Question

Show that points (3, 2), (-2, -3) and (2.3) are vertices of right angle triangle.​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• We are given (3, 2), (-2, -3) and (2,3).

• Let A=(3, 2); B=(-2 ,-3) and C=(2, 3).

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• We need to show that they are the vertices of right-angled triangle.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

A triangle is a right angled triangle if it follows the pythagoras property.

Let's find the distance between the points using the distance formula:

$\boxed{\sf{\blue{d = \sqrt {(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2} }}}$

Distance between A and B:

$:\to \sf AB = \sqrt {(-2 - 3)^2 + (-3 - 2)^2}$

$:\to \sf AB=\sqrt{50} \ units.$

Distance between B and C:

$: \to \sf BC= \sqrt {(2 - (-2))^2 + (3 - (-3))^2}$

$: \to \sf BC=\sqrt{52} \ units.$

Distance between A and C:

$: \to \sf AC = \sqrt {(3 - 2)^2 + (2 - 3)^2}$

$: \to \sf AC=\sqrt{2} \ units.$

Now, use pythagoras theorem to prove the equality.

$:\to \sf AB^2+AC^2=BC^2$

$:\to \Big( \sqrt{50} \Big)^2 +\Big( \sqrt{2} \Big)^2=\Big( \sqrt{52} \Big)^2$

$: \to \sf 50+2=52$

$: \to \sf{\red{52=52}}$

Since LHS = RHS, it shows pythagoras property, thus it is a right-angled triangle.