Question

The equations of the sides ab bc and ca are 2x+y=0

Answer 1

Step-by-step explanation:

We have:AB:2x+y=0 ;(i)BC:x+py=q ;(ii)AC:x−y=3 :(iii)(i)+(iii), we get,2x+y+x−y=0+3⇒3x=3⇒x=1Put x=1 in (iii), we get,1−y=3⇒y=−2⇒A(x1,y1)=(1,−2)(ii)−(iii), we get,x+py−x+y=q−3⇒y(p+1)=q−3⇒y=q−3p+1(ii)+p×(iii)x+py+px−py=q+3p⇒x(p+1)=q+3p⇒x=3p+qp+1⇒C(x2,y2)=(3p+qp+1,q−3p+1)2(ii)−(i)2x+2py−2x−y=2q⇒y(2p−1)=2q⇒y=2q2p−1p×(i)−(ii)2px+py−x−py=0−q⇒x(2p−1)=−q⇒x=−q2p−1⇒B(x3,y3)=(−q2p−1,2q2p−1)Case(i) When P(2,3) is centroid.A(x1,y1)=(1,−2) C(x2,y2)=(3p+qp+1,q−3p+1) B(x3,y3)=(−q2p−1,2q2p−1)⇒(x1+x2+x33,y1+y2+y33)=(2,3)⇒x1+x2+x33=3⇒x1+x2+x3=6⇒1+3p+qp+1−q2p−1=6⇒3p+qp+1−q2p−1=5⇒(3p+q)(2p−1)−q(p+1)=5(p+1)(2p−1)⇒6p2−3p+2pq−q−pq−q=10p2−5p+10p−5⇒4p2−3p−2q+pq=10p2+5p−5⇒6p2+8p+2q−pq=5 ;(iv)andy1+y2+y33=3⇒y1+y2+y3=9⇒−2+q−3p+1+2q2p−1=9⇒q−3p+1+2q2p−1=11⇒(q−3)(2p−1)+2q(p+1)=11(p+1)(2p−1)⇒3pq−q−6p+3+2pq+2q=22p2−11p+22p−11⇒5pq−6p+q+3=22p2+11p−11⇒22p2+17p−q−5pq=11 ;(v)As (iv) and (v) contains terms like p2, pq, p and q hence it is irreducible top+q. So we cannot proceed further.

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