Question

The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a  right traingle ,right-angled at B. Find the values of p.

Answer 1

ΔABC is right triangle at B.

AC²+ AB²= BC² ....(1)

Also, A=( 4,7) , B=( p,3) and C=( 7,3)

Now, AC2=97-4)2+(3-7_2=(3)2+(-4)2=9+16=25

AB2=(p-4)2+(3-7)2=p2-8p+16+(-4)2

=p2-8p+16+16

=p2-8p+32

ΔABC is right triangle at B.

AC²+ AB²= BC² ....(1)

Also, A=( 4,7) , B=( p,3) and C=( 7,3)

Now, AC²=(7-4)²+(3-7)²=(3)²+(-4)²=9+16=25

AB²=(p-4)²+(3-7)²=p²-8p+16+(-4)²

=p²-8p+16+16

=p²-8p+32

BC²=(7-p)²+(3-3)²=49-14p+p2+0

=p²-14p+49

bc²=(7-p)2+(3-3)2=49-14p+p2+0

=p²-14p+49

From (1), we have

25= (p²- 8p +32)+( p²-14p+49)

25=2p²-22p+81

2p²-22p+56=0

p²-11p+28=0

p²-7p-4p+28=0

(p-4)(p-7)=0

p=4,p=7