A) Right angled
B) Equilateral
C) Isosceles
D) Both and
Correct Answer: D
Solution :
(d): Let\[A=\left( 7,9 \right),B=\left( 3,-7 \right)and\,C=\left( -3,3 \right)\] Then \[AB=\sqrt{{{(3-7)}^{2}}+{{(-7-9)}^{2}}}=\sqrt{272}\] \[BC=\sqrt{{{(-3-3)}^{2}}+{{(3+7)}^{2}}}=\sqrt{136}\] \[AC=\sqrt{{{(7+3)}^{2}}+{{(9-3)}^{2}}}=\sqrt{136}\] \[\therefore \]\[A{{B}^{2}}=\text{ }B{{C}^{2}}+\text{ }A{{C}^{2}}\] Hence by Pythagoras theorem, triangle ABC is a right angled triangle. Also, Since \[BC=CA\], hence ABC is an isosceles triangle. Thus ABC is a right angled isosceles triangle.
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