Question

state if points A(4,7), B(8,4),C(7,11) are vertices of a right angled or not​

Answer 1

SOLUTION :-

In order to check whether the given points are vertices of a right angled triangle or not , first we should find the length of the sides of the triangle.

$\bf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\bullet \sf \ AB = \sqrt{(8-4)^2+(4-7)^2}$

        $= \sf \sqrt{4^2+3^2} \\\\= \sqrt{16+9} \\\\= \sqrt{25} \\\\= 5 \ units$

$\bullet \sf \ BC = \sqrt{(7-8)^2+(11-4)^2}$

        $= \sf \sqrt{(-1)^2+7^2} \\\\= \sqrt{1+49} \\\\= \sqrt{50} \ units$

$\bullet \ \sf AC =\sqrt{(7-4)^2+(11-7)^2}$

        $= \sf \sqrt{3^2+4^2} \\\\= \sqrt{9+16} \\\\= \sqrt{25} \\\\= 5 \ units$

According to the Pythagoras theorem,

Δ ABC is said to be right angled triangle if BC² = AC² + AB² .

$\longrightarrow\sf AC^2+AB^2= 5^2+5^2 \\\\\longrightarrow AC^2+AB^2 = 25+2 5\\\\\longrightarrow AC^2+AB^2= 50$

$\longrightarrow\sf BC^2= (\sqrt{50} )^2 \\\\\longrightarrow BC^2= 50$

∴ BC² = AC² + AB²

Hence the given points are the vertices of a right angled triangle.