{y_1+y_2}{2} \)
\( -3 = \frac{y+4}{2} \)
\( -3 \times 2 = y+4 \)
\( -6 = y+4 \)
\( y = -6-4 \)
\( \implies y = -10 \)
Hence, the coordinates of point A are \( (3, -10) \).
In simple words: The center of a circle is always the middle point of its diameter. We have the center and one end of the diameter. Using the midpoint formula in reverse, we calculated the coordinates for the other end of the diameter, which is point A.

Exam Tip: The center of a circle is always the midpoint of any diameter. If you have the center and one endpoint of a diameter, you can easily find the other endpoint using the midpoint formula principles.

Question 8. If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that \( AP = \frac {3}{7}AB \) and P lies on the line segment AB.
Answer: We are given points \( A(-2, -2) \) and \( B(2, -4) \).
We are told that \( AP = \frac {3}{7}AB \).
This implies that the length of AP is 3 parts, while the total length of AB is 7 parts.
If AP is 3 parts of AB, then PB must be the remaining parts of AB, i.e., \( PB = AB - AP \).
\( PB = 7 - 3 = 4 \) parts.
So, point P divides the line segment AB in the ratio \( AP:PB = 3:4 \).
Thus, \( m_1 = 3 \) and \( m_2 = 4 \).
Here, \( x_1 = -2 \), \( y_1 = -2 \), \( x_2 = 2 \), \( y_2 = -4 \).
Using the section formula to find the coordinates of P \( (x, y) \):
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \)
\( x = \frac{3(2)+4(-2)}{3+4} \)
\( x = \frac{6-8}{7} \)
\( x = \frac{-2}{7} \)

\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \)
\( y = \frac{3(-4)+4(-2)}{3+4} \)
\( y = \frac{-12-8}{7} \)
\( y = \frac{-20}{7} \)
Hence, the coordinates of point P are \( \left( -\frac{2}{7}, -\frac{20}{7} \right) \).
In simple words: The problem states that AP is 3/7 of the total length of AB. This means P divides AB in a 3:4 ratio (since 7-3 = 4 parts remain). We then applied the section formula with this ratio to find the exact x and y coordinates of point P.

Exam Tip: When given a fractional relationship like \( AP = \frac{a}{b}AB \), understand that this implies the ratio \( AP:PB = a:(b-a) \). This crucial step helps in applying the section formula correctly.

Question 9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer: Let the line segment be AB, with \( A(-2, 2) \) and \( B(2, 8) \).
Let P, Q, and R be the three points that divide the line segment into four equal parts. So, \( AP = PQ = QR = RB \).

Case I: For point P
Point P divides AB in the ratio \( 1:3 \) (1 part from A, 3 parts from B). So, \( m_1 = 1 \) and \( m_2 = 3 \).
We have \( x_1 = -2 \), \( y_1 = 2 \), \( x_2 = 2 \), \( y_2 = 8 \).
The coordinates of P are given by:
\( P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \)
\( P(x, y) = \left( \frac{1(2)+3(-2)}{1+3}, \frac{1(8)+3(2)}{1+3} \right) \)
\( P(x, y) = \left( \frac{2-6}{4}, \frac{8+6}{4} \right) \)
\( P(x, y) = \left( \frac{-4}{4}, \frac{14}{4} \right) \)
\( \implies P(x, y) = \left( -1, \frac{7}{2} \right) \)

Case II: For point Q
Point Q is the midpoint of AB, as it divides the segment into two equal halves (2 parts from A, 2 parts from B), meaning a ratio of \( 2:2 \), which simplifies to \( 1:1 \).
Using the midpoint formula for Q:
\( Q(x, y) = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\( Q(x, y) = \left( \frac{-2+2}{2}, \frac{2+8}{2} \right) \)
\( Q(x, y) = \left( \frac{0}{2}, \frac{10}{2} \right) \)
\( \implies Q(x, y) = (0, 5) \)

Case III: For point R
Point R divides AB in the ratio \( 3:1 \) (3 parts from A, 1 part from B). So, \( m_1 = 3 \) and \( m_2 = 1 \).
We have \( x_1 = -2 \), \( y_1 = 2 \), \( x_2 = 2 \), \( y_2 = 8 \).
The coordinates of R are given by:
\( R(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \)
\( R(x, y) = \left( \frac{3(2)+1(-2)}{3+1}, \frac{3(8)+1(2)}{3+1} \right) \)
\( R(x, y) = \left( \frac{6-2}{4}, \frac{24+2}{4} \right) \)
\( R(x, y) = \left( \frac{4}{4}, \frac{26}{4} \right) \)
\( \implies R(x, y) = \left( 1, \frac{13}{2} \right) \)
Thus, the coordinates of the points dividing the line segment into four equal parts are \( P\left(-1, \frac{7}{2}\right) \), \( Q(0, 5) \), and \( R\left(1, \frac{13}{2}\right) \).
In simple words: To divide a line into four equal parts, we need three points: P, Q, and R. P divides the line in a 1:3 ratio, Q divides it in a 2:2 (or 1:1) ratio, and R divides it in a 3:1 ratio. We applied the section formula for P and R, and the simpler midpoint formula for Q, to find their respective coordinates.

Exam Tip: For problems involving division into multiple equal parts, identify the ratio for each point of division. For 'n' equal parts, there are 'n-1' division points. The midpoint formula can be a shortcut for the central point(s) if applicable.

Question 10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Answer: Let the given vertices of the rhombus be \( A(3, 0) \), \( B(4, 5) \), \( C(-1, 4) \), and \( D(-2, -1) \).
The area of a rhombus can be found using the formula: Area \( = \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of its diagonals.
First, we calculate the lengths of the diagonals AC and BD using the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).

Length of diagonal AC (\( d_1 \)):
\( A(3, 0) \) and \( C(-1, 4) \)
\( d_1 = \sqrt{(-1-3)^2 + (4-0)^2} \)
\( d_1 = \sqrt{(-4)^2 + (4)^2} \)
\( d_1 = \sqrt{16 + 16} \)
\( d_1 = \sqrt{32} \)
\( d_1 = \sqrt{16 \times 2} \)
\( \implies d_1 = 4\sqrt{2} \) units.

Length of diagonal BD (\( d_2 \)):
\( B(4, 5) \) and \( D(-2, -1) \)
\( d_2 = \sqrt{(-2-4)^2 + (-1-5)^2} \)
\( d_2 = \sqrt{(-6)^2 + (-6)^2} \)
\( d_2 = \sqrt{36 + 36} \)
\( d_2 = \sqrt{72} \)
\( d_2 = \sqrt{36 \times 2} \)
\( \implies d_2 = 6\sqrt{2} \) units.

Area of rhombus ABCD:
Area \( = \frac{1}{2} \times d_1 \times d_2 \)
Area \( = \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2}) \)
Area \( = \frac{1}{2} \times 24 \times (\sqrt{2} \times \sqrt{2}) \)
Area \( = \frac{1}{2} \times 24 \times 2 \)
Area \( = 24 \) square units.
In simple words: To find the area of the rhombus, we need the lengths of its two diagonals. We used the distance formula to calculate the length of diagonal AC and diagonal BD. Once we had these lengths, we applied the specific formula for the area of a rhombus, which is half the product of its diagonals, to get the final area.

Exam Tip: Always remember the formula for the area of a rhombus: \( \frac{1}{2} \times d_1 \times d_2 \). Carefully calculate the lengths of both diagonals using the distance formula, and simplify any square roots if possible, to avoid errors in the final area calculation.