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# 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.

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

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

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)

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)