Question

what are the coordinates of the orthocenter of △ABC with vertices at A(1, 2), B(1, 6), and C(5, 6)?

Answer 1

(1,6)
This should help for future problems
https://www.mathportal.org/calculators/analytic-geometry/triangle-calculator.php

Answer 2

Answer:

(1,6)

Step-by-step explanation:

We are given that the vertices of triangle ABC are  A at (1,2),B at (1,6) and C at (5,6).

We have to find the coordinates of the orthocenter.

Distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula

$AB=(1-1)^2+(6-2)^2=16$

$BC=(5-1)^2+(6-6)^2=16$

$AC^2=(1-5)^2+(2-6)^2=32$

$AC^2=AB^2+BC^2$

Hence, the triangle is a right triangle because it satisfied Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2$

The orthocenter is the intersection of three  altitudes of triangle .The orthocenter of right triangle is the vertex of triangle .

The vertex of triangle is at B.

Therefore, the ortho-center of triangle ABC is B(1,6).