# What is the orthocenter of the triangle formed by the intersection of the lines x = 4 , y = 1/2x + 7 , and y = -x + 1?

Mar 22, 2016

The Orthocenter, O is$\implies O \left(0 , 5\right)$ #### Explanation:

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side. So we need to find lines that are perpendicular to the give equation and where at least 2 of them intersect:

A perpendicular line to any line, $y = m x + b$ is give by:
y_(per) = m_(per)x+ b_(per); m_(per)= -1/m =====> (1)
First we need to get the coordinates of the vertices,
Let's solve equations:
Given ${y}_{1} = \frac{1}{2} x + 7$ and $x = 4$ simultaneously to get $y = 11$
${y}_{2} = - x + 1$ and $x = 4$ simultaneously to get $y = 3$
y=1/2x+7, y=-x+1; 1/2x+7=-x6+1; x=-4; y = 5
Thus the vertices of the triangle are given by:
$A \left(- 4 , 5\right) , B \left(4 , 11\right)$ and $C \left(4 , - 3\right)$

Now let find the perpendicular line to ${y}_{1}$ that passes through point $C \left(4 , - 3\right)$ from (1) ${m}_{p e r} = - \frac{1}{\frac{1}{2}} = - 2$
${y}_{p e {r}_{1}} = - 2 x + {b}_{p e r}$ let x=4; y = -3 " and solve for " b_(per)
-3=-2(4)+ b_(per); b_(per)=5
So what we have is ${y}_{p e {r}_{1}} = - 2 x + 5$
and the line perpendicular to x=4 passing through A(-4,5) is $y = 5$

Thus the Orthocenter is ${y}_{p e {r}_{1}} = - 2 x + 5 = 5$ Solve for $x$
$x = 0$ $\therefore \text{the Orthocenter,} \implies O \left(0 , 5\right)$