3. Prove that the points (4,3), (7,-1) and (93)
are the veretices of an isosceles triangle.
Answers
Answer:
now,the points would be the vertices of a right angled triangle.. if ..any one side is equal to the root of sum of other two sides...
\underline{\large\mathcal\red{!! solution!!}
\begin{gathered}ab = \sqrt{(7 - 3) {}^{2} + (9 + 7) {}^{2} } = \sqrt{4 {}^{2} + 16 {}^{2} } = 4 \sqrt{17} \\ \\ bc = \sqrt{(3 + 3) {}^{2} + ( - 7 - 3) {}^{2} } = \sqrt{6 {}^{2} + 10 {}^{2}} = 2 \sqrt{34} \\ \\ ca = \sqrt{( - 3 - 7) {}^{2} + (3 - 9) {}^{2} } = \sqrt{10 {}^{2} + 6 {}^{2} } = 2 \sqrt{34} \\ \\ \end{gathered}
ab=
(7−3)
2
+(9+7)
2
=
4
2
+16
2
=4
17
bc=
(3+3)
2
+(−7−3)
2
=
6
2
+10
2
=2
34
ca=
(−3−7)
2
+(3−9)
2
=
10
2
+6
2
=2
34
Now ,from Pythagorean theorem....for a right angled triangle...
\begin{gathered}bc {}^{2} + ca {}^{2} = (2 \sqrt{34} ) {}^{2} + (2 \sqrt{34} ) {}^{2} = 272 \\ = (4 \sqrt{17} ) {}^{2} = ab {}^{2} \end{gathered}
bc
2
+ca
2
=(2
34
)
2
+(2
34
)
2
=272
=(4
17
)
2
=ab
2