# If the coordinates of A= (0,0), B= (3,-3) and C= (-2,-8), help me to answer below questions?

## a) Prove that ABC is a right angle triangle b)Find the value of BAC and the value of BCA.Then, prove that BAC + BCA = 90 degree Thank you! :)

Mar 16, 2017

Refer to the Explanation Section.

#### Explanation:

Part (a) :-

Slope of $A B = \frac{0 - \left(- 3\right)}{0 - 3} = \frac{3}{-} 3 = - 1 = {m}_{1} , s a y .$

Slope of $B C = \frac{- 3 - \left(- 8\right)}{3 - \left(- 2\right)} = \frac{5}{5} = 1 = {m}_{2} , s a y .$

Slope of $C A = \frac{- 8 - 0}{- 2 - 0} = 4 = {m}_{3} , s a y .$

Since, ${m}_{1} \cdot {m}_{2} = - 1 \Rightarrow A B \bot B C \ldots \ldots \ldots \ldots \ldots \left(1\right)$

Also, as ${m}_{1} \ne {m}_{2} \ne {m}_{3} \Rightarrow A , B , C \text{ are non-collinear.} \ldots \left(2\right)$

From $\left(1\right) \mathmr{and} \left(2\right) ,$ we conclude that the points $A , B , C$ form a

right-angled $\Delta ,$ having right-angle at the vertex $B .$

This proves $\left(a\right) .$

Part (b):-

To find the measure of $\angle B A C ,$ let us note that, this angle is the

angle btwn. lines $B A \mathmr{and} A C ,$ having slopes, ${m}_{1} \mathmr{and} {m}_{3} ,$

resp.

$\therefore \tan \angle B A C = | \frac{{m}_{1} - {m}_{3}}{1 + {m}_{1} {m}_{3}} | = | \frac{- 1 - 4}{1 + \left(- 1\right) \left(4\right)} |$

$\Rightarrow \tan \angle B A C = \frac{5}{3.} \ldots \ldots \ldots . \left(3\right) .$

Similarly, $\tan \angle B C A = | \frac{1 - 4}{1 + \left(1\right) \left(4\right)} = \frac{3}{5.} \ldots \ldots \ldots \ldots \ldots \ldots . . \left(4\right) .$

$\therefore \left(\tan \angle B A C\right) \left(\tan \angle B C A\right) = \frac{5}{3} \cdot \frac{3}{5} = 1.$

$\therefore \tan \angle B A C = \frac{1}{\tan \angle B C A} = \cot \angle B C A = \tan \left({90}^{\circ} - \angle B C A\right) .$

$\Rightarrow \angle B A C + \angle B C A = {90}^{\circ} .$

This proves $\left(b\right) .$

Enjoy Maths.!

Mar 16, 2017

#### Explanation:

In the triangle we have $A B = \sqrt{{\left(3 - 0\right)}^{2} + {\left(- 3 - 0\right)}^{2}} = \sqrt{9 + 9} = \sqrt{18}$

$A C = \sqrt{{\left(- 2 - 0\right)}^{2} + {\left(- 8 - 0\right)}^{2}} = \sqrt{4 + 64} = \sqrt{68}$

$B C = \sqrt{{\left(- 2 - 3\right)}^{2} + {\left(- 8 - \left(- 3\right)\right)}^{2}} = \sqrt{25 + 25} = \sqrt{50}$

As $A {C}^{2} = A {B}^{2} + B {C}^{2}$,

$\Delta A B C$ is right angled at $\angle A B C$

and hence $\angle B A C + \angle B C A = {90}^{\circ}$