TS ICET 23 May 2019 Shift 2 Paper

Question : 139 of 200

Marks: +1, -0

The points A (5, 2), B (6, -15), C (0, 0) are vertices of the triangle which is

Solution:

A(5, 2), B(6,–15),C(0, 0)

A B=√{(6-5)^{2}+(-15-2)^{2}}=√{290}

B C=√{(6-0)^{2}+(0+15)^{2}}=√{261}

AC=√{5^{2}+2^{2}}=√{29}

Clearly

BC^2+AC^2=AB^2

∴ ABC is a right-angle triangle.

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