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.
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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)
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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)
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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
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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
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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
Answers & Solutions:
1) Answer (D)
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.
2) Answer (B)
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.
3) Answer (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.
4) Answer (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.
5) Answer (C)
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.
6) Answer (B)
$P(x, y) = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$
$P(x, 3) = (\frac{-1m+3n}{m+n}, \frac{2m-n}{m+n})$
$\frac{2m-n}{m+n} = 3$
$2m-n = 3m+3n$
$-m = 4n$
$m/n = -4/1$
So the answer is option B.
7) Answer (D)
(-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.
8) Answer (A)
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.
9) Answer (B)
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.
10) Answer (D)
$P(x, y) = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$
$P(x, y) = (\frac{2(4)+3(2)}{2+3}, \frac{2(1)+3(3)}{2+3})$
$P(x, y) = (\frac{14}{5}, \frac{11}{5})$
So the answer is option D.
11) Answer (C)
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.
12) Answer (A)
(-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.
13) Answer (B)
Put y = 0 in 3x + 2y = 12,
3x = 12
x = 4
So the point is (4,0)
So the answer is option B.
14) Answer (D)
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.
15) Answer (B)
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.
16) Answer (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.
17) Answer (C)
From the diagram answer is (-1,-1)
So the answer is option C.
18) Answer (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.
19) Answer (A)
($\frac{2-1+x}{3} , \frac{3-3+y}{3}$) = (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.
20) Answer (D)
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.
21) Answer (C)
(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.
22) Answer (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,distance = $y_{2}-y_{1} = 4-3 = 1$
So the answer is option A.
23) Answer (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.
24) Answer (C)
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.
25) Answer (A)
The point (2, -2) lies on the line y = -2, so reflection is itself
So the answer is option A.
26) Answer (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.
27) Answer (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.
28) Answer (D)
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.
29) Answer (C)
From the diagram reflection of (5 , 2) is (5 , -4)
So the answer is option C.
30) Answer (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.
31) Answer (B)
from the diagram answer is (-7 , 2)
So the answer is option B.
32) Answer (B)
from the diagram answer is (-7 , 2)
So the answer is option B.
33) Answer (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.
34) Answer (D)
slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{1-1}=\infty$
So the answer is option D.
35) Answer (C)
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.
36) Answer (B)
from the diagram answer is (-1 , 2)
So the answer is option B.
37) Answer (C)
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.
38) Answer (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.
39) Answer (A)
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.
40) Answer (B)
A(1,2) , B(2,3) and C(3,1)
D = mid point of B and C = $(\frac{2+3}{2} , \frac{3+1}{2})$ = $(\frac{5}{2} , \frac{4}{2})$ = $(5/2 , 2)$
A(1 , 2) , D(5/2 , 2)
Equation of AD is y = 2 ($\because$ y value is same for both points)
So the answer is option B.
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