SSC CGL 2025 PYQ — Question 57

SSC CGL - 17 Sep 2025 - Shift 3 • Quantitative Aptitude • Q57

What is the equation of the perpendicular bisector of the line segment connecting the points (2, 8) and (6, 4)?

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e.  y = x + 2.

Midpoint (M)=(x1+x22,y1+y22)(M)= \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

For (2, 8) and (6, 4): M=(2+62,8+42)M = \left( \frac{2+6}{2}, \frac{8+4}{2} \right) = (8/2, 12/2) = (4, 6)

Slope of line through two points (m)=y2−y1x2−x1=4−86−2=−44=−1(m) = \frac{y_2 - y_1}{x_2 - x_1}= \frac{4-8}{6-2} = -\frac{4}{4} = -1

Perpendicular slope =−1m=−1−1=1= -\frac{1}{m} = -\frac{1}{-1} = 1

Equation of line (point-slope form): y - y1 = m(x - x1)

Using (4, 6) and m = 1: y - 6 = 1(x - 4)

⇒ y = x + 2