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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.
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
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