Question

Two vertices an isosceles triangle are $(2,0)$ and $(2,5)$. Find the third vertex if the length of the equal sides is $3$.

Answer 1

$PA=PB=3$

by distance formula

$\sqrt{(X_1-X_2{)}^2+(Y_1-Y_2{)}^2}$

equating PA and PB, we get

$\sqrt{(x-2{)}^2+(y-0{)}^2}=\sqrt{(x-2{)}^2+(y-5{)}^2}$

squaring both sides , we get

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

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

$y^2=y^2+5^2-10y$

$y=\frac{5}{2}$

$v(x-2)62+y^2=3$

squaring both we get

$(x-2{)}^2+y^2=9$

$x^2+2^2-4x+y^2=9$

$x^2-5-4x+y^2=0$

putting the value of y

$x^2-4x-5(5/2{)}^2=0$

$x^2-4x+5/4=0$

on comparing

$a=1$ , $b=-4$, $c=5/4$

by quadratic formula

$x=(-b\pm \sqrt{(b^2-4ac)})/2a$

putting valueof a, b , c and then solving we get

$x=-(4\pm \sqrt{11})/2$

$x=2\pm \sqrt{11}/2$

hence the third vertex P(x ,y )

$=(2\pm \sqrt{11}/2,5/2)$