Exams › SSC CGL (Prelims) › General

Find the equation of the perpendicular bisector of the line segment joining the points $(1,5)$ and $(5,9)$.

1. $y=-x+10$
2. $y=x+4$
3. $y=-x+14$
4. $y=x-4$

Correct answer: $y=-x+10$

Solution

The midpoint of $(1,5)$ and $(5,9)$ is $(3,7)$. The slope of the segment is $(9-5)/(5-1)=1$, so the perpendicular slope is $-1$. Using point-slope form through $(3,7)$ gives $y-7=-1(x-3)$, i.e. $y=-x+10$.

Related SSC CGL (Prelims) General questions

• What is the equation of a line passing through \((0,4)\) with slope \(-2\)?
• What is the slope of the line perpendicular to $y = 5x - 8$?
• Which line has slope \(-2\) and passes through \((3,5)\)?
• A line \(L\) is the perpendicular bisector of the line segment joining points \(P(4,6)\) and \(Q(10,2)\). What is the y-intercept of line \(L\)?
• Find the slope of the line $4x - 5y = 10$.
• Determine the slope of a straight line that passes through the coordinates \((5,-2)\) and \((8,7)\).

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