Question

Find the point on X-axis which is equidistant from $A\left(-3,4\right)$ and $B\left(1,-4\right)$.

• A. $(1,-1)$
• B. $(-1,1)$
• C. $(-1,0)$
• D. $(1,0)$

Answer 1

Answer: (C)

$(-1,0)$

Let the point be $c(x,0)$

$\therefore d\left(AC\right)=\sqrt{(x-(-3){)}^2+(0-4{)}^2}$

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

$\therefore d\left(AB\right)=\sqrt{(x-1{)}^2+(0-(-4){)}^2}$

$=\sqrt{(x-1{)}^2+(4{)}^2}$......(2)

$\therefore d\left(AC\right)=d\left(AB\right)$

$\therefore \sqrt{(x+3{)}^2+16}=\sqrt{(x-1{)}^2+16}$

$\therefore (x+3{)}^2+16=(x-1{)}^2+16$.......squaring

$\therefore x^2+6x+9=x^2-2x+1$

$\therefore 8=-8x$

$\therefore x=-1$

Hence the point is $(-1,0)$