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

Part (a) :-

Slope of `AB={0-(-3)}/(0-3)=3/-3=-1=m_1, say.`

Slope of `BC={-3-(-8)}/{3-(-2)}=5/5=1=m_2, say.`

Slope of `CA={-8-0}/{-2-0}=4=m_3, say.`

Since, `m_1*m_2=-1 rArr AB bot BC...............(1)`

Also, as `m_1nem_2nem_3 rArr A, B, C" are non-collinear."...(2)`

From `(1) and (2),` we conclude that the points `A, B, C` form a

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

This proves `(a).`

Part (b):-

To find the measure of `/_BAC,` let us note that, this angle is the

angle btwn. lines `BA and AC,` having slopes, `m_1 and m_3,`

resp.

`:. tan/_BAC=|(m_1-m_3)/(1+m_1m_3)|=|(-1-4)/{1+(-1)(4)}|`

`rArr tan/_BAC=5/3...........(3).`

Similarly, `tan/_BCA=|(1-4)/{1+(1)(4)}=3/5.....................(4).`

`:. (tan/_BAC)(tan/_BCA)=5/3*3/5=1.`

`:. tan/_BAC=1/(tan/_BCA)=cot/_BCA=tan(90^@-/_BCA).`

`rArr /_BAC+/_BCA=90^@.`

This proves `(b).`

Enjoy Maths.!

In the triangle we have `AB=sqrt((3-0)^2+(-3-0)^2)=sqrt(9+9)=sqrt18`

`AC=sqrt((-2-0)^2+(-8-0)^2)=sqrt(4+64)=sqrt68`

`BC=sqrt((-2-3)^2+(-8-(-3))^2)=sqrt(25+25)=sqrt50`

As `AC^2=AB^2+BC^2`,

`DeltaABC` is right angled at `/_ABC`

and hence `/_BAC+/_BCA=90^@`