Question

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

• A. $(1,6)$
• B. $(-\frac{1}{2},5)$
• C. $(\frac{5}{6},6)$
• D. $(-7,\frac{1}{9})$

Answer 1

The correct option is C $(\frac{5}{6},6)$

Since the triangle is supposed to be isosceles, hence the third vertex, A should lie on the perpendicular bisector of BC.

The perpendicular from vertex A to the base BC will bisect BC.

In other words, the perpendicular from A will pass through the midpoint of BC.

Now slope of BC

$=\frac{7-3}{-2-1}$

$=\frac{-4}{3}$

Hence slope of altitude from A is $=\frac{3}{4}$.

The midpoint of BC $=\frac{1-2}{2},\frac{7+3}{2}$

$=(\frac{-1}{2},5)$

Thus equation of the altitude is

$\frac{y-5}{(x+\frac{1}{2})}=\frac{3}{4}$

$y-5=\frac{3x}{4}+\frac{3}{8}$

$y=\frac{3x}{4}+\frac{43}{8}$

Hence vertex A will be of the form $(x,\frac{3x}{4}+\frac{43}{8})$

Considering the options

If we substitute $x=\frac{5}{6}$

We get

$y=\frac{3\left(\frac{5}{6}\right)}{4}+\frac{43}{8}$
$=6$

Hence one possible vertex is $(\frac{5}{6},6)$

If we substitute $x=-7$

$y=\frac{3(-7)}{4}+\frac{43}{8}$

$=\frac{1}{8}$

Hence another possible vertex is $(-7,\frac{1}{8})$.

Thus we get two positions for the vertex A.