Question

The vertices of a Triangle ABC , right angled at B , are A(-2,2) B(2,-2) and C(k,2) find the value of k .

Answer 1

Question -

The vertices of a triangle ABC, right angled at B, are A(-2,2) B(2,-2) and C(k, 2) find the value of k

Solution -

We know that,

$distance \: formula = \sqrt{( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{1} ) {}^{2} }$

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

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

Squaring both sides,

${AB }^{2} = {(4)}^{2} + {( - 4)}^{2}$

${AB }^{2} = 16 + 16$

${AB }^{2} = 32$

And,

$BC = \sqrt{( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{2} ) {}^{2} }$

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

Squaring both sides,

${BC }^{2} = {(k - 2)}^{2} + 16$

And,

$AC = \sqrt{( x_{2} - x_{1}) {}^{2} + ( y_{2} - y_{1} {}^{2}) }$

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

Squaring both sides,

$AC {}^{2} = {(k + 2)}^{2}$

Now,

By applying Pythagoras theorem in right triangle ABC we will get -

h²= b²+p²

AC² = BC² + AB²

By putting the values we will get -

${(k + 2)}^{2} = {(k - 2)}^{2} + 16 + 32$

${k}^{2} + 4 + 4k = {k}^{2} + 4 - 4k + 48$

$4k + 4k = 48$

$8k = 48$

$k = \frac{48}{8}$

$k = 6$

Therefore the value of k is 6