Class 10 Maths: Coordinate Geometry – Extra Questions with Answers

Ans: A

Solution: The base of the triangle lies on the line y=1, with length 2 – (-2) = 4. The height of the triangle is the distance from (2,5) to the line y=1, which is 5-1 = 4. The area is (1/2) * base * height = (1/2) * 4 * 4 = 8.
Answer: 8

Ans: D

Solution:
(i)
Distance between (2, -1) and (6, 2): √((6-2)² + (2-(-1))²) = √(16 + 9) = 5
Distance between (6, 2) and (10, 5): √((10-6)² + (5-2)²) = √(16 + 9) = 5
Distance between (2, -1) and (10, 5): √((10-2)² + (5-(-1))²) = √(64 + 36) = 10
Since 5 + 5 = 10, these points are collinear.

(ii)
Distance between (7, 9) and (1, 1): √((1-7)² + (1-9)²) = √(36 + 64) = 10
Distance between (1, 1) and (-5, -7): √((-5-1)² + (-7-1)²) = √(36 + 64) = 10
Distance between (7, 9) and (-5, -7): √((-5-7)² + (-7-9)²) = √(144 + 256) = 20
Since 10 + 10 = 20, these points are collinear.

Answer: (i) and (ii)

Ans: D

Solution:
DE = DF
sqrt((x-5)^2 + (3-(-1))^2) = sqrt((x-(-1))^2 + (3-(-1))^2)
(x-5)^2 + 16 = (x+1)^2 + 16
x^2 – 10x + 25 = x^2 + 2x + 1
-12x = -24
x = 2

Answer: 2

Ans: C

Solution:
The distance between Q and C is $\sqrt{(x-2)^2 + (y-3)^2}$.
The distance between Q and D is $\sqrt{(x-6)^2 + (y-1)^2}$.
Since Q is equidistant from C and D, these distances are equal:
$\sqrt{(x-2)^2 + (y-3)^2} = \sqrt{(x-6)^2 + (y-1)^2}$
Square both sides:
$(x-2)^2 + (y-3)^2 = (x-6)^2 + (y-1)^2$
$x^2 – 4x + 4 + y^2 – 6y + 9 = x^2 – 12x + 36 + y^2 – 2y + 1$
$-4x – 6y + 13 = -12x – 2y + 37$
$8x – 4y – 24 = 0$
$2x – y – 6 = 0$

Answer: 2x – y – 6 = 0

Ans: C

Solution:
Let the y-axis divide the line segment in the ratio k:1. The x-coordinate of the point of division is 0. Using the section formula:
(k*6 + 1*(-2)) / (k+1) = 0
6k – 2 = 0
6k = 2
k = 1/3
Answer: 1:3

Ans: D

Solution:
Let the point $P\left(\frac{17}{5}, y\right)$ divide the line segment joining $A(1, -3)$ and $B(4, 6)$ in the ratio $k:1$.
Using the section formula, we have:
$\frac{17}{5} = \frac{4k + 1}{k+1}$
$17(k+1) = 5(4k+1)$
$17k + 17 = 20k + 5$
$3k = 12$
$k = 4$
So the ratio is $4:1$.
Now,
$y = \frac{6k + (-3)}{k+1} = \frac{6(4) – 3}{4+1} = \frac{24-3}{5} = \frac{21}{5}$

Answer: $4:1, \frac{21}{5}$

Ans: D

Solution:
The length of segment PQ is $|5-1| = 4$. Since the x-coordinates of P and Q are the same, the line segment PQ is vertical. The horizontal distance from the line containing segment PQ to point R is $|2 – (-2)| = 4$. The area of the triangle is $\frac{1}{2} \times base \times height = \frac{1}{2} \times 4 \times 4 = 8$.
Answer: 8

Ans: B

Solution:
1. Calculate the slopes of the sides:
* Slope of AB: (5-1)/(4-1) = 4/3
* Slope of BC: (9-5)/(1-4) = -4/3
* Slope of CD: (5-9)/(-2-1) = 4/3
* Slope of DA: (1-5)/(1-(-2)) = -4/3
2. Since the slopes of AB and CD are equal and the slopes of BC and DA are equal, the opposite sides are parallel.
3. Also, none of the adjacent sides are perpendicular because (4/3) * (-4/3) != -1. Therefore it is a parallelogram.
Answer: Parallelogram

Ans: A

Solution: Calculate the slopes of the sides to determine if any sides are parallel or perpendicular.
Slope of (6,8) to (3,7): (7-8)/(3-6) = 1/3
Slope of (3,7) to (-2,-2): (-2-7)/(-2-3) = 9/5
Slope of (-2,-2) to (1,-1): (-1-(-2))/(1-(-2)) = 1/3
Slope of (1,-1) to (6,8): (8-(-1))/(6-1) = 9/5
Since opposite sides have the same slopes, it is a parallelogram. None of the slopes are negative reciprocals of each other, so it is not a rectangle.

Answer: Parallelogram

Ans: C

Solution:
Let the center of the circle be $(h, k)$. Then the distance from the center to each of the three points is the same (the radius). Therefore,
$(6-h)^2 + (-6-k)^2 = (2-h)^2 + (-7-k)^2$ and $(6-h)^2 + (-6-k)^2 = (3-h)^2 + (3-k)^2$.
Expanding the first equation, we get
$36 – 12h + h^2 + 36 + 12k + k^2 = 4 – 4h + h^2 + 49 + 14k + k^2$
$72 – 12h + 12k = 53 – 4h + 14k$
$19 – 8h – 2k = 0$
$8h + 2k = 19$ (1)
Expanding the second equation, we get
$36 – 12h + h^2 + 36 + 12k + k^2 = 9 – 6h + h^2 + 9 – 6k + k^2$
$72 – 12h + 12k = 18 – 6h – 6k$
$54 – 6h + 18k = 0$
$6h – 18k = 54$
$h – 3k = 9$ (2)
From (2), $h = 3k + 9$. Substituting this into (1), we have
$8(3k + 9) + 2k = 19$
$24k + 72 + 2k = 19$
$26k = -53$
$k = -\frac{53}{26}$
$h = 3(-\frac{53}{26}) + 9 = \frac{-159}{26} + \frac{234}{26} = \frac{75}{26}$
Thus, the center is $(\frac{75}{26}, -\frac{53}{26})$.

Answer: $(\frac{75}{26}, -\frac{53}{26})$