Question

show that the points A(-4,10) and B(5,2) and origin O(0,0) form a right anglex triangle​

Answer 1

Solution :-

Given :-

The points A ( -4 , 10 ), B ( 5 , 2 ) and O ( 0 , 0 ) are the vertices of a triangle.

We have to show that :-

These points form a right angled triangle.

First lets find AB, BO and AO.

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

$\bullet\sf \ AB = \sqrt{(5-(-4))^2+(2-10)^2}$

        $= \sf \sqrt{9^2+(-8)^2} \\\\= \sqrt{81+64} \\\\= \sqrt{145} \ units$

$\bullet \sf \ BO = \sqrt{(0-5)^2+(0-2)^2}$

        $= \sf \sqrt{(-5)^2+(-2)^2} \\\\= \sqrt{25+4} \\\\= \sqrt{29} \ units$

$\bullet \sf \ AO = \sqrt{(0-(-4)))^2+(0-10)^2}$

        $= \sf \sqrt{4^2+(-10)^2} \\\\= \sqrt{16+100} \\\\= \sqrt{116} \ units$

According to the Pythagorean theorem,

ΔABO is said to be right angled triangle if AB² = BO² + AO²

$\longrightarrow \sf AB^2 = \sqrt{145} \times \sqrt{145} \\\\\longrightarrow AB^2=145$

$\longrightarrow \sf BO^2+AO^2=( \sqrt{116} )^2+(\sqrt{29} )^2\\\\\longrightarrow BO^2+AO^2 = 116+29 \\\\\longrightarrow BO^2+AO^2 = 145$

∴ AB² = BO² + AO²

The given points form a right angled triangle.