SSC CGL Coordinate Geometry Questions with Solutions PDF:
Download SSC CGL Coordinate Geometry questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Coordinate Geometry objective questions for SSC exams.
Download SSC CGL Coordinate Geometry Questions PDF
Question 1: Find equation of the perpendicular bisector of segment joining the points (2,-5) and (0,7)?
a) x – 6y = 5
b) x + 6y = -5
c) x – 6y = -5
d) x + 6y = 5
Question 2: Find equation of the perpendicular to segment joining the points A(0,4) and B(-5,9) and passing through the point P. Point P divides segment AB in the ratio 2:3.
a) x – y = 8
b) x – y = -8
c) x + y = -8
d) x + y = 8
Question 3: The co-ordinates of the centroid of a triangle ABC are (-1,-2) what are the co-ordinates of vertex C, if co-ordinates of A and B are (6,-4) and (-2,2) respectively?
a) (-7,-4)
b) (7,4)
c) (7,-4)
d) (-7,4)
Question 4: What is the slope of the line parallel to the line passing through the points (4,-2) and (-3,5)?
a) 3/7
b) 1
c) -3/7
d) -1
Question 5: The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?
a) -7
b) 4
c) 7
d) -4
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Question 6: Find k, if the line 4x+y = 1 is perpendicular to the line 5x+ky = 2?
a) 20
b) -20
c) 4
d) -4
Question 7: Find equation of the perpendicular bisector of segment joining the points (2,-6) and (4,0)?
a) x + 3y = 6
b) x + 3y = -6
c) x – 3y = -6
d) x – 3y = 6
Question 8: In what ratio is the segment joining (12,1) and (3,4) divided by the Y axis?
a) 4:1
b) 1:4
c) 4:3
d) 3:4
Question 9: The line passing through (4,3) and (y,0) is parallel to the line passing through (1,2) and (3,0). Find y?
a) 1
b) 7
c) 2
d) 5
Question 10: What is the slope of the line perpendicular to the line passing through the points (8,2) and (3,1)?
a) -5
b) 3/5
c) 5/3
d) 1/5
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Question 11: The line passing through (2,-1) and (y,-2) is parallel to the line passing through (-3,4) and (0,3) Find y?
a) -2
b) 2
c) 5
d) -5
Question 12: The point P(3,-2) divides the segment joining the points (x,0) and (0,y) in the ratio 1:3. Find x and y?
a) x = 4; y = -8
b) x = -3; y = -8
c) x = 3; y = 8
d) x = -3; y = 8
Question 13: What is the equation of the line if its slope is 1/4 and y-intercept is -3?
a) x – 4y = 12
b) x + 4y = 12
c) x – 4y = -12
d) x + 4y = -12
Question 14: What is the slope of the line parallel to the line passing through the points (6,3) and (2,1)?
a) 1/2
b) 1
c) 2
d) 1
Question 15: In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?
a) 1:3
b) 3:2
c) 3:1
d) 2:3
Question 16: At what point does the line 4x – 3y = -6 intercept the y-axis?
a) (0,2)
b) (0,3/2)
c) (2,0)
d) (3/2,0)
Question 17: The slope of the line passing through the points (7,-2) and (x,1) is -3/10. Find x.
a) 4
b) 2
c) -4
d) -3
Question 18: The co-ordinates of the centroid of a triangle ABC are (3,2). What are the co-ordinates of vertex C if co-ordinates of A and B are (-2,5) and (6,2) respectively?
a) (-5,-1)
b) (5,-1)
c) (5,1)
d) (-5,1)
Question 19: The slope of the line passing through the points (-5,1) and (x,-4) is -5/8. Find x.
a) 4
b) 3
c) 2
d) -1
Question 20: The point P(a,b) is first reflected in origin to P1 and P1 is reflected in y-axis to (6,-5). The co-ordinates of point P are
a) (-6,-5)
b) (6,5)
c) (-6,5)
d) (6,-5)
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Question 21: The line passing through (2,-1) and (y,-2) is parallel to the line passing through (-3,4) and (0,3) Find y?
a) -2
b) 2
c) 5
d) -5
Question 22: The point P(3,-2) divides the segment joining the points (x,0) and (0,y) in the ratio 1:3. Find x and y?
a) x = 4; y = -8
b) x = -3; y = -8
c) x = 3; y = 8
d) x = -3; y = 8
Question 23: What is the equation of the line if its slope is 1/4 and y-intercept is -3?
a) x – 4y = 12
b) x + 4y = 12
c) x – 4y = -12
d) x + 4y = -12
Question 24: What is the slope of the line parallel to the line passing through the points (6,3) and (2,1)?
a) 1/2
b) 1
c) 2
d) 1
Question 25: In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?
a) 1:3
b) 3:2
c) 3:1
d) 2:3
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Answers & Solutions:
1) Answer (C)
Let line $l$ perpendicularly bisects line joining A(2,-5) and B(0,7) at C, thus C is the mid point of AB.
=> Coordinates of C = $(\frac{2 + 0}{2} , \frac{-5 + 7}{2})$
= $(\frac{2}{2} , \frac{2}{2}) = (1,1)$
Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(7 + 5)}{(0 – 2)}$
= $\frac{12}{-2} = -6$
Let slope of line $l = m$
Product of slopes of two perpendicular lines = -1
=> $m \times -6 = -1$
=> $m = \frac{1}{6}$
Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$
$\therefore$ Equation of line $l$
=> $(y – 1) = \frac{1}{6}(x – 1)$
=> $6y – 6 = x – 1$
=> $x – 6y = 1 – 6 = -5$
=> Ans – (C)
2) Answer (B)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Coordinates of A(0,4) and B(-5,9). Let coordinates of P = (x,y) which divides AB in ratio = 2 : 3
=> $x = \frac{(2 \times -5) + (3 \times 0)}{2 + 3}$
=> $5x = -10$
=> $x = \frac{-10}{5} = -2$
Similarly, $y = \frac{(2 \times 9) + (3 \times 4)}{2 + 3}$
=> $5y = 18 + 12 = 30$
=> $y = \frac{30}{5} = 6$
=> Point P = (-2,6)
Slope of AB = $\frac{9 – 4}{-5 – 0} = \frac{5}{-5} = -1$
Let slope of line perpendicular to AB = $m$
Also, product of slopes of two perpendicular lines is -1
=> $m \times -1 = -1$
=> $m = 1$
Equation of lines having slope $m$ and passing through point P(-2,6) is
=> $(y – 6) = 1(x + 2)$
=> $y – 6 = x + 2$
=> $x – y = -8$
=> Ans – (B)
3) Answer (A)
Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$
Let coordinates of vertex C = $(x , y)$
Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)
=> $-1 = \frac{-2 + 6 + x}{3}$
=> $x + 4 = -1 \times 3 = -3$
=> $x = -3 – 4 = -7$
Similarly, => $-2 = \frac{-4 + 2 + y}{3}$
=> $y – 2 = -2 \times 3 = -6$
=> $y = -6 + 2 = -4$
$\therefore$ Coordinates of vertex C = (-7,-4)
=> Ans – (A)
4) Answer (D)
Slope of line passing through points (4,-2) and (-3,5)
= $\frac{5 + 2}{-3 – 4} = \frac{7}{-7} = -1$
Slope of two parallel lines is always equal.
=> Slope of the line parallel to the line having slope -1 = $-1$
=> Ans – (D)
5) Answer (C)
Slope of line having equation : $ax + by + c = 0$ is $\frac{-a}{b}$
=> Slope of line $20x + 5y = 3$ is $\frac{-20}{5} = -4$
Slope line passing through (-2,5) and (6,b) = $\frac{b – 5}{6 + 2} = \frac{(b – 5)}{8}$
Also, product of slopes of two perpendicular lines is -1
=> $\frac{(b – 5)}{8} \times -4 = -1$
=> $b – 5 = \frac{8}{4} = 2$
=> $b = 2 + 5 = 7$
=> Ans – (C)
6) Answer (B)
Slope of line having equation : $ax + by + c = 0$ is $\frac{-a}{b}$
Thus, slope of line $4x + y = 1$ is $\frac{-4}{1} = -4$
Similarly, slope of line $5x + ky = 2$ is $\frac{-5}{k}$
Also, product of slopes of two perpendicular lines is -1
=> $\frac{-5}{k} \times -4 = -1$
=> $\frac{20}{k} = -1$
=> $k = -20$
=> Ans – (B)
7) Answer (B)
Let line $l$ perpendicularly bisects line joining A(2,-6) and B(4,0) at C, thus C is the mid point of AB.
=> Coordinates of C = $(\frac{2 + 4}{2} , \frac{-6 + 0}{2})$
= $(\frac{6}{2} , \frac{-6}{2}) = (3,-3)$
Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(0 + 6)}{(4 – 2)}$
= $\frac{6}{2} = 3$
Let slope of line $l = m$
Product of slopes of two perpendicular lines = -1
=> $m \times 3 = -1$
=> $m = \frac{-1}{3}$
Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$
$\therefore$ Equation of line $l$
=> $(y + 3) = \frac{-1}{3}(x – 3)$
=> $3y + 9 = -x + 3$
=> $x + 3y = 3 – 9 = -6$
=> Ans – (B)
8) Answer (A)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Let the ratio in which the segment joining (12,1) and (3,4) divided by the y-axis = $k$ : $1$
Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $(0,y)$
Now, point P (0,y) divides (12,1) and (3,4) in ratio = k : 1
=> $0 = \frac{(3 \times k) + (12 \times 1)}{k + 1}$
=> $3k + 12 = 0$
=> $k = \frac{-12}{3} = -4$
$\therefore$ Line segment joining (12,1) and (3,4) is divided by the Y axis in the ratio = 4 : 1 externally
=> Ans – (A)
9) Answer (B)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of line passing through (1,2) and (3,0) = $\frac{0 – 2}{3 – 1} = \frac{-2}{2} = -1$
Slope of line passing through (4,3) and (y,0) = $\frac{0 – 3}{y – 4} = \frac{-3}{(y – 4)}$
Also, slopes of parallel lines are equal.
=> $\frac{-3}{y – 4} = -1$
=> $y – 4 = 3$
=> $y = 3 + 4 = 7$
=> Ans – (B)
10) Answer (A)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of line passing through (3,1) and (8,2) = $\frac{2 – 1}{8 – 3} = \frac{1}{5}$
Let slope of line perpendicular to it = $m$
Also, product of slopes of two perpendicular lines = -1
=> $m \times \frac{1}{5} = -1$
=> $m = -5$
=> Ans – (A)
11) Answer (C)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of line passing through (-3,4) and (0,3) = $\frac{3 – 4}{0 + 3} = \frac{-1}{3}$
Slope of line passing through (2,-1) and (y,-2) = $\frac{-2 + 1}{y – 2} = \frac{-1}{(y – 2)}$
Also, slopes of parallel lines are equal.
=> $\frac{-1}{y – 2} = \frac{-1}{3}$
=> $y – 2 = 3$
=> $y = 3 + 2 = 5$
=> Ans – (C)
12) Answer (A)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Now, point P (3,-2) divides (x,0) and (0,y) in ratio = 1 : 3
=> $3 = \frac{(1 \times 0) + (3 \times x)}{1 + 3}$
=> $x = \frac{3 \times 4}{3} = 4$
Similarly, $-2 = \frac{(1 \times y) + (3 \times 0)}{1 + 3}$
=> $y = -2 \times 4 = -8$
=> Ans – (A)
13) Answer (A)
Slope, $m = \frac{1}{4}$ and y-intercept, $c = -3$
Equation of line = $y = mx + c$
=> $y = \frac{1}{4} x + -3$
=> $y = \frac{x – 12}{4}$
=> $4y = x – 12$
=> $x – 4y = 12$
=> Ans – (A)
14) Answer (A)
Slope of line passing through points (2,1) and (6,3)
= $\frac{3 – 1}{6 – 2} = \frac{2}{4} = \frac{1}{2}$
Slope of two parallel lines is always equal.
=> Slope of the line parallel to the line having slope 1/2 = $\frac{1}{2}$
=> Ans – (A)
15) Answer (C)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Let the ratio in which the segment joining (-1,-12) and (3,4) divided by the x-axis = $k$ : $1$
Since, the line segment is divided by x-axis, thus y coordinate of the point will be zero, let the point of intersection = $(x,0)$
Now, point P (x,0) divides (-1,-12) and (3,4) in ratio = k : 1
=> $0 = \frac{(4 \times k) + (-12 \times 1)}{k + 1}$
=> $4k – 12 = 0$
=> $k = \frac{12}{4} = 3$
$\therefore$ Required ratio = 3 : 1
=> Ans – (C)
16) Answer (A)
When a line intercepts y-axis at a point, then x-coordinate of that point is 0.
Let the line intercepts y-axis at $(0,y)$
Equation of line = $4x – 3y = -6$
Putting $x = 0$ in above equation, we get :
=> $(4 \times 0) – 3y = -6$
=> $3y = 6$
=> $y = \frac{6}{3} = 2$
$\therefore$ The line 4x – 3y = -6 will intercept the y-axis at = (0,2)
=> Ans – (A)
17) Answer (D)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of the line passing through the points (7,-2) and (x,1)
= $\frac{1 + 2}{x – 7} = \frac{-3}{10}$
=> $\frac{3}{x – 7} = \frac{-3}{10}$
=> $x – 7 = -10$
=> $x = -10 + 7 = -3$
=> Ans – (D)
18) Answer (B)
Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$
Let coordinates of vertex C = $(x , y)$
Vertex A(-2,5) and Vertex B(6,2) and Centroid = (3,2)
=> $3 = \frac{-2 + 6 + x}{3}$
=> $x + 4 = 3 \times 3 = 9$
=> $x = 9 – 4 = 5$
Similarly, => $2 = \frac{5 + 2 + y}{3}$
=> $y + 7 = 2 \times 3 = 6$
=> $y = 6 – 7 = -1$
$\therefore$ Coordinates of vertex C = (5,-1)
=> Ans – (B)
19) Answer (B)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of the line passing through the points (-5,1) and (x,-4)
= $\frac{-4 – 1}{x + 5} = \frac{-5}{8}$
=> $\frac{-5}{x + 5} = \frac{-5}{8}$
=> $x + 5 = 8$
=> $x = 8 – 5 = 3$
=> Ans – (B)
20) Answer (B)
P(a,b) after reflection at the origin = (-a,-b)
Reflection of point (-a,-b) in the y-axis is (a,-b)
According to ques,
=> $(a,-b) = (6,-5)$
=> $a = 6$ and $-b = -5$
$\therefore$ Coordinates of Point P = (6,5)
=> Ans – (B)
21) Answer (C)
Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 – y_1}{x_2 – x_1}$
=> Slope of line passing through (-3,4) and (0,3) = $\frac{3 – 4}{0 + 3} = \frac{-1}{3}$
Slope of line passing through (2,-1) and (y,-2) = $\frac{-2 + 1}{y – 2} = \frac{-1}{(y – 2)}$
Also, slopes of parallel lines are equal.
=> $\frac{-1}{y – 2} = \frac{-1}{3}$
=> $y – 2 = 3$
=> $y = 3 + 2 = 5$
=> Ans – (C)
22) Answer (A)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Now, point P (3,-2) divides (x,0) and (0,y) in ratio = 1 : 3
=> $3 = \frac{(1 \times 0) + (3 \times x)}{1 + 3}$
=> $x = \frac{3 \times 4}{3} = 4$
Similarly, $-2 = \frac{(1 \times y) + (3 \times 0)}{1 + 3}$
=> $y = -2 \times 4 = -8$
=> Ans – (A)
23) Answer (A)
Slope, $m = \frac{1}{4}$ and y-intercept, $c = -3$
Equation of line = $y = mx + c$
=> $y = \frac{1}{4} x + -3$
=> $y = \frac{x – 12}{4}$
=> $4y = x – 12$
=> $x – 4y = 12$
=> Ans – (A)
24) Answer (A)
Slope of line passing through points (2,1) and (6,3)
= $\frac{3 – 1}{6 – 2} = \frac{2}{4} = \frac{1}{2}$
Slope of two parallel lines is always equal.
=> Slope of the line parallel to the line having slope 1/2 = $\frac{1}{2}$
=> Ans – (A)
25) Answer (C)
Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b
= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$
Let the ratio in which the segment joining (-1,-12) and (3,4) divided by the x-axis = $k$ : $1$
Since, the line segment is divided by x-axis, thus y coordinate of the point will be zero, let the point of intersection = $(x,0)$
Now, point P (x,0) divides (-1,-12) and (3,4) in ratio = k : 1
=> $0 = \frac{(4 \times k) + (-12 \times 1)}{k + 1}$
=> $4k – 12 = 0$
=> $k = \frac{12}{4} = 3$
$\therefore$ Required ratio = 3 : 1
=> Ans – (C)
