Question

The perpendicular bisector of the line segment joining P(1,2) and Q (k, 3) has y- intercept -4. Then a possible value of k is -

Answer 1

Let $A(x,y)$ be any point on the perpendicular bisector of PQ. Then,

$AP=AQ$

$\sqrt{(x-1{)}^2+(y-2{)}^2}=\sqrt{(x-k{)}^2+(y-3{)}^2}$

As the point have $y-intercept$$=-4$

Hence $x=0$

Therefore,

$\sqrt{(x-1{)}^2+(y-2{)}^2}=\sqrt{(x-k{)}^2+(y-3{)}^2}$

$=\sqrt{(-1{)}^2+(-4-2{)}^2}=\sqrt{(-k{)}^2+(-4-3{)}^2}$

Squaring Both Sides We Get,

$=37=k^2+49$

$k^2=12$

$k=\sqrt{12}$