{"id":26693,"date":"2019-03-29T13:37:40","date_gmt":"2019-03-29T08:07:40","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26693"},"modified":"2019-04-03T15:38:53","modified_gmt":"2019-04-03T10:08:53","slug":"ssc-cgl-coordinate-geometry-questions-with-solutions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-cgl-coordinate-geometry-questions-with-solutions-pdf\/","title":{"rendered":"SSC CGL Coordinate Geometry Questions with Solutions PDF"},"content":{"rendered":"

SSC CGL Coordinate Geometry Questions with Solutions PDF:<\/strong><\/span><\/h1>\n

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.<\/p>\n

Download SSC CGL Coordinate Geometry Questions PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>Find equation of the perpendicular bisector of segment joining the points (2,-5) and (0,7)?<\/p>\n

a)\u00a0x – 6y = 5<\/p>\n

b)\u00a0x + 6y = -5<\/p>\n

c)\u00a0x – 6y = -5<\/p>\n

d)\u00a0x + 6y = 5<\/p>\n

Question 2:\u00a0<\/b>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.<\/p>\n

a)\u00a0x – y = 8<\/p>\n

b)\u00a0x – y = -8<\/p>\n

c)\u00a0x + y = -8<\/p>\n

d)\u00a0x + y = 8<\/p>\n

Question 3:\u00a0<\/b>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)\u00a0 and (-2,2) respectively?<\/p>\n

a)\u00a0(-7,-4)<\/p>\n

b)\u00a0(7,4)<\/p>\n

c)\u00a0(7,-4)<\/p>\n

d)\u00a0(-7,4)<\/p>\n

Question 4:\u00a0<\/b>What is the slope of the line parallel to the line passing through the points (4,-2) and (-3,5)?<\/p>\n

a)\u00a03\/7<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0-3\/7<\/p>\n

d)\u00a0-1<\/p>\n

Question 5:\u00a0<\/b>The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?<\/p>\n

a)\u00a0-7<\/p>\n

b)\u00a04<\/p>\n

c)\u00a07<\/p>\n

d)\u00a0-4<\/p>\n

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Question 6:\u00a0<\/b>Find k, if the line 4x+y = 1 is perpendicular to the line 5x+ky = 2?<\/p>\n

a)\u00a020<\/p>\n

b)\u00a0-20<\/p>\n

c)\u00a04<\/p>\n

d)\u00a0-4<\/p>\n

Question 7:\u00a0<\/b>Find equation of the perpendicular bisector of segment joining the points (2,-6) and (4,0)?<\/p>\n

a)\u00a0x + 3y = 6<\/p>\n

b)\u00a0x + 3y = -6<\/p>\n

c)\u00a0x – 3y = -6<\/p>\n

d)\u00a0x – 3y = 6<\/p>\n

Question 8:\u00a0<\/b> \u00ad In what ratio is the segment joining (12,\u00ad1) and (\u00ad3,4) divided by the Y \u00adaxis?<\/p>\n

a)\u00a04:1<\/p>\n

b)\u00a01:4<\/p>\n

c)\u00a04:3<\/p>\n

d)\u00a03:4<\/p>\n

Question 9:\u00a0<\/b>The line passing through (4,3) and (y,0) is parallel to the line passing through (\u00ad1,2) and (3,0). Find y?<\/p>\n

a)\u00a0\u00ad1<\/p>\n

b)\u00a07<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0\u00ad5<\/p>\n

Question 10:\u00a0<\/b>What is the slope of the line perpendicular to the line passing through the points (8,\u00ad2) and (3,\u00ad1)?<\/p>\n

a)\u00a0-5<\/p>\n

b)\u00a03\/5<\/p>\n

c)\u00a05\/3<\/p>\n

d)\u00a01\/5<\/p>\n

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Question 11:\u00a0<\/b>The line passing through (2,-1) and (y,-2) is parallel to the line passing through (-3,4) and (0,3) Find y?<\/p>\n

a)\u00a0-2<\/p>\n

b)\u00a02<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0-5<\/p>\n

Question 12:\u00a0<\/b>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?<\/p>\n

a)\u00a0x = 4; y = -8<\/p>\n

b)\u00a0x = -3; y = -8<\/p>\n

c)\u00a0x = 3; y = 8<\/p>\n

d)\u00a0x = -3; y = 8<\/p>\n

Question 13:\u00a0<\/b>What is the equation of the line if its slope is 1\/4 and y-intercept is -3?<\/p>\n

a)\u00a0x – 4y = 12<\/p>\n

b)\u00a0x + 4y = 12<\/p>\n

c)\u00a0x – 4y = -12<\/p>\n

d)\u00a0x + 4y = -12<\/p>\n

Question 14:\u00a0<\/b>What is the slope of the line parallel to the line passing through the points (6,\u00ad3) and (2,\u00ad1)?<\/p>\n

a)\u00a0\u00ad1\/2<\/p>\n

b)\u00a0\u00ad1<\/p>\n

c)\u00a02<\/p>\n

d)\u00a01<\/p>\n

Question 15:\u00a0<\/b>In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?<\/p>\n

a)\u00a01:3<\/p>\n

b)\u00a03:2<\/p>\n

c)\u00a03:1<\/p>\n

d)\u00a02:3<\/p>\n

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Question 16:\u00a0<\/b>At what point does the line 4x – 3y = -6 intercept the y-axis?<\/p>\n

a)\u00a0(0,2)<\/p>\n

b)\u00a0(0,3\/2)<\/p>\n

c)\u00a0(2,0)<\/p>\n

d)\u00a0(3\/2,0)<\/p>\n

Question 17:\u00a0<\/b>The slope of the line passing through the points \u00a0(7,-2) and (x,1) is -3\/10. Find x.<\/p>\n

a)\u00a04<\/p>\n

b)\u00a02<\/p>\n

c)\u00a0-4<\/p>\n

d)\u00a0-3<\/p>\n

Question 18:\u00a0<\/b>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?<\/p>\n

a)\u00a0(-5,-1)<\/p>\n

b)\u00a0(5,-1)<\/p>\n

c)\u00a0(5,1)<\/p>\n

d)\u00a0(-5,1)<\/p>\n

Question 19:\u00a0<\/b>The slope of the line passing through the points (-5,1) and (x,-4) is -5\/8. Find x.<\/p>\n

a)\u00a04<\/p>\n

b)\u00a03<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0-1<\/p>\n

Question 20:\u00a0<\/b>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<\/p>\n

a)\u00a0(-6,-5)<\/p>\n

b)\u00a0(6,5)<\/p>\n

c)\u00a0(-6,5)<\/p>\n

d)\u00a0(6,-5)<\/p>\n

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Question 21:\u00a0<\/b>The line passing through (2,-1) and (y,-2) is parallel to the line passing through (-3,4) and (0,3) Find y?<\/p>\n

a)\u00a0-2<\/p>\n

b)\u00a02<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0-5<\/p>\n

Question 22:\u00a0<\/b>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?<\/p>\n

a)\u00a0x = 4; y = -8<\/p>\n

b)\u00a0x = -3; y = -8<\/p>\n

c)\u00a0x = 3; y = 8<\/p>\n

d)\u00a0x = -3; y = 8<\/p>\n

Question 23:\u00a0<\/b>What is the equation of the line if its slope is 1\/4 and y-intercept is -3?<\/p>\n

a)\u00a0x – 4y = 12<\/p>\n

b)\u00a0x + 4y = 12<\/p>\n

c)\u00a0x – 4y = -12<\/p>\n

d)\u00a0x + 4y = -12<\/p>\n

Question 24:\u00a0<\/b>What is the slope of the line parallel to the line passing through the points (6,\u00ad3) and (2,\u00ad1)?<\/p>\n

a)\u00a0\u00ad1\/2<\/p>\n

b)\u00a0\u00ad1<\/p>\n

c)\u00a02<\/p>\n

d)\u00a01<\/p>\n

Question 25:\u00a0<\/b>In what ratio is the segment joining (-1,-12) and (3,4) divided by the x-axis?<\/p>\n

a)\u00a01:3<\/p>\n

b)\u00a03:2<\/p>\n

c)\u00a03:1<\/p>\n

d)\u00a02:3<\/p>\n

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 <\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let line $l$ perpendicularly bisects line joining \u00a0A(2,-5) and B(0,7) at C, thus C is the mid point of AB.<\/p>\n

=> Coordinates of C = $(\\frac{2 + 0}{2} , \\frac{-5 + 7}{2})$<\/p>\n

= $(\\frac{2}{2} , \\frac{2}{2}) = (1,1)$<\/p>\n

Now, slope of AB = $\\frac{y_2 – y_1}{x_2 – x_1} = \\frac{(7 + 5)}{(0 – 2)}$<\/p>\n

= $\\frac{12}{-2} = -6$<\/p>\n

Let slope of line $l = m$<\/p>\n

Product of slopes of two perpendicular lines = -1<\/p>\n

=> $m \\times -6 = -1$<\/p>\n

=> $m = \\frac{1}{6}$<\/p>\n

Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$<\/p>\n

$\\therefore$ Equation of line $l$<\/p>\n

=> $(y – 1) = \\frac{1}{6}(x – 1)$<\/p>\n

=> $6y – 6 = x – 1$<\/p>\n

=> $x – 6y = 1 – 6 = -5$<\/p>\n

=> Ans – (C)<\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Coordinates of A(0,4) and B(-5,9). Let coordinates of\u00a0P = (x,y) which divides AB in ratio = 2\u00a0: 3<\/p>\n

=> $x = \\frac{(2 \\times -5) + (3 \\times 0)}{2 + 3}$<\/p>\n

=> $5x = -10$<\/p>\n

=> $x = \\frac{-10}{5} = -2$<\/p>\n

Similarly, $y = \\frac{(2 \\times 9) + (3 \\times 4)}{2 + 3}$<\/p>\n

=> $5y = 18 + 12 = 30$<\/p>\n

=> $y = \\frac{30}{5} = 6$<\/p>\n

=> Point P = (-2,6)<\/p>\n

Slope of AB = $\\frac{9 – 4}{-5 – 0} = \\frac{5}{-5} = -1$<\/p>\n

Let slope of line perpendicular to AB = $m$<\/p>\n

Also, product of slopes of two perpendicular lines is -1<\/p>\n

=> $m \\times -1 = -1$<\/p>\n

=> $m = 1$<\/p>\n

Equation of lines having slope $m$ and passing through point P(-2,6) is<\/p>\n

=> $(y – 6) = 1(x + 2)$<\/p>\n

=> $y – 6 = x + 2$<\/p>\n

=> $x – y = -8$<\/p>\n

=> Ans – (B)<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\\frac{x_1 + x_2 + x_3}{3} , \\frac{y_1 + y_2 + y_3}{3})$<\/p>\n

Let coordinates of vertex C = $(x , y)$<\/p>\n

Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)<\/p>\n

=> $-1 = \\frac{-2 + 6 + x}{3}$<\/p>\n

=> $x + 4 = -1 \\times 3 = -3$<\/p>\n

=> $x = -3 – 4 = -7$<\/p>\n

Similarly, => $-2 = \\frac{-4 + 2 + y}{3}$<\/p>\n

=> $y – 2 = -2 \\times 3 = -6$<\/p>\n

=> $y = -6 + 2 = -4$<\/p>\n

$\\therefore$ Coordinates of vertex C = (-7,-4)<\/p>\n

=> Ans – (A)<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Slope of line passing through points\u00a0(4,-2) and (-3,5)<\/p>\n

= $\\frac{5 + 2}{-3 – 4} = \\frac{7}{-7} = -1$<\/p>\n

Slope of two parallel lines is always equal.<\/p>\n

=> Slope of the line parallel to the line having slope -1 = $-1$<\/p>\n

=> Ans – (D)<\/p>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Slope of line having equation : $ax + by + c = 0$ is $\\frac{-a}{b}$<\/p>\n

=> Slope of line $20x + 5y = 3$ is $\\frac{-20}{5} = -4$<\/p>\n

Slope\u00a0line passing through (-2,5) and (6,b) = $\\frac{b – 5}{6 + 2} = \\frac{(b – 5)}{8}$<\/p>\n

Also, product of slopes of two perpendicular lines is -1<\/p>\n

=> $\\frac{(b – 5)}{8} \\times -4 = -1$<\/p>\n

=> $b – 5 = \\frac{8}{4} = 2$<\/p>\n

=> $b = 2 + 5 = 7$<\/p>\n

=> Ans – (C)<\/p>\n\n

6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Slope of line having equation : $ax + by + c = 0$ is $\\frac{-a}{b}$<\/p>\n

Thus, slope of line $4x + y = 1$ is $\\frac{-4}{1} = -4$<\/p>\n

Similarly, slope of line $5x + ky = 2$ is $\\frac{-5}{k}$<\/p>\n

Also, product of slopes of two perpendicular lines is -1<\/p>\n

=> $\\frac{-5}{k} \\times -4 = -1$<\/p>\n

=> $\\frac{20}{k} = -1$<\/p>\n

=> $k = -20$<\/p>\n

=> Ans – (B)<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let line $l$ perpendicularly bisects line joining \u00a0A(2,-6) and B(4,0) at C, thus C is the mid point of AB.<\/p>\n

=> Coordinates of C = $(\\frac{2 + 4}{2} , \\frac{-6 + 0}{2})$<\/p>\n

= $(\\frac{6}{2} , \\frac{-6}{2}) = (3,-3)$<\/p>\n

Now, slope of AB = $\\frac{y_2 – y_1}{x_2 – x_1} = \\frac{(0 + 6)}{(4 – 2)}$<\/p>\n

= $\\frac{6}{2} = 3$<\/p>\n

Let slope of line $l = m$<\/p>\n

Product of slopes of two perpendicular lines = -1<\/p>\n

=> $m \\times 3 = -1$<\/p>\n

=> $m = \\frac{-1}{3}$<\/p>\n

Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$<\/p>\n

$\\therefore$ Equation of line $l$<\/p>\n

=> $(y + 3) = \\frac{-1}{3}(x – 3)$<\/p>\n

=> $3y + 9 = -x + 3$<\/p>\n

=> $x + 3y = 3 – 9 = -6$<\/p>\n

=> Ans – (B)<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Let the ratio in which\u00a0the segment joining (12,1) and (3,4) divided by the y-axis = $k$ : $1$<\/p>\n

Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $(0,y)$<\/p>\n

Now, point P (0,y) divides (12,1) and (3,4) in ratio = k : 1<\/p>\n

=> $0 = \\frac{(3 \\times k) + (12 \\times 1)}{k + 1}$<\/p>\n

=> $3k + 12 = 0$<\/p>\n

=> $k = \\frac{-12}{3} = -4$<\/p>\n

$\\therefore$ Line segment joining (12,\u00ad1) and (\u00ad3,4) is divided by the Y \u00adaxis in the ratio = 4 : 1 externally<\/p>\n

=> Ans – (A)<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=>\u00a0Slope of line passing through\u00a0(1,2) and (3,0) = $\\frac{0 – 2}{3 – 1} = \\frac{-2}{2} = -1$<\/p>\n

Slope of line passing through\u00a0(4,3) and (y,0) = $\\frac{0 – 3}{y – 4} = \\frac{-3}{(y – 4)}$<\/p>\n

Also, slopes of parallel lines are equal.<\/p>\n

=> $\\frac{-3}{y – 4} = -1$<\/p>\n

=> $y – 4 = 3$<\/p>\n

=> $y = 3 + 4 = 7$<\/p>\n

=> Ans – (B)<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=>\u00a0Slope of line passing through\u00a0(3,1) and (8,2) = $\\frac{2 – 1}{8 – 3} = \\frac{1}{5}$<\/p>\n

Let slope of line perpendicular to it = $m$<\/p>\n

Also, product of slopes of two perpendicular lines = -1<\/p>\n

=> $m \\times \\frac{1}{5} = -1$<\/p>\n

=> $m = -5$<\/p>\n

=> Ans – (A)<\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=>\u00a0Slope of line passing through\u00a0(-3,4) and (0,3) = $\\frac{3 – 4}{0 + 3} = \\frac{-1}{3}$<\/p>\n

Slope of line passing through\u00a0(2,-1) and (y,-2) = $\\frac{-2 + 1}{y – 2} = \\frac{-1}{(y – 2)}$<\/p>\n

Also, slopes of parallel lines are equal.<\/p>\n

=> $\\frac{-1}{y – 2} = \\frac{-1}{3}$<\/p>\n

=> $y – 2 = 3$<\/p>\n

=> $y = 3 + 2 = 5$<\/p>\n

=> Ans – (C)<\/p>\n

12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Now, point P (3,-2) divides (x,0) and (0,y) in ratio = 1 : 3<\/p>\n

=> $3 = \\frac{(1 \\times 0) + (3 \\times x)}{1 + 3}$<\/p>\n

=> $x = \\frac{3 \\times 4}{3} = 4$<\/p>\n

Similarly, $-2 = \\frac{(1 \\times y) + (3 \\times 0)}{1 + 3}$<\/p>\n

=> $y = -2 \\times 4 = -8$<\/p>\n

=> Ans – (A)<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Slope, $m = \\frac{1}{4}$ and y-intercept, $c = -3$<\/p>\n

Equation of line = $y = mx + c$<\/p>\n

=> $y = \\frac{1}{4} x + -3$<\/p>\n

=> $y = \\frac{x – 12}{4}$<\/p>\n

=> $4y = x – 12$<\/p>\n

=> $x – 4y = 12$<\/p>\n

=> Ans – (A)<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Slope of line passing through points\u00a0(2,1) and (6,\u00ad3)<\/p>\n

= $\\frac{3 – 1}{6 – 2} = \\frac{2}{4} = \\frac{1}{2}$<\/p>\n

Slope of two parallel lines is always equal.<\/p>\n

=> Slope of the line parallel to the line having slope 1\/2 = $\\frac{1}{2}$<\/p>\n

=> Ans – (A)<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Let the ratio in which\u00a0the segment joining (-1,-12) and (3,4) divided by the x-axis = $k$ : $1$<\/p>\n

Since, the line segment is divided by x-axis, thus y coordinate of the point will be zero, let the point of intersection = $(x,0)$<\/p>\n

Now, point P (x,0) divides (-1,-12) and (3,4) in ratio = k : 1<\/p>\n

=> $0 = \\frac{(4 \\times k) + (-12 \\times 1)}{k + 1}$<\/p>\n

=> $4k – 12 = 0$<\/p>\n

=> $k = \\frac{12}{4} = 3$<\/p>\n

$\\therefore$ Required ratio = 3 : 1<\/p>\n

=> Ans – (C)<\/p>\n

Free SSC Daily Practice Set<\/a><\/p>\n

16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

When a line intercepts y-axis at a point, then x-coordinate of that point is 0.<\/p>\n

Let the line intercepts y-axis at $(0,y)$<\/p>\n

Equation of line = $4x – 3y = -6$<\/p>\n

Putting $x = 0$ in above equation, we get\u00a0:<\/p>\n

=> $(4 \\times 0) – 3y = -6$<\/p>\n

=> $3y = 6$<\/p>\n

=> $y = \\frac{6}{3} = 2$<\/p>\n

$\\therefore$ The line 4x – 3y = -6 will intercept the y-axis at = (0,2)<\/p>\n

=> Ans – (A)<\/p>\n

17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=> Slope of the line passing through the points (7,-2) and (x,1)<\/p>\n

= $\\frac{1 + 2}{x – 7} = \\frac{-3}{10}$<\/p>\n

=> $\\frac{3}{x – 7} = \\frac{-3}{10}$<\/p>\n

=> $x – 7 = -10$<\/p>\n

=> $x = -10 + 7 = -3$<\/p>\n

=> Ans – (D)<\/p>\n

18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\\frac{x_1 + x_2 + x_3}{3} , \\frac{y_1 + y_2 + y_3}{3})$<\/p>\n

Let coordinates of vertex C = $(x , y)$<\/p>\n

Vertex A(-2,5) and Vertex B(6,2) and Centroid = (3,2)<\/p>\n

=> $3 = \\frac{-2 + 6 + x}{3}$<\/p>\n

=> $x + 4 = 3 \\times 3 = 9$<\/p>\n

=> $x = 9 – 4 = 5$<\/p>\n

Similarly, => $2 = \\frac{5 + 2 + y}{3}$<\/p>\n

=> $y + 7 = 2 \\times 3 = 6$<\/p>\n

=> $y = 6 – 7 = -1$<\/p>\n

$\\therefore$ Coordinates of vertex C = (5,-1)<\/p>\n

=> Ans – (B)<\/p>\n

19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=> Slope of the line passing through the points (-5,1) and (x,-4)<\/p>\n

= $\\frac{-4 – 1}{x + 5} = \\frac{-5}{8}$<\/p>\n

=> $\\frac{-5}{x + 5} = \\frac{-5}{8}$<\/p>\n

=> $x + 5 = 8$<\/p>\n

=> $x = 8 – 5 = 3$<\/p>\n

=> Ans – (B)<\/p>\n

20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

P(a,b) after reflection at the origin = (-a,-b)<\/p>\n

Reflection of point (-a,-b) in the y-axis is (a,-b)<\/p>\n

According to ques,<\/p>\n

=> $(a,-b) = (6,-5)$<\/p>\n

=> $a = 6$ and $-b = -5$<\/p>\n

$\\therefore$ Coordinates of Point P = (6,5)<\/p>\n

=> Ans – (B)<\/p>\n\n

21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Slope of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\\frac{y_2 – y_1}{x_2 – x_1}$<\/p>\n

=>\u00a0Slope of line passing through\u00a0(-3,4) and (0,3) = $\\frac{3 – 4}{0 + 3} = \\frac{-1}{3}$<\/p>\n

Slope of line passing through\u00a0(2,-1) and (y,-2) = $\\frac{-2 + 1}{y – 2} = \\frac{-1}{(y – 2)}$<\/p>\n

Also, slopes of parallel lines are equal.<\/p>\n

=> $\\frac{-1}{y – 2} = \\frac{-1}{3}$<\/p>\n

=> $y – 2 = 3$<\/p>\n

=> $y = 3 + 2 = 5$<\/p>\n

=> Ans – (C)<\/p>\n

22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Now, point P (3,-2) divides (x,0) and (0,y) in ratio = 1 : 3<\/p>\n

=> $3 = \\frac{(1 \\times 0) + (3 \\times x)}{1 + 3}$<\/p>\n

=> $x = \\frac{3 \\times 4}{3} = 4$<\/p>\n

Similarly, $-2 = \\frac{(1 \\times y) + (3 \\times 0)}{1 + 3}$<\/p>\n

=> $y = -2 \\times 4 = -8$<\/p>\n

=> Ans – (A)<\/p>\n

23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Slope, $m = \\frac{1}{4}$ and y-intercept, $c = -3$<\/p>\n

Equation of line = $y = mx + c$<\/p>\n

=> $y = \\frac{1}{4} x + -3$<\/p>\n

=> $y = \\frac{x – 12}{4}$<\/p>\n

=> $4y = x – 12$<\/p>\n

=> $x – 4y = 12$<\/p>\n

=> Ans – (A)<\/p>\n

24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Slope of line passing through points\u00a0(2,1) and (6,\u00ad3)<\/p>\n

= $\\frac{3 – 1}{6 – 2} = \\frac{2}{4} = \\frac{1}{2}$<\/p>\n

Slope of two parallel lines is always equal.<\/p>\n

=> Slope of the line parallel to the line having slope 1\/2 = $\\frac{1}{2}$<\/p>\n

=> Ans – (A)<\/p>\n

25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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\u00a0: b<\/p>\n

= $(\\frac{a x_2 + b x_1}{a + b} , \\frac{a y_2 + b y_1}{a + b})$<\/p>\n

Let the ratio in which\u00a0the segment joining (-1,-12) and (3,4) divided by the x-axis = $k$ : $1$<\/p>\n

Since, the line segment is divided by x-axis, thus y coordinate of the point will be zero, let the point of intersection = $(x,0)$<\/p>\n

Now, point P (x,0) divides (-1,-12) and (3,4) in ratio = k : 1<\/p>\n

=> $0 = \\frac{(4 \\times k) + (-12 \\times 1)}{k + 1}$<\/p>\n

=> $4k – 12 = 0$<\/p>\n

=> $k = \\frac{12}{4} = 3$<\/p>\n

$\\therefore$ Required ratio = 3 : 1<\/p>\n

=> Ans – (C)<\/p>\n\n

SSC Free Previous Papers App<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

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. 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