CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 26

Previous Year Problems (CBSE Board)

Question 1. Find the ratio in which the y-axis divides the line segment joining the points (−4, −6) and (10, 12). Also find the coordinates of the point of division. [3 Marks, CBSE 2012]
Answer:
Let the ratio be k : 1. The x-coordinate of the point is \( \dfrac{10k - 4}{k + 1} \). On the y-axis it is 0: \( 10k = 4 \Rightarrow k = \dfrac{2}{5} \). The ratio is 2 : 5.
The y-coordinate is \( \dfrac{12k - 6}{k + 1} = \dfrac{\frac{24}{5} - 6}{\frac{7}{5}} = -\dfrac{6}{7} \). The point is \( \left(0, -\dfrac{6}{7}\right) \).
In simple words: On the y-axis the x-coordinate is zero.

Teacher's Note:
1. Take the ratio k : 1.
2. Use the x-coordinate to find k.

Question 2. Prove that the points A(0, −1), B(−2, 3), C(6, 7) and D(8, 3) are the vertices of a rectangle ABCD. [4 Marks, CBSE 2012]
Answer:
\( AB = \sqrt{4 + 16} = 2\sqrt{5} \), \( BC = \sqrt{64 + 16} = 4\sqrt{5} \), \( CD = \sqrt{4 + 16} = 2\sqrt{5} \), \( DA = \sqrt{64 + 16} = 4\sqrt{5} \).
\( AC = \sqrt{36 + 64} = 10 \), \( BD = \sqrt{100 + 0} = 10 \).
Opposite sides are equal and the diagonals are equal, so ABCD is a rectangle.
In simple words: Opposite sides equal and diagonals equal means a rectangle.

Teacher's Note:
1. Find all four sides and both diagonals.
2. Equal diagonals separate a rectangle from a parallelogram.

Question 3. Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear. [CBSE 2013]
Answer:
Let the point (2, 1) divide the segment joining (x, −1) and (4, 5) in the ratio k : 1. Comparing y-coordinates: \( \dfrac{5k - 1}{k + 1} = 1 \Rightarrow 4k = 2 \Rightarrow k = \dfrac{1}{2} \).
Comparing x-coordinates: \( \dfrac{\frac{1}{2}(4) + x}{\frac{3}{2}} = 2 \Rightarrow 2 + x = 3 \Rightarrow \) x = 1.
Check: the points (1, −1), (2, 1), (4, 5) satisfy \( y = 2x - 3 \).
In simple words: A middle point of three collinear points divides the segment in some ratio.

Teacher's Note:
1. Use the y-coordinates to find the ratio first.
2. Then use it for x.

Question 4. The mid-point P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, y). Find the ratio in which P divides CD. Also find the value of y. [4 Marks, CBSE 2013]
Answer:
\( P = \left(\dfrac{-10 - 2}{2}, \dfrac{4 + 0}{2}\right) = (-6, 2) \).
Let P divide CD in the ratio k : 1. Then \( \dfrac{-4k - 9}{k + 1} = -6 \Rightarrow -4k - 9 = -6k - 6 \Rightarrow k = \dfrac{3}{2} \). The ratio is 3 : 2.
\( \dfrac{ky - 4}{k + 1} = 2 \Rightarrow \dfrac{3}{2}y - 4 = 5 \Rightarrow \) y = 6.
In simple words: Find P first, then use the section formula on CD.

Teacher's Note:
1. Use the x-coordinate to find the ratio.
2. Then use the y-coordinate to find y.

Question 5. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is: (A) \( 7 + \sqrt{5} \) (B) 5 (C) 10 (D) 12 [1 Mark, CBSE 2013]
Answer:
The sides are 4, 3 and \( \sqrt{9 + 16} = 5 \). Perimeter \( = 12 \). Answer: (D) 12.
In simple words: Right triangle with legs 3 and 4.

Teacher's Note:
1. This is the 3-4-5 triangle.

Question 6. Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts. Find the coordinates of the points P, Q and R. [3 Marks, CBSE 2014]
Answer:
Each part is \( \dfrac{1}{5} \) of AB. Moving from A to B, x increases by 5 in total and y increases by 5 in total, so each step adds 1 to both coordinates.
P = (2, 3), Q = (3, 4), R = (4, 5) (and S = (5, 6)). Check with the section formula: P divides AB in 1 : 4, so \( P = \left(\dfrac{1 \cdot 6 + 4 \cdot 1}{5}, \dfrac{1 \cdot 7 + 4 \cdot 2}{5}\right) = (2, 3) \).
P(2, 3), Q(3, 4), R(4, 5).
In simple words: Divide the total change in each coordinate by 5.

Teacher's Note:
1. Use ratios 1:4, 2:3, 3:2 for P, Q, R.
2. Equal steps make the points easy.

Question 7. If the coordinates of points A and B are (−2, −2) and (2, −4) respectively, find the coordinates of P such that \( AP = \dfrac{3}{7}AB \), where P lies on the line segment AB. [2 Marks, CBSE 2014]
Answer:
\( AP = \dfrac{3}{7}AB \) means \( AP : PB = 3 : 4 \).
\( P = \left(\dfrac{3(2) + 4(-2)}{7}, \dfrac{3(-4) + 4(-2)}{7}\right) = \left(\dfrac{-2}{7}, \dfrac{-20}{7}\right) \). The point is \( \left(-\dfrac{2}{7}, -\dfrac{20}{7}\right) \).
In simple words: AP = 3/7 AB gives the ratio AP : PB = 3 : 4.

Teacher's Note:
1. PB = AB − AP = 4/7 AB.
2. Then apply the section formula.

Question 8. The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p. [2 Marks, CBSE 2014]
Answer:
By Pythagoras, \( AB^2 + BC^2 = AC^2 \): \( (p - 4)^2 + 16 + (7 - p)^2 = 9 + 16 \Rightarrow (p - 4)^2 + (7 - p)^2 = 9 \).
\( 2p^2 - 22p + 65 = 9 \Rightarrow p^2 - 11p + 28 = 0 \Rightarrow (p - 4)(p - 7) = 0 \).
If p = 7, then B = C(7, 3), which is not allowed. So p = 4.
In simple words: Use Pythagoras with the right angle at B.

Teacher's Note:
1. Reject the value that makes two vertices coincide.

Question 9. Find the relation between x and y if the points A(x, y), B(−5, 7) and C(−4, 5) are collinear. [2 Marks, CBSE 2015]
Answer:
A lies on the line BC, so A divides BC in some ratio k : 1: \( x = \dfrac{-4k - 5}{k + 1} \) and \( y = \dfrac{5k + 7}{k + 1} \).
Then \( 2x + y = \dfrac{-8k - 10 + 5k + 7}{k + 1} = \dfrac{-3(k + 1)}{k + 1} = -3 \).
So the relation is 2x + y + 3 = 0.
In simple words: Eliminate k from the two section-formula equations.

Teacher's Note:
1. Eliminate the ratio.
2. Check with B: 2(−5) + 7 + 3 = 0.

Question 10. Prove that the points (3, 0), (6, 4) and (−1, 3) are the vertices of a right-angled isosceles triangle. [2 Marks, CBSE 2015]
Answer:
Let A(3, 0), B(6, 4), C(−1, 3). \( AB = \sqrt{9 + 16} = 5 \), \( AC = \sqrt{16 + 9} = 5 \), \( BC = \sqrt{49 + 1} = \sqrt{50} \).
AB = AC, and \( AB^2 + AC^2 = 25 + 25 = 50 = BC^2 \). So the triangle is right-angled isosceles, right-angled at A.
In simple words: Two equal sides and Pythagoras.

Teacher's Note:
1. Check the equal sides first.
2. Then check the sum of their squares.

Question 11. Let P and Q be the points of trisection of the line segment joining the points A(2, −2) and B(−7, 4) such that P is nearer to A. Find the coordinates of P and Q. [2 Marks, CBSE 2015]
Answer:
P divides AB in 1 : 2 and Q divides AB in 2 : 1.
\( P = \left(\dfrac{1(-7) + 2(2)}{3}, \dfrac{1(4) + 2(-2)}{3}\right) = (-1, 0) \), \( Q = \left(\dfrac{2(-7) + 1(2)}{3}, \dfrac{2(4) + 1(-2)}{3}\right) = (-4, 2) \).
P(−1, 0) and Q(−4, 2).
In simple words: Trisection points divide the segment in the ratios 1 : 2 and 2 : 1.

Teacher's Note:
1. The point nearer A has the smaller first number.

Question 12. If the point P(x, y) is equidistant from the points A(a + b, b − a) and B(a − b, a + b), prove that \( bx = ay \). [3 Marks, CBSE 2015]
Answer:
\( PA^2 = PB^2 \Rightarrow (x - a - b)^2 + (y - b + a)^2 = (x - a + b)^2 + (y - a - b)^2 \).
Expanding and cancelling the squares: \( -2x(a + b) - 2y(b - a) = -2x(a - b) - 2y(a + b) \), that is \( -2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2ay - 2by \).
This gives \( -4bx + 4ay = 0 \), so bx = ay. Hence proved.
In simple words: Equal distances mean equal squared distances.

Teacher's Note:
1. Square both distances and subtract.
2. Many terms cancel.

Question 13. In what ratio does the point \( \left(\dfrac{24}{11}, y\right) \) divide the line segment joining the points P(2, −2) and Q(3, 7)? Also find the value of y. [2 Marks, CBSE 2016]
Answer:
Let the ratio be k : 1. \( \dfrac{3k + 2}{k + 1} = \dfrac{24}{11} \Rightarrow 33k + 22 = 24k + 24 \Rightarrow k = \dfrac{2}{9} \). The ratio is 2 : 9.
\( y = \dfrac{7k - 2}{k + 1} = \dfrac{\frac{14}{9} - 2}{\frac{11}{9}} = \dfrac{-4}{11} \). So y = −4/11.
In simple words: Find k from x, then y from k.

Teacher's Note:
1. Always use the x-coordinate (known) first.

Question 14. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, −5) is the mid-point of PQ, then find the coordinates of P and Q. [4 Marks, CBSE 2017]
Answer:
P is on the y-axis: P(0, b). Q is on the x-axis: Q(a, 0). Midpoint \( = \left(\dfrac{a}{2}, \dfrac{b}{2}\right) = (2, -5) \Rightarrow a = 4,\ b = -10 \).
P(0, −10) and Q(4, 0).
In simple words: A point on the y-axis has x = 0; on the x-axis it has y = 0.

Teacher's Note:
1. Write the unknown coordinates first.
2. Then use the midpoint.