Question

Orthocenter of the triangle whose vertices are (4,5) (4,7) (8,5) is

Answer 1

Solution!!

The vertices of a triangle are given in the question. We are asked to find the orthocenter.

All we have to do is prove that it is a right angled triangle. We'll use the distance formula and Pythagoras theorem to do so. If it's a right angled triangle then we can easily calculate the orthocenter.

A = (4,5)

B = (4,7)

C = (8,5)

Distance Formula = $\sf \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

AB = $\sf \sqrt{(4-4)^{2}+(7-5)^{2}}$

AB = $\sf \sqrt{(0)^{2}+(2)^{2}}$

AB = $\sf \sqrt{0+4}$

AB = $\sf \sqrt{4}$

AB = $\sf 2$

BC = $\sf \sqrt{(8-4)^{2}+(5-7)^{2}}$

BC = $\sf \sqrt{(4)^{2}+(-2)^{2}}$

BC = $\sf \sqrt{16+4}$

BC = $\sf \sqrt{20}$

AC = $\sf \sqrt{(8-4)^{2}+(5-5)^{2}}$

AC = $\sf \sqrt{(4)^{2}+(0)^{2}}$

AC = $\sf \sqrt{16}$

AC = $\sf 4$

Now, let's use the Pythagoras theorem and check if it is a triplet. If it is a triplet, then it is a right angled triangle.

(AB)² + (AC)² = (BC)²

(2)² + (4)² = (√20)²

4 + 16 = 20

20 = 20

Hence, the triangle is a right angled triangle.

Now, we know that the orthocenter is the point where all altitudes of a triangle intersect.

Construct a right angled triangle as shown in the attachment. Draw an altitude from point A to D on BC.

If we observe carefully, we will notice that AB, AD and AC are altitudes of the triangle and they all meet at point A. So, point A is the orthocenter of the triangle.

Orthocenter = (4,5)