Coordinate Geometry Test Pdf

Coordinate geometry is one of the heavy-hitter topics on the SAT, and you'll need to be able to maneuver your way through its many facets in order to take on the variety of questions you'll see on the test. Luckily, though, coordinate geometry is not difficult to visualize or wrap your head around once you know the basics. COORDINATE GEOMETRY Mathematics 1 MM1G1a,b,c,d,e 1. Algebraic formulas can be used to find measures of distance on the coordinate plane. The coordinate plane allows precise communication about graphical representations. The coordinate plane permits use of algebraic methods to obtain geometric results. Equations can be identities that express. (a) Lines 1 and 2 are perpendicular (b) Lines 1 and 4 are parallel (c) Lines 2 and 4 are perpendicular (d) Lines 2 and 3 are parallel I don’t know 3. A straight line has equation 10y = 3x + 15. Which of the following is true? Y-intercept is 0.3 (e) I don’t know 4. Students review and then take the unit assessment, hopefully demonstrating their knowledge of coordinate geometry, as well as their ability to organize and justify their work. Plan your 60-minute lesson in Math or Geometry with helpful tips from Beth Menzie. The Mathematics Test is a 60-question, 60-minute examination that measures mathematics reasoning abilities. The test focuses on the solution of practical quantitative problems that are encountered in high school and some college courses. The test uses a work-sample approach that measures mathematical skills in the context of.

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Related Topics

• 
  What is the area of triangle QRS, in square units given the vertices of parallelogram QRST in the standard (x, y) coordinate plane below?
  The correct answer to the above question is option A.24 To calculate the area of triangle QRS, in the vertices of parallelogram in the image above, we need to determine the base and height of triangle QRS. The base QR = 6 (from -3 to 3) The height RS = 8 (from -5 to 3) Area of a triangle = ½ base height Therefore the area of triangle QRS = ½ 68 = 24 The area of QRS is 24 square units Hope this helps.
  What is the area of triangle QRS, in square units given the vertices of parallelogram QRST in the standard (x, y) coordinate plane below?

  The correct answer to the above question is option A.24 To calculate the area of triangle QRS, in the vertices of parallelogram in the image above, we need to determine the base and height of triangle QRS. The base QR = 6 (from -3 to 3) The height RS = 8 (from -5 to 3) Area of a triangle = ½ base height Therefore the area of triangle QRS = ½ 68 = 24 The area of QRS is 24 square units Hope this helps.
• 
  What is the slope of the straight line 7y-2x=11 equal to?
  THE ANSWER IS2/7 IN THE QUESTION BOTH THE ANSWERS ARE SIMILAR BUT IT SHOWING WRONG PLEASE CHECK IT
  What is the slope of the straight line 7y-2x=11 equal to?

  THE ANSWER IS2/7 IN THE QUESTION BOTH THE ANSWERS ARE SIMILAR BUT IT SHOWING WRONG PLEASE CHECK IT
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  What is the value of k? A straight line y=2x +k passes through the point (1,2).
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  What is the distance in the standard (x, y) coordinate plane between points (0, 1) and (4, 4)?
  My answer is the fourth one, which is D. I solved it using the distance formula.
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• 
  What is the area of the triangle formed by the points (1,1), (0,1) (0,0) and (1,0) is?
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• 
  What is the equation of the line that passes through the origin and the point (3, 4) in the standard (x, y) coordinate plane?
  1. Y = 4/3xSlope = 3/4. remember the origin is just the point (0, 0) so you can use the slope formula followed by point slope form: y - y1 = m(x - x1) but if you think about it, b has to be 0 if it goes through the origin!
  What is the equation of the line that passes through the origin and the point (3, 4) in the standard (x, y) coordinate plane?

  1. Y = 4/3xSlope = 3/4. remember the origin is just the point (0, 0) so you can use the slope formula followed by point slope form: y - y1 = m(x - x1) but if you think about it, b has to be 0 if it goes through the origin!
• 
  What is the approximate distance between the points (4, -3) and (-6, 5) in the standard (x, y) coordinate plane?
  What is the approximate distance between the points (4, -3) and (-6, 5) in the standard (x, y) coordinate plane?

• What point is the origin?
  (1, 0)
  (0, 0)
  (0, 1)
  None of the above

• What is the point below?
  (-2, 5)
  (5, 2)
  (5, -2)
  (2, 5)

• What point is the origin?
  (1, 0)
  (0, 0)
  (0, 1)
  None of the above

• What point is the origin?
  (1, 0)
  (0, 0)
  (0, 1)
  None of the above

• What is a property of a rectangle (check all that apply)
  180 degree angles
  80 degree angles
  90 degree angles
  One set of parallel sides
  Two sets of parallel sides
  No parallel sides
  Closed quadrilateral
  Can be open on one side
  Curved lines
  Can be a square but does not have to be

• Given the vertices of parallelogram QRST in the standard (x, y) coordinate plane below, what is the area of triangle QRS, in square units?
  24
  28
  48
  60
  80

• Point A is to be graphed in a quadrant, not on an axis, of the standard (x, y) coordinate plane below. If the x-coordinate and the y-coordinate of point A are to have the same signs, then point A must be located in:
  Quadrant I only
  Quadrant II only
  Quadrant III only
  Quadrant I or II only
  Quadrant I or III only

• Find the midpoint between (3,5) and (9,1)
  (12,6)
  (6,6)
  (6,3)
  (3,2)

• What is the distance formula?A. B. C. y=mx+b
  This one.
  This one.
  Or this one.

• If (4,1) is the midpoint of the interval from (x,-2) to (5,y), what is the value of y?
  0
  6
  -3
  -10
  4

• What letter has a reflection with a vertical line?
  The letter T
  The letter P

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

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

=> $m times 3 = -1$

=> $m = frac{-1}{3}$

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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

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=> $3 = frac{(1 times 0) + (3 times x)}{1 + 3}$

Coordinate Geometry Test Pdf Software

=> $x = frac{3 times 4}{3} = 4$

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

=> $0 = frac{(4 times k) + (-12 times 1)}{k + 1}$

=> $4k – 12 = 0$

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=> $k = frac{12}{4} = 3$

Coordinate Geometry Questions And Answers

$therefore$ Required ratio = 3 : 1

=> Ans – (C)