Question

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

• A. (1, 6)
• B. $\left( - \frac{1}{2},5 \right)$
• C. $\left( \frac{5}{6},6 \right)$
• D. None of these

Answer 1

Answer: (C)

$\left( \frac{5}{6},6 \right)$

Answer 2

Let the vertex of triangle be $A(x,y)$.

Then the vertex $A(x,y)$ is equidistant from B and C because ABC is an isosceles triangle, therefore

$(x - 1)^{2} + (y - 3)^{2}$= $(x + 2)^{2} + (y - 7)^{2}$ ⇒ $6x - 8y + 43 = 0$

Thus, any point lying on this line can be the vertex A except the mid point $\left( - \frac{1}{2},5 \right)$ of BC. Hence vertex A is $\left( \frac{5}{6},6 \right)$