Question

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

• A. $(1,6)$
• B. $(-\frac{1}{2},5)$
• C. $(\frac{5}{6},6)$
• D. $(-7,\frac{1}{8})$

Answer 1

Answer: (C), (D)

The correct options are
C $(\frac{5}{6},6)$
D $(-7,\frac{1}{8})$

The vertex $A(x,y)$ is equidistant from $B$ and $C.$
$\therefore (x-1{)}^2+(y-3{)}^2=(x+2{)}^2+(y-7{)}^2$

$\Rightarrow x^2+1-2x+y^2-6y+9=x^2+4x+4+y^2-14y+49$
$\Rightarrow -2x-6y+10=4x-14y+53$
$\Rightarrow 6x-8y+43=0$ (1)

Any point satisfying this can be the vertex $A,$ except the mid-point $(-1/2,5)$ of $BC$.

Since the coordinates are given in (c) and (d) satisfy Equation (1),

these are therefore the correct answers.