0
1067

Coordinate Geometry Questions for SSC GD PDF Set – 2

SSC GD Constable Coordinate Geometry Question paper with answers download PDF based on SSC GD exam previous papers. 40 Very important Coordinate Geometry questions for GD Constable.

1500+ Must Solve Questions for SSC Exams (Question bank)

Question 1: The coordinates of the vertices of a right angled triangle are A(2,3) B(8,3) and C(2, -2). The triangle is right angled at A. Find the orthocenter of the triangle ABC ?

a) (2,-2)

b) (8,3)

c) (6,3)

d) (2,3)

Question 2: What is the equation of a circle of radius 4 units centered at (1, 2) ?

a) $x^2+y^2-2x-4y+5=0$

b)  $x^2+y^2-2x-4y-11 = 0$

c)  $x^2+y^2+2x-4y-11 = 0$

d)  $x^2+y^2-2x+4y-11 = 0$

Question 3: What is the reflection of the point ( -4, 2) in the line x = -3?

a) (-2, -2)

b) (-2, 2)

c) (1, -2)

d) (1, 2)

Question 4: Find the slope of the line joining A(5,-1) & B(2, 2) ?

a) 1

b) -1

c) 2

d) -2

Question 5: Point P is the midpoint of segment AB. Coordinates of P are (-1 , -1) and B are (-5, -4) What are the coordinates of point A?

a) (3,-2)

b) (-3,-2)

c) (3,2)

d) (-3,2)

Question 6: In what ratio does the point T(x,3) divide the segment joining the points S(3, -1) and U(-1, 2)?

a) 4:1

b) -4:1

c) 1:4

d) -1:4

Question 7: What is the reflection of the point (-4 , 3) in the line x = -1?

a) (-2 , 3)

b) (2 , -3)

c) (-2 , -3)

d) (2 , 3)

Question 8: At what point does the line 3x – 2y = 12 intersect the Y-axis?

a) (0,-6)

b) (-6,0)

c) (0,6)

d) (6,0)

Question 9: Find the slope of the line parallel to x-3y+2=0 ?

a) -1/3

b) 1/3

c) 3

d) -3

Question 10: Find the point which divides the line joining A(2,3) B(4,1) in the ratio 2:3 internally?

a) (-14/5, -11/5)

b) (14/5, -11/5)

c) (-14/5, 11/5)

d) (14/5, 11/5)

Question 11: If P(x,2) is the centroid of the triangle formed by A(2,y), B(-1,-3) & C(2,6). Then find x*y ?

a) 13

b) 31

c) 3

d) -3

Question 12: What is the reflection of the point (-2,-2) in the line y = -1?

a) (-2 , 0)

b) (-2, -1)

c) (0 , -2)

d) (-2, 1)

Question 13: At what point does the line 3x + 2y = 12 cuts the X-axis?

a) (0,6)

b) (4,0)

c) (-4,0)

d) (0,-6)

Question 14: What is the slope of the line perpendicular to the line 2x+2y=3?

a) -2

b) 2

c) -1

d) 1

Question 15: Find the ratio in which P(3,Y) divides the line joining A(1,-3) and B(4,2) ?

a) 1:2

b) 2:1

c) -2:1

d) -1:2

Question 16: Point P is the midpoint of segment AB. Coordinates of P are (2 , -1) and B are (5, -4) What are the coordinates of point A?

a) (-1, -2)

b) (-1, 2)

c) (1, 2)

d) (1, -2)

Question 17: What is the reflection of the point (-1 , 1) in the line y = 0?

a) (-2, 1)

b) (-2, -1)

c) (-1, -1)

d) (1, -1)

Question 18: What is the slope of the line perpendicular to the line passing through the points (-3 , 1) and (-2 , 2)?

a) 0

b) 1

c) -1

d) 2

Question 19: If P(1,2) is the centroid of the triangle formed by A(2,3), B(-1,-3) & C(x,y). Then find x-y ?

a) -4

b) 4

c) 3

d) -3

Question 20: In what ratio does the point T(1, y) divide the segment joining the points S(3, -1) and U(-1, 2)?

a) 2 : 1

b) 1 : 3

c) 3 : 2

d) 1 : 1

Question 21: What is the reflection of the point (4 , 3) in the line y = -1?

a) (4, 5)

b) (-4, 5)

c) (4, -5)

d) (-4, -5)

Question 22: Find the length of the median BE of a triangle ABC, with vertices A(1,2) , B(2,3) and C(3,6) ?

a) 1 unit

b) 2 units

c) 3 units

d) 4 units

Question 23: Find the slope of the line joining A(-1,2) and B(1, -2) ?

a) -2

b) 0

c) 2

d) undefined

Question 24: If P(2, 3) is the midpoint of A(4, y) and B(x, 1), then find x ?

a) 1

b) 2

c) 0

d) -1

Question 25: What is the reflection of the point ( 2, -2) in the line y = -2?

a) (2, -2)

b) (-2, 2)

c) (-2, -2)

d) (2, 2)

Question 26: Find the equation of median BE of the triangle ABC, with vertices A(1,2) , B(2,3) and C(3,6) ?

a) x=2

b) x=3

c) y=2

d) y=3

Question 27: What is the slope of the line perpendicular to the line 4x-3y=5 ?

a) -3/4

b) 4/3

c) 3/4

d) -4/3

Question 28: If P(2, 3) is the midpoint of A(4, y) and B(x, 1), then find y ?

a) 2

b) 3

c) 4

d) 5

Question 29: What is the reflection of the point ( 5, 2) in the line y = -1?

a) (-5, 4)

b) (-5, -4)

c) (5, -4)

d) (5, 4)

Question 30: Find the slope of the line joining A(-3, 2) and B(1, 4) ?

a) -1/2

b) 2

c) 1/2

d) -1

Question 31: What is the reflection of the point ( 1, 2) in the line x = -3?

a) (-7, -2)

b) (-7, 2)

c) (7, -2)

d) (7, 2)

Question 32: What is the reflection of the point ( 1, 2) in the line x = -3?

a) (-7, -2)

b) (-7, 2)

c) (7, -2)

d) (7, 2)

Question 33: Find the length of the median AD of a triangle ABC, with vertices A(1,2) , B(2,3) and C(0,5) ?

a) 1 unit

b) 2 units

c) 3 units

d) 4 units

Question 34: Find the slope of the line joining A(1,2) and B(1, -2) ?

a) -4

b) 0

c) 2

d) undefined

Question 35: Find the mid-point of the line joining A(3, -2) and B(-3, -2) ?

a) (3, -2)

b) (-3, 0)

c) (0, -2)

d) (0, 2)

Question 36: What is the reflection of the point ( -3, 2) in the line x = -2?

a) (-1, -2)

b) (-1, 2)

c) (1, -2)

d) (1, 2)

Question 37: Find the slope of the line joining A(-1, -5) and B(2, 4)

a) -3

b) 0

c) 3

d) -1

Question 38: Find the centroid of the triangle with vertices A(2,3), B(-3,2) & C(1,4) ?

a) (0,0)

b) (2,-3)

c) (0,3)

d) (3,0)

Question 39: What is the reflection of the point (4,2) about origin ?

a) (-4, -2)

b) (4, 2)

c) (-4, 2)

d) (4, -2)

Question 40: : Find the equation of the median AD of a triangle ABC, with vertices A(1,2) , B(2,3) and C(3,1) ?

a) y=1

b) y=2

c) y=3

d) y=4

Orthocentre of a right angled triangle is the vertex where the angle is 90°

Here, it is right angled at A, so A(2, 3) is the orthocentre.

So the answer is option D.

Eq.of a circle of radius r and having centre at ($x_1, y_1$) is $(x-x_1)^2+(y-y_1)^2=r^2$

Hence, $(x-1)^2+(y-2)^2=4^2$

==> $x^2-2x+1+y^2-4y+4 = 16$

==> $x^2+y^2-2x-4y-11 = 0$

So the answer is option B.

(-4, 2) lies on left to the line x=-3

Distance between the line & point is -3-(-4) = 1

So its reflection must be 1 unit to the right side the line , i.e; (-3+1, 2) = (-2, 2)

So the answer is option B.

Slope = $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{2+1}{2-5} = 3/-3 = -1$

So the answer is option B.

Let (x, y) be the coordinates of A, then

$(\frac{x-5}{2} , \frac{y-4}{2}) = (-1, -1)$

x-5 = -2 & y-4 = -2

x = 3 & y = 2

A = (x, y) = (3,2)

So the answer is option C.

$P(x, y) = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$

$P(x, 3) = (\frac{-1m+3n}{m+n}, \frac{2m-n}{m+n})$

$\frac{2m-n}{m+n} = 3$

$2m-n = 3m+3n$

$-m = 4n$

$m/n = -4/1$

So the answer is option B.

(-4, 3) lies 3 units left to the line x = -1

So its reflection must lie 3 unit right to the line,

So its reflection is (-1+3, 3) = (2, 3)

So the answer is option D.

3x – 2y = 12 lines cuts Y-axis, where x = 0

3(0)-2y = 12

-2y = 12

y = -6

So the point is (0, -6)

So the answer is option A.

Slope of a line ax+by+c = 0 is -a/b

Slope of line x-3y+2=0 = -1/-3 = 1/3

Slope of parallel line = 1/3

So the answer is option B.

$P(x, y) = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$

$P(x, y) = (\frac{2(4)+3(2)}{2+3}, \frac{2(1)+3(3)}{2+3})$

$P(x, y) = (\frac{14}{5}, \frac{11}{5})$

So the answer is option D.

Centroid $= (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$

$(x, 2) = (\frac{2-1+2}{3}, \frac{y-3+6}{3})$

$(x, 2) = (1, \frac{y+3}{3})$

=> $x = 1$ , & $2 = (y+3)/3$

=> $x = 1$, & $y = 3$

$x*y = 3$

So the answer is option C.

(-2, -2) lies 1 unit below the line y = -1

So its reflection must lie 1 unit above the line,

So its reflection is (-2, -1+1) = (-2, 0)

So the answer is option A.

Put y = 0 in 3x + 2y = 12,

3x = 12

x = 4

So the point is (4,0)

So the answer is option B.

slope of the line perpendicular to the line ax+by+c=0 is, b/a

slope of the line perpendicular to the line 2x+2y=3 is 2/2=1

So the answer is option D.

Let the ratio be m:n

(3, Y) = ($\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}$)

3 = $\frac{4m+n}{m+n}$

3m+3n = 4m+n

2n = m

m/n = 2/1

So the answer is option B.

Let A be (x, y)

(2,-1) = ($\frac{x+5}{2} , \frac{y-4}{2}$)

x+5 = 2*2 and y-4 = -1*2

x = 4-5 and y = -2+4

x = -1 and y = 2

A = (x,y) = (-1,2)

So the answer is option B.

From the diagram answer is (-1,-1)

So the answer is option C.

Slope = m = $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{2-1}{-2+3}$ = $1$

Slope of perpendicular line = -1/m = -1/1 = -1

So the answer is option C.

($\frac{2-1+x}{3} , \frac{3-3+y}{3}$) = (1, 2)

($\frac{1+x}{3} , \frac{y}{3}$) = (1, 2)

1+x = 3 and y = 6

x = 2 and y = 6

x-y = 2-6 = -4

So the answer is option A.

Let the ratio is m:n

1 = $\frac{m(-1)+n(3)}{m+n}$

$m+n = -m+3n$

$2m = 2n$

$m:n = 1:1$

So the answer is option D.

(4, 3) lies above the line y=-1

Distance b/w the line & point is 3-(-1) = 4

So its reflection must be 4 units below the line , i.e; (4, -1-4) = (4, -5)

So the answer is option C.

E = mid point of AC = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ = $(\frac{1+3}{2} , \frac{2+6}{2})$ = $(2, 4)$

B = (2 , 3) & E = ( 2 , 4)

Since x values are same,distance = $y_{2}-y_{1} = 4-3 = 1$

So the answer is option A.

slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{1+1}=-4/2=-2$

So the answer is option A.

Mid point = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$

(2, 3) = $(\frac{4+x}{2} , \frac{1+y}{2})$

$\frac{4+x}{2}$ = 2

4+x = 4

x = 0

So the answer is option C.

The point (2, -2) lies on the line y = -2, so reflection is itself

So the answer is option A.

E = mid point of AC = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ = $(\frac{1+3}{2} , \frac{2+6}{2})$ = $(2, 4)$

B = (2 , 3) & E = ( 2 , 4)

Since x values are same, Equation of BE is x = 2

So the answer is option A.

Slope of a line ax+by+c = 0 is $\frac{-a}{b}$

Slope of the line 4x-3y=5 is m =-4/-3 = 4/3

Slope of a line perpendicular to this line = -1/m = -3/4

So the answer is option A.

Mid point = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$

(2, 3) = $(\frac{4+x}{2} , \frac{1+y}{2})$

$\frac{1+y}{2}$ = 3

1+y = 6

y = 5

So the answer is option D.

From the diagram reflection of (5 , 2) is (5 , -4)

So the answer is option C.

slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4-2}{1+3}=2/4=1/2$

So the answer is option C.

from the diagram answer is (-7 , 2)

So the answer is option B.

from the diagram answer is (-7 , 2)

So the answer is option B.

D = mid point of BC = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ = $(\frac{2+0}{2} , \frac{3+5}{2})$ = $(1, 4)$

A = (1 , 2) & D = ( 1 , 4)

Since x values are same, Distance = $y_{2}-y_{1}=4-2=2$

So the answer is option B.

slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{1-1}=\infty$

So the answer is option D.

Mid-point = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ = $(\frac{3-3}{2} , \frac{-2-2}{2})$ = $( 0 , -2)$

So the answer is option C.

from the diagram answer is (-1 , 2)

So the answer is option B.

slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4+5}{2+1}=9/3=3$

So the answer is option C.

Centroid = $(\frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3})$ = $(\frac{2-3+1}{3} , \frac{3+2+4}{3})$ = $(\frac{0}{3} , \frac{9}{3})$ = (0, 3)

So the answer is option C.

To find reflection about origin, just change the signs.

reflection of the point (4,2) about origin is (-4, -2)

So the answer is option A.

D = mid point of B and C = $(\frac{2+3}{2} , \frac{3+1}{2})$ = $(\frac{5}{2} , \frac{4}{2})$ = $(5/2 , 2)$
Equation of AD is y = 2 ($\because$ y value is same for both points)