Question

The point on the line $4x-y-2=0$ which is equidistant from the points $(-5,6)$ and $(3,2)$ is

• A. $(2,6)$
• B. $(4,14)$
• C. $(1,2)$
• D. $(3,10)$

Answer 1

Answer: (B)

$(4,14)$

Let the point on line $4x-y-2=0$ be $P(x,y)$.

Let $A\equiv (-5,6)$ and $B\equiv (3,2)$

$4x-y-2=0$ ....(1)

Point P is equidistant from points A and B ....Given

$\therefore AP=PB$

By distance formula,

$(x+5{)}^2+(y-6{)}^2=(x-3{)}^2+(y-2{)}^2$

$x^2+10x+25+y^2-12y+36=x^2-6x+9+y^2-4y+4$

$16x-8y+48=0$

$4x-2y+12=0$ ....(2)

Subtract eq(2) from eq (1), we get

$y=14$

Substitute in eq(1), we get

$x=4$

So, the point on the line is $(4,14)$.