Question

if B(2,3) and C(5,1) are the vertices ofΔABC such that AB=5 and AC=7then orthocenter of ΔABC lies on​

Answer 1

Given:

B(2,3) and C(5,1) are the vertices ofΔABC

AB=5

AC=7

To find:

Where the orthocenter of the triangle lies.

Solution:

1) The length of the side BC is given by

• $\sqrt{(5-2)^{2}+(1-3)^{2} }$  (By the distance formula)

• $\sqrt{3^{2}+2^{2} }$

• $\sqrt{13}$

2) So the sides of the given triangle are 5, 7, and √13. The triangle of these sides would be an obtuse triangle.

• $7^2> 5^2+\sqrt{13}^2$

• 49 > 25+13

• 49 > 38

3) The orthocenter in an obtuse triangle lies outside the triangle.

The orthocenter of the triangle ABC lies Outside the triangle.