Question

A triangle has corners A, B, and C located at `(4 ,5 )`, `(3 ,6 )`, and `(2 ,9 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

The two end points of altitude are `(2,9) **and** (1,8)`
Length of altitude is `sqrt2`

Explanation:

CD is the altitude to be drawn from corner on line AB. Lines CD and AB are perpendicular to each other . Slope of the line AB is `m_1=(y_2-y_1)/(x_2-x_1) = (6-5)/(3-4) = -1`. Slope of the line CD is `m_2= (-1)/m_1=1`.Since product of slopes of perpendicular lines is `-1`.

Equation of line AB is `y-y_1=m_1(x-x_1) or y-5 = -1(x-4) or y+x=9 (1)`
Equation of line CD is `y-y_1=m_2(x-x_1) or y-9 = 1(x-2) or y-x=7 (2)`

Solving equation (1) and equation (2) we get the intersecting point (x,y) of line AB and line CD. Adding equation (1) & (2) we get `2y=16 or y=8 and x= 9-8=1 :. (x,y)=(1,8)`.

The two end points of altitude CD are `(2,9) and (1,8)`

Length of altitude CD is `d=sqrt((2-1)^2+(9-8)^2)=sqrt2` [Ans]