Question

What is the orthocenter of a triangle with corners at `(4 ,3 )`, `(5 ,4 )`, and (2 ,8 )#?

Answer 1

`(40/7,30/7)` is the intersection point of altitudes and is the orthcenter of the triangle.

Explanation:

Orthocenter of a triangle is the point of intersection of all the altitudes of the triangle. Let A(4,3) , B(5,4) and C(2,8,) are the vertices of the triangle.
Let AD be the altitude drawn from A perpendiclar to BC and CE be the altitude drawn fromC on AB.
Slope of the line BC is `(8-4)/(2-5)= -4/3 :.`Slope of AD is `-1/(-4/3) = 3/4`The equation of altitude AD is `y-3= 3/4(x-4) or 4y-12=3x-12 or 4y-3x=0 (1)`
Now Slope of the line AB is `(4-3)/(5-4)=1 :.`Slope of CE is `-1/1 = -1`The equation of altitude CE is `y-8= -1(x-2) or y+x=10(2)`
Solving `4y-3x =0 (1)`and `y+x=10 (2)` we get `x = 40/7 ; y =30/7 :. (40/7,30/7)` is the intersection point of two altitudes and is the orthcenter of the triangle.[Ans]