CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 29

Exercise 2 Objective Questions

Question 1. The distance between the points A(−7, 7) and B(−2, −5) is:
(A) 13 units   (B) 14 units   (C) 15 units   (D) None of these
Answer:
Ans. (A) 13 units
\( AB = \sqrt{(-2 + 7)^2 + (-5 - 7)^2} = \sqrt{25 + 144} = 13 \).
In simple words: Distance formula.

Teacher's Note:
1. 5-12-13 triple.

Question 2. If the points (0, 4), (4, 0) and (6, 2P) are collinear, the value of P is:
(A) −1   (B) 7   (C) 6   (D) 4
Answer:
Ans. (A) −1
Let (6, 2P) lie on the line through (0, 4) and (4, 0). The point (4, 0) divides the segment from (0, 4) to (6, 2P) in the ratio k : 1: \( \dfrac{6k}{k + 1} = 4 \Rightarrow k = 2 \). Then \( \dfrac{2(2P) + 4}{3} = 0 \Rightarrow P = -1 \). (Check: the line is x + y = 4, and 6 + 2P = 4 gives P = −1.)
In simple words: Use the section formula to express collinearity.

Teacher's Note:
1. The points (0, 4) and (4, 0) lie on x + y = 4.

Question 3. Consider the triangle OAB in the xy-plane where O = (0, 0), A = (6, 0), B = \( (\sqrt{2}, 3) \). A square PQRS is inscribed in the triangle with P and Q on OA, R on AB and S on BO. The side of the square equals:
(A) \( 3/\sqrt{2} \)   (B) \( \dfrac{9}{4} \)   (C) \( \dfrac{3}{2}\sqrt{\dfrac{5}{2}} \)   (D) 2
Answer:
Ans. (D) 2
The base OA = 6 and the height of B above OA is 3. Let the side be s. The line SR is parallel to OA at height s, so by similar triangles \( \dfrac{SR}{OA} = \dfrac{3 - s}{3} \). Since SR = s: \( \dfrac{s}{6} = \dfrac{3 - s}{3} \Rightarrow s = 6 - 2s \Rightarrow s = 2 \).
In simple words: Use similar triangles above the square.

Teacher's Note:
1. The x-coordinate of B does not matter.

Question 4. The point on the y-axis which is equidistant from A(−5, −2) and B(3, 2) is:
(A) (−4, 0)   (B) (0, −2)   (C) (−2, 0)   (D) (0, −4)
Answer:
Ans. (B) (0, −2)
Let the point be (0, y). \( 25 + (y + 2)^2 = 9 + (y - 2)^2 \Rightarrow 29 + 4y = 13 - 4y \Rightarrow y = -2 \). The point is (0, −2).
In simple words: x = 0 on the y-axis.

Teacher's Note:
1. Solve the linear equation in y.

Question 5. If two opposite vertices of a square are (5, 4) and (1, −6), then the coordinates of its remaining two vertices are:
(A) (−2, 2) and (5, 3)   (B) (8, −3) and (−2, 1)   (C) (8, 6) and (3, 5)   (D) (1, −3) and (2, 5)
Answer:
Ans. (B) (8, −3) and (−2, 1)
The centre is the midpoint \( (3, -1) \). The half-diagonal from the centre to (5, 4) is (2, 5); turning it through 90 degrees gives (−5, 2) or (5, −2). The other vertices are \( (3 + 5, -1 - 2) = (8, -3) \) and \( (3 - 5, -1 + 2) = (-2, 1) \). Check: the distance from (8, −3) to (5, 4) is \( \sqrt{58} \), to (1, −6) is \( \sqrt{58} \).
In simple words: The diagonals of a square bisect each other at right angles and are equal.

Teacher's Note:
1. Rotate the half-diagonal by 90 degrees.

Question 6. The value of p for which the points (−5, 1), (1, p) and (4, −2) are collinear is:
(A) 0   (B) 2   (C) −1   (D) None of these
Answer:
Ans. (C) −1
Let (1, p) divide the segment joining (−5, 1) and (4, −2) in the ratio k : 1. \( \dfrac{4k - 5}{k + 1} = 1 \Rightarrow k = 2 \). Then \( p = \dfrac{2(-2) + 1}{3} = -1 \).
In simple words: Find the ratio from the x-coordinates.

Teacher's Note:
1. Then use it for the y-coordinate.

Exercise 2 Objective Questions: Car Instruction Questions

A special fully automatic car follows only the following instructions. G₁(x): the car shall move forward x metres. G₂(x): the car shall turn right and move x metres. G₃(x): the car shall turn left and move x metres. G₄(x): the car shall move backward x metres (moving back along the direction it is currently facing). The car starts at the origin.

Question 7. The car is given instructions G₁(100), G₃(50), G₄(10). Assume that the car was initially at the origin. Find the shortest distance of the car from the original position.
(A) \( 10\sqrt{115} \)   (B) \( 10\sqrt{116} \)   (C) \( 8\sqrt{116} \)   (D) \( 5\sqrt{116} \)
Answer:
Ans. (B) \( 10\sqrt{116} \)
Take the initial direction as the +x axis. G₁(100): the car is at (100, 0). G₃(50): it turns left (now facing +y) and moves 50: (100, 50). G₄(10): it moves back 10 along +y: (100, 40).
Distance \( = \sqrt{100^2 + 40^2} = \sqrt{11600} = 10\sqrt{116} \).
In simple words: Track the position and the facing direction step by step.

Teacher's Note:
1. Backward means opposite to the current facing direction.

Question 8. The car is given instructions G₂(50), G₃(30), G₄(20). Find the shortest distance of the car from the original position. Assume that the car was initially at the origin and facing the negative x-axis.
(A) \( 10\sqrt{25} \)   (B) \( 10\sqrt{26} \)   (C) \( 10\sqrt{30} \)   (D) \( 10\sqrt{15} \)
Answer:
Ans. (B) \( 10\sqrt{26} \)
The car faces −x (west). G₂(50): it turns right (facing north, +y) and moves 50: (0, 50). G₃(30): it turns left (facing west, −x) and moves 30: (−30, 50). G₄(20): it moves back 20, that is towards +x: (−10, 50).
Distance \( = \sqrt{10^2 + 50^2} = \sqrt{2600} = 10\sqrt{26} \).
In simple words: Turning right from west faces north.

Teacher's Note:
1. Moving backward reverses the facing direction.

Exercise 2 Objective Questions: Section Formula and Geometry

Question 9. If A(2, 2), B(−4, −4), C(5, −8) are the vertices of any triangle, the length of the median passing through C will be:
(A) \( \sqrt{65} \)   (B) \( \sqrt{117} \)   (C) \( \sqrt{85} \)   (D) \( \sqrt{113} \)
Answer:
Ans. (C) \( \sqrt{85} \)
The midpoint of AB is \( (-1, -1) \). The median is \( \sqrt{(5 + 1)^2 + (-8 + 1)^2} = \sqrt{36 + 49} = \sqrt{85} \).
In simple words: A median goes to the midpoint of the opposite side.

Teacher's Note:
1. Find the midpoint of AB first.

Question 10. The coordinates of the point which divides the line segment joining (−7, 4) and (−6, −5) internally in the ratio 7 : 2 is:
(A) \( \left(\dfrac{56}{9}, 3\right) \)   (B) \( \left(-\dfrac{56}{9}, -3\right) \)   (C) \( \left(-\dfrac{61}{9}, 2\right) \)   (D) \( \left(\dfrac{61}{9}, -2\right) \)
Answer:
Ans. (B) \( \left(-\dfrac{56}{9}, -3\right) \)
\( x = \dfrac{7(-6) + 2(-7)}{9} = -\dfrac{56}{9} \), \( y = \dfrac{7(-5) + 2(4)}{9} = -3 \).
In simple words: Apply the section formula.

Teacher's Note:
1. The nearer end gets the smaller multiplier.

Question 11. The co-ordinates of one end of a diameter of a circle are (5, −7). If the centre is (7, 3), the co-ordinates of the other end of the diameter are:
(A) (6, −2)   (B) (9, 13)   (C) (−2, 6)   (D) (13, 9)
Answer:
Ans. (B) (9, 13)
The centre is the midpoint: \( \dfrac{5 + x}{2} = 7 \Rightarrow x = 9 \), \( \dfrac{-7 + y}{2} = 3 \Rightarrow y = 13 \). The other end is (9, 13).
In simple words: Centre = midpoint of the diameter.

Teacher's Note:
1. Double the centre and subtract the known end.

Question 12. If A and B are the points (−3, 4) and (2, 1), then the co-ordinates of the point C on AB produced such that AC = 2BC are:
(A) (2, 4)   (B) (3, 7)   (C) (7, −2)   (D) \( \left(-\dfrac{1}{2}, \dfrac{5}{2}\right) \)
Answer:
Ans. (C) (7, −2)
C lies on AB produced beyond B with AC = 2BC, so B is the midpoint of AC. \( \dfrac{-3 + x}{2} = 2 \Rightarrow x = 7 \) and \( \dfrac{4 + y}{2} = 1 \Rightarrow y = -2 \). C = (7, −2).
In simple words: If AC = 2BC and B is between A and C, then B is the midpoint of AC.

Teacher's Note:
1. Use the midpoint formula in reverse.

Question 13. The orthocentre of the triangle ABC is B and its circumcentre is S(a, b). If A is the origin, then the co-ordinates of C are:
(A) (2a, 2b)   (B) \( \left(\dfrac{a}{2}, \dfrac{b}{2}\right) \)   (C) \( \left(\sqrt{a^2 + b^2}, 0\right) \)   (D) None
Answer:
Ans. (A) (2a, 2b)
The orthocentre of a triangle coincides with a vertex only when the triangle is right-angled at that vertex, so \( \angle B = 90^\circ \). The circumcentre of a right triangle is the midpoint of the hypotenuse AC. So \( \dfrac{0 + x}{2} = a \) and \( \dfrac{0 + y}{2} = b \), giving C = (2a, 2b).
In simple words: Orthocentre at B means a right angle at B.

Teacher's Note:
1. The circumcentre is the midpoint of the hypotenuse.

Question 14. If the middle points of the sides of a triangle are (−2, 3), (4, −3) and (4, 5), then the centroid of the triangle is:
(A) \( \left(\dfrac{5}{3}, 2\right) \)   (B) \( \left(\dfrac{5}{6}, 1\right) \)   (C) \( \left(1, \dfrac{5}{6}\right) \)   (D) \( \left(2, \dfrac{5}{3}\right) \)
Answer:
Ans. (D) \( \left(2, \dfrac{5}{3}\right) \)
The centroid of the triangle equals the centroid of its medial triangle: \( \left(\dfrac{-2 + 4 + 4}{3}, \dfrac{3 - 3 + 5}{3}\right) = \left(2, \dfrac{5}{3}\right) \).
In simple words: The vertices' average equals the midpoints' average.

Teacher's Note:
1. The sum of the vertices equals the sum of the midpoints.