Question

The vertices of triangle abc, right angled at b are a(-2,1), b(2,-2), c(k,2). find the value of k

Answer 1

sorry for the scribles

Answer 2

Answer:

$Value \: of \: k = 5$

Step-by-step explanation:

Given A(-2,1),B(2,-2) and C(k,2)

are vertices of right triangle ∆ABC .

$Distance \: between \\joining \: two \: points \\(x_{1},y_{1})\: and \: (x_{2},y_{2})\\= \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$i) AC^{2}\\=\left(k+2\right)^{2}+\left(2-1\right)^{2}\\=k^{2}+4k+4+1\\=k^{2}+4k+5\: ---(1)$

$ii) AB^{2}\\=\left(2+2\right)^{2}+\left(-2-1\right)^{2}\\=4^{2}+3^{2}\\=16+9\\=25 ---(2)$

$iii) BC^{2}\\=\left(k-2\right)^{2}+\left(2+2\right)^{2}\\=k^{2}-4k+4+16\\=k^{2}-4k+20\: ---(3)$

/* According to the problem given,

$AC^{2}=AB^{2}+BC^{2}\\(By \: Phythagorean \:theorem)$

$k^{2}+4k+5=25+k^{2}-4k+20$

$\implies k^{2}+4k+5=45+k^{2}-4k$

$\implies k^{2}+4k-k^{2}+4k=45-5$

$\implies 8k = 40$

$\implies k =\frac{40}{8}$

$\implies k = 5$

Therefore,

$Value \: of \: k = 5$

•••♪