The points (- 4, 0), (4, 0) and (0, 3)

(b) Let A(- 4, 0), B(4, 0), C(0, 3) are the given vertices.

Now, distance between A (-4, 0) and B (4, 0),

$A B=\sqrt{[4-(-4)]^{2}+(0-0)^{2}}$

$\left[\because\right.$ distance between two points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

$=\sqrt{(4+4)^{2}}=\sqrt{8^{2}}=8$

Distance between $B(4,0)$ and $C(0,3)$.

$B C=\sqrt{(0-4)^{2}+(3-0)^{2}}=\sqrt{16+9}=\sqrt{25}=5$

Distance between $A(-4,0)$ and $C(0,3)$,

$A C=\sqrt{[0-(-4)]^{2}+(3-0)^{2}}=\sqrt{16+9}=\sqrt{25}=5$

$\because$ $B C=A C$

Hence, ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.