Question

If $B(1,3)$ is equidistant form $A(6,-1)$ and $C(x,8)$, then $x=$

• A. $-2,5$
• B. $2,-5$
• C. $-3,5$
• D. $3,-5$

Answer 1

Answer: (C)

$-3,5$

Given Points

$A(6,-1)B(1,3)C(x,8)$

Distance from $A$ to $B$

$AB=\sqrt{(1-6{)}^2+(3+1{)}^2}$

$AB=\sqrt{5^2+4^2}$

$AB=\sqrt{25+16}$

$AB=\sqrt{41}$

So the Distance from $B$ to $C$ is $\sqrt{41}$

$⟹BC=\sqrt{(x-1{)}^2+(8-3{)}^2}$

$⟹\sqrt{41}=\sqrt{x^2-2x+1+25}$

$⟹41=x^2-2x+26$

$⟹x^2-2x-15=0$

$⟹(x+3)(x-5)=0$

$⟹x=-3,5$