#### 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.

#### Solution

ΔABC is right triangle at B.

AC^{2}+ AB^{2}= BC^{2} ....(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.

AC^{2}+ AB^{2}= BC^{2} ....(1)

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

Now, AC^{2}=(7-4)^{2}+(3-7)^{2}=(3)^{2}+(-4)^{2}=9+16=25

AB^{2}=(p-4)^{2}+(3-7)^{2}=p^{2}-8p+16+(-4)^{2}

=p^{2}-8p+16+16

=p^{2}-8p+32

BC^{2}=(7-p)^{2}+(3-3)^{2}=49-14p+p2+0

=p^{2}-14p+49

bc^{2}=(7-p)2+(3-3)2=49-14p+p2+0

=p^{2}-14p+49

From (1), we have

25= (p^{2}- 8p +32)+( p^{2}-14p+49)

25=2p^{2}-22p+81

2p^{2}-22p+56=0

p^{2}-11p+28=0

p^{2}-7p-4p+28=0

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

p=4,p=7