Question

Show that the point (8, -10), (7, -3) and (0, -4 ) are the vertices of a right triangle.

Answer 1

Proof :

Let, the points are A (8, -10), B (7, -3) and C (0, -4)

• AB = $\sqrt{(8-7)^{2}+(-10+3)^{2}}$ units

$=\sqrt{1^{2}+7^{2}}$ units

$=\sqrt{1+49}$ units

$=\sqrt{50}$ units

• BC = $\sqrt{(7-0)^{2}+(-3+4)^{2}}$ units

$=\sqrt{7^{2}+1^{2}}$ units

$=\sqrt{49+1}$ units

$=\sqrt{50}$ units

• CA = $\sqrt{(0-8)^{2}+(-4+10)^{2}}$ units

$=\sqrt{8^{2}+6^{2}}$ units

$=\sqrt{64+36}$ units

$=\sqrt{100}$ units

We see that, $AB^{2}+BC^{2}=CA^{2}$

where $(\sqrt{50})^{2}+(\sqrt{50})^{2}=(\sqrt{100})^{2}$

By Pythagorian theorem, we can say that the triangle ABC is a right angled triangle.

Hence, proved.