The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is

Solution:

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,

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

$=\sqrt{(8)^{2}+(0)^{2}}$

$=\sqrt{64+0}$

$=\sqrt{64}$

$=8$ units

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

$=\sqrt{(-4)^{2}+(3)^{2}}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$ units

$A C=\sqrt{(0+4)^{2}+(3-0)^{2}}$

$=\sqrt{(4)^{2}+(3)^{2}}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$ units

BC = AC = 5 units

Therefore, $\Delta A B C$ is isosceles.