Question

The perpendicular bisector of the line segment joining $\text{P}\left(\text{1,}\text{4}\right)\text{and}\text{Q}\left(\text{K,}\text{3}\right)$ has y-intercept -4 Then a possible value of K is

• A. -4
• B. 1
• C. 2
• D. -2

Answer 1

Answer: (A)

-4

Slope of $PQ=\frac{3-4}{k-1}=\frac{-1}{k-1}=\frac{1}{1-k}$

Slope of perpendicular bisector of $PQ=\frac{-1}{1-k}$ $=k-1$

$\Rightarrow$ $(\frac{1+k}{2},\frac{7}{2})(k-1)$

$\Rightarrow$$y-\frac{7}{2}=(k-1)(x-\frac{1+k}{2})$

$\Rightarrow$$\frac{2y-7}{2}=(k-1)\left(\frac{2x-1-k}{2}\right)$

$\Rightarrow$$(2y-k)=(k-1)(2x-1-k)$

$\Rightarrow$$(2y-7)=2xk-k-k^2-2x+1+k$

$\Rightarrow$$2x(1-k)+2y+k^2-8=0$

For y-intercept $x=0$,

$\Rightarrow$$2y+k^2-8=0$

$\Rightarrow$$2y=8-k^2$

$\Rightarrow$$y=4-\frac{k^2}{2}$

Given y-intercept$=-4$

$\Rightarrow$$4-\frac{k^2}{2}=-4$

$\Rightarrow$$\frac{k^2}{2}=8$

$\Rightarrow$$k^2=16$

$\Rightarrow$$k=\pm 4$