Question

The equation to the perpendicular bisector of the line segment joining (5,7) (3,1) is

• A. $3x+y=16$
• B. $x-3y=12$
• C. $x+3y=16$
• D. $3x-y=12$

Answer 1

The correct option is C $x+3y=16$

Here, $A=(5,7)$ and $B=(3,1)$

Here, $x_1=5,y_1=7,x_2=3$ and $y_2=1$

$\Rightarrow$ Mid-point of $A$ and $B=$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{5+3}{2},\frac{7+1}{2})$

$=(4,4)$

$\Rightarrow$ Slope of $AB\left(m\right)=\frac{1-7}{3-5}=\frac{-6}{-2}=3$

$\Rightarrow$ Slope of the perpendicular $\left(m_1\right)$ $=\frac{-1}{m}=\frac{-1}{3}$

The perpendicular passes through $(4,4)$.

The equation is,

$\Rightarrow$ $y-4=\frac{-1}{3}(x-4)$

$\Rightarrow$ $3y-12=-x+4$

$\Rightarrow$ $x+3y=16$