CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 28

Exercise 1 Objective Questions: Section A (Introduction and Distance Formula)

Question 1. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is:
(A) 5   (B) 3   (C) 4   (D) \( \sqrt{34} \)
Answer:
Ans. (D) \( \sqrt{34} \)
The diagonals of rectangle AOBC are AB and OC (equal in length). \( AB = \sqrt{25 + 9} = \sqrt{34} \).

Teacher's Note:
1. A rectangle has equal diagonals.
2. Join the two opposite vertices A and B.

Question 2. The points (−a, −b), (0, 0), (a, b) and (a², ab) are:
(A) Collinear   (B) Vertices of a parallelogram   (C) Vertices of a rectangle   (D) None of these
Answer:
Ans. (A) Collinear
The points (−a, −b), (0, 0) and (a, b) lie on the line through the origin with direction (a, b), and (a², ab) = a(a, b) also lies on it. In distances: \( OA = \sqrt{a^2 + b^2} \), \( OB = |a|\sqrt{a^2 + b^2} \), and the sums of two distances equal the third along the line. All four points are collinear.

Teacher's Note:
1. A point t(a, b) lies on the line through the origin and (a, b).
2. Here t = −1, 0, 1 and a.

Question 3. The points (0, −1), (−2, 3), (6, 7) and (8, 3) are:
(A) Collinear   (B) Vertices of a parallelogram which is not a rectangle   (C) Vertices of a rectangle which is not a square   (D) None of these
Answer:
Ans. (C) Vertices of a rectangle which is not a square
With A(0, −1), B(−2, 3), C(6, 7), D(8, 3): \( AB = CD = 2\sqrt{5} \), \( BC = DA = 4\sqrt{5} \), and the diagonals \( AC = BD = 10 \). So it is a rectangle, and since AB ≠ BC it is not a square.
In simple words: Equal diagonals mean a rectangle.

Teacher's Note:
1. Unequal adjacent sides: not a square.

Question 4. The distance between the points A(cos α, −sin α) and B(−cos α, sin α) is:
(A) 1   (B) 2   (C) 3   (D) 4
Answer:
Ans. (B) 2
\( AB = \sqrt{(2\cos\alpha)^2 + (2\sin\alpha)^2} = \sqrt{4} = 2 \).
In simple words: Use sin² + cos² = 1.

Teacher's Note:
1. The two points are opposite ends of a diameter of the unit circle.

Question 5. A point on the y-axis which is equidistant from A(6, 5) and B(−4, 3) is:
(A) (0, 9)   (B) (9, 0)   (C) (0, −9)   (D) (0, 3)
Answer:
Ans. (A) (0, 9)
Let the point be (0, y). \( 36 + (y - 5)^2 = 16 + (y - 3)^2 \Rightarrow 61 - 10y = 25 - 6y \Rightarrow y = 9 \). The point is (0, 9).
In simple words: A point on the y-axis has x = 0.

Teacher's Note:
1. Equate the squared distances.

Question 6. The points on the x-axis which are at a distance of \( 2\sqrt{5} \) from the point (7, −4) are:
(A) (5, 0)   (B) (9, 0)   (C) (A) and (B) both   (D) None of these
Answer:
Ans. (C) (A) and (B) both
For (x, 0): \( (x - 7)^2 + 16 = 20 \Rightarrow x = 5 \) or 9. Both (5, 0) and (9, 0) satisfy the condition.
In simple words: Take the point as (x, 0).

Teacher's Note:
1. A quadratic gives two values.

Exercise 1 Objective Questions: Section B (Section Formula)

Question 7. If (3, −4) and (−6, 5) are the extremities of a diagonal of a parallelogram and (−2, 1) is its third vertex, then its fourth vertex is:
(A) (−1, 0)   (B) (0, −1)   (C) (−1, 1)   (D) None of these
Answer:
Ans. (A) (−1, 0)
The diagonals bisect each other. Midpoint of the given diagonal \( = \left(-\dfrac{3}{2}, \dfrac{1}{2}\right) \). The fourth vertex \( = 2(\text{midpoint}) - (-2, 1) = (-3 + 2, 1 - 1) = (-1, 0) \).
In simple words: The diagonals share a midpoint.

Teacher's Note:
1. Fourth vertex = 2 × midpoint − third vertex.

Question 8. The coordinates of the points P and Q are respectively (4, −3) and (−1, 7). Then the abscissa of a point R on the line segment PQ such that \( \dfrac{PR}{PQ} = \dfrac{3}{5} \) is:
(A) \( \dfrac{18}{5} \)   (B) \( \dfrac{17}{5} \)   (C) \( \dfrac{17}{8} \)   (D) 1
Answer:
Ans. (D) 1
\( PR : RQ = 3 : 2 \). Abscissa \( = \dfrac{3(-1) + 2(4)}{5} = \dfrac{5}{5} = 1 \).
In simple words: PR/PQ = 3/5 means PR : RQ = 3 : 2.

Teacher's Note:
1. The ratio uses the whole segment as the denominator.

Question 9. If the point P(6, 3) lies on the line segment joining A(4, 2) and B(8, 4), then:
(A) \( AP = \tfrac{1}{3}AB \)   (B) \( \tfrac{1}{2}AP = PB \)   (C) \( PB = \tfrac{1}{3}AB \)   (D) \( AP = \tfrac{1}{2}AB \)
Answer:
Ans. (D) \( AP = \tfrac{1}{2}AB \)
The midpoint of AB is \( \left(\dfrac{4 + 8}{2}, \dfrac{2 + 4}{2}\right) = (6, 3) = P \). So P is the midpoint and \( AP = \tfrac{1}{2}AB \).
In simple words: Compute the midpoint of AB.

Teacher's Note:
1. P is the midpoint, so AP = PB.

Question 10. The fourth vertex D of a parallelogram ABCD whose three vertices are A(−2, 3), B(6, 7) and C(8, 3) is:
(A) (0, 1)   (B) (0, −1)   (C) (−1, 0)   (D) (1, 0)
Answer:
Ans. (B) (0, −1)
Diagonals AC and BD bisect each other: \( \dfrac{-2 + 8}{2} = \dfrac{6 + x}{2} \Rightarrow x = 0 \) and \( \dfrac{3 + 3}{2} = \dfrac{7 + y}{2} \Rightarrow y = -1 \). D = (0, −1).
In simple words: A and C are opposite vertices.

Teacher's Note:
1. Equal midpoints of AC and BD.

Question 11. If \( P\left(\dfrac{a}{3}, 4\right) \) is the mid point of the line segment joining the points Q(−6, 5) and R(−2, 3), then the value of a is:
(A) −4   (B) −12   (C) 12   (D) −6
Answer:
Ans. (B) −12
The midpoint is \( (-4, 4) \), so \( \dfrac{a}{3} = -4 \Rightarrow a = -12 \).
In simple words: Use the midpoint formula for x.

Teacher's Note:
1. The y-coordinate 4 matches.

Question 12. If the centroid of a triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then \( a^3 + b^3 + c^3 \) is:
(A) abc   (B) a + b + c   (C) 3abc   (D) 0
Answer:
Ans. (C) 3abc
The centroid is \( \left(\dfrac{a + b + c}{3}, \dfrac{a + b + c}{3}\right) = (0, 0) \), so \( a + b + c = 0 \). Then \( a^3 + b^3 + c^3 = 3abc \).
In simple words: If a + b + c = 0 then a³ + b³ + c³ = 3abc.

Teacher's Note:
1. Use a³ + b³ + c³ − 3abc = (a + b + c)(…).

Question 13. The ratio in which the point \( P\left(\dfrac{-2}{5}, 6\right) \) divides the join of A(−4, 3) and B(2, 8) is:
(A) 2 : 3   (B) 1 : 3   (C) 3 : 2   (D) 3 : 1
Answer:
Ans. (C) 3 : 2
Let the ratio be k : 1. \( \dfrac{8k + 3}{k + 1} = 6 \Rightarrow 2k = 3 \Rightarrow k = \dfrac{3}{2} \), so the ratio is 3 : 2. Check: \( x = \dfrac{\frac{3}{2}(2) - 4}{\frac{5}{2}} = -\dfrac{2}{5} \).
In simple words: Use the y-coordinate to find the ratio.

Teacher's Note:
1. Verify with x.

Question 14. The point which lies on the perpendicular bisector of the line segment joining the points A(−2, −5) and B(2, 5) is:
(A) (0, 0)   (B) (0, 2)   (C) (2, 0)   (D) (−2, 0)
Answer:
Ans. (A) (0, 0)
A point on the perpendicular bisector is equidistant from A and B. For (0, 0): \( OA = OB = \sqrt{29} \). For (0, 2): \( \sqrt{4 + 49} \ne \sqrt{4 + 9} \); for (2, 0): \( \sqrt{41} \ne 5 \); for (−2, 0): \( 5 \ne \sqrt{41} \). So (0, 0).
In simple words: The perpendicular bisector is the set of points equidistant from A and B.

Teacher's Note:
1. The midpoint (0, 0) always lies on it.

Question 15. The line segment joining the points (−3, −4) and (1, −2) is divided by the y-axis in the ratio:
(A) 1 : 3   (B) 3 : 1   (C) 2 : 3   (D) 3 : 2
Answer:
Ans. (B) 3 : 1
Let the ratio be k : 1. The x-coordinate is \( \dfrac{k - 3}{k + 1} = 0 \Rightarrow k = 3 \). The ratio is 3 : 1.
In simple words: The x-coordinate is 0 on the y-axis.

Teacher's Note:
1. Solve for k.

Question 16. The point which divides the line segment joining the points (7, −6) and (3, 4) in the ratio 1 : 2 internally lies in the:
(A) I quadrant   (B) II quadrant   (C) III quadrant   (D) IV quadrant
Answer:
Ans. (D) IV quadrant
\( \left(\dfrac{1(3) + 2(7)}{3}, \dfrac{1(4) + 2(-6)}{3}\right) = \left(\dfrac{17}{3}, -\dfrac{8}{3}\right) \). x > 0 and y < 0, so the point lies in the IV quadrant.
In simple words: Find the point, then check the signs.

Teacher's Note:
1. (+, −) is the fourth quadrant.