Question

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

• A. $(-\frac{1}{2},-4)$
• B. $(1,6)$
• C. $(\frac{5}{6},6)$
• D. None of these

Answer 1

The correct option is D None of these

Given that ABC is an isosceles triangle.

The $2$ vertices are $B(1,3)=(x_2,y_2)$ & $C(-2,7)={\frac{(}{x}}_2,y_2)$

Since the triangle is Isosceles. So, AB$=$AC …………$\left(1\right)$

And also vertex lies on y-axis (A)

Let A be $(0,y)=(x_1,y_1)$

Equation $\left(1\right)$ AB$=$AC

$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\sqrt{(1-0{)}^2+(3-y{)}^2}=\sqrt{(-2-0{)}^2+(7-y{)}^2}$

$1+(3-y{)}^2=4+(7-y{)}^2$

$y^2-6y+10=y^2-14y+53$

$y=\frac{41}{9}$

$y=4.6$

So, the co-ordinates of vertex A are $A(0,4.6)$.