Question

If the points A(1, 2), B(k+1, 2) and C(6, 7) are the vertices of a triangle having B=90degree, find the value of k.

Answer 1

$\underline{\large{\pink{\bf{QuEstion}}}}$

If the points A(1,2) , B(K+1 , 2) and C( 6 , 7) are the vertices of a triangle having ∠ B = 90° , find the value of 'k'.

$\underline {\large{\pink{\bf{SoluTion}}}}$

Finding distances AB , BC , AC

using distance formula

$\tt distance = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\\\\\tt ( where (x_1,y_1) and (x_2,y_2)\\\tt are\:coordinates\: of\: two\: distinct\: points )$

$AB = \sqrt{(k+1-1)^2 + (2-2)^2} \\AB = \sqrt{k^2} = k$

$BC = \sqrt{(k+1-6)^2+(7-2)^2} \\\\BC = \sqrt{k^2+25-10k+25} \\\\BC = \sqrt{k^2-10k+50} \:$

$AC = \sqrt{(1-6)^2+(2-7)^2} \\\\AC = \sqrt{50}$

By pythagorean theorem in Δ ABC

AC² = AB² + BC²

$(\sqrt{50} )^2 = (k)^2 + (\sqrt{k^2-10k+50} )^2\\\\50 = k^2 + k^2 - 10 k + 50\\\\2 k^2 - 10 k = 0 \\\\k ( 2 k - 10 ) = 0\\\\2k= 10\\\\\boxed{k = 5}$

HENCE,

the value of k is 5.