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! :)

2 Answers
Mar 16, 2017

Answer:

Refer to the Explanation Section.

Explanation:

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.!

Mar 16, 2017

Answer:

Please see below.

Explanation:

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^@#