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Question
If A(4,3),B(6,−2) and C(a,−3) are the vertices of a triangle right angled at A, find a .
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Solution
△ABCisarightangledtrianglehenceAB2+AC2=BC2(4−6)2+[3−(−2)]2+(4−a)2+[3−(−3)]2=(6−a)2+[−2−(−3)]2=4+25+(4−a)2+36=(6−a)2+1(6−a)2−(4−a)2=64(6−a−4+a)(6−a+4−a)=642(10−2a)=6410−2a=322a=−22a=−11
△ABCisarightangledtrianglehenceAB2+AC2=BC2(4−6)2+[3−(−2)]2+(4−a)2+[3−(−3)]2=(6−a)2+[−2−(−3)]2=4+25+(4−a)2+36=(6−a)2+1(6−a)2−(4−a)2=64(6−a−4+a)(6−a+4−a)=642(10−2a)=6410−2a=322a=−22a=−11
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