UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The right bisector (perpendicular bisector) of a line segment is the line that is perpendicular to the segment and passes through its midpoint.Explanation:Let the given points be $A(1,1)$ and $B(2,3)$.Step 1: Find the slope of $AB$.Slope $= \frac{3 - 1}{2 - 1} = 2$.Step 2: Slope of a line perpendicular to $AB$ is the negative reciprocal: $-\frac{1}{2}$.Step 3: Find the midpoint of $AB$.Midpoint $= \left( \frac{1+2}{2}, \frac{1+3}{2} \right) = \left( \frac{3}{2}, 2 \right)$.Step 4: Use point-slope form $y - y_1 = m(x - x_1)$ with $m = -\frac{1}{2}$ and point $\left( \frac{3}{2}, 2 \right)$.Equation: $y - 2 = -\frac{1}{2} \left( x - \frac{3}{2} \right)$.Multiply both sides by 2: $2(y - 2) = -1 \left( x - \frac{3}{2} \right)$.Simplify: $2y - 4 = -x + \frac{3}{2}$.Multiply by 2 again: $4y - 8 = -2x + 3$.Rearrange: $2x + 4y - 11 = 0$.Answer:The equation of the right bisector is $2x + 4y - 11 = 0$, which corresponds to option A.