Question

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1,3)$ and $C(-2,7)$, the coordinates of vertex $A$ can be

• A. $(1,6)$
• B. $(-1/2,5)$
• C. $(5/6,6)$
• D. none of these

Answer 1

The correct option is A $(1,6)$

$AB=AC$

$\sqrt{(x-1{)}^2+(y-3{)}^2}=\sqrt{(x-(-2){)}^2+(y-7{)}^2}$

$(x-1{)}^2+(y-3{)}^2=(x+2{)}^2+(y-7{)}^2$

$=2x+1-6y+9=4x+4-14y+49$

$\Rightarrow 6x-8y+43=0$

By substituting $(\frac{5}{6},6)$

$\Rightarrow \text{/}6\times \frac{5}{\text{/}6}-8\times 6+43$

$\Rightarrow 0$

Hence $(\frac{5}{6},6)$ is answer.