Exercise 26 Page 417

Let's examine the given diagram.

Given that $p,$ $q,$ and $r$ are altitudes of $△ABC,$ we will show that $p,$ $q,$ and $r$ intersect at a point called the orthocenter of the triangle. To do so, we should first determine the slopes of the altitudes and then write their equations. Therefore, we will be able to determine their point of intersection.

Slope of line $p$

Looking at the diagram, we see that line $p$ is perpendicular to $\overline{BC},$ which means that their slopes are opposite reciprocals of each other. Since the slope of $\overline{BC}$ is $\frac{c}{\text{-}b},$ let's find the slope of line $p.$

Equation of line $p$

We have found that the slope of line $p$ is $\frac{b}{c}$ and looking at the diagram we see that it passes through $A(a,0).$ Therefore, its equation can be written in point-slope form.

Equation of line $q$

Let's examine line $q$ on the diagram.

Notice that line $q$ lies on the $y\text{-}$axis, which means that we can immediately determine its equation as $x=0.$

Point of intersection of lines $p$ and $q$

To find the point of intersection of lines $p$ and $q,$ we will solve the system formed by the equations of lines $p$ and $q.$ As we can see, the $x\text{-}$coordinate of the point is $0.$ To find its $y\text{-}$coordinate, we will use the Substitution Method. Therefore, the point of intersection is $\left(0,\text{-}\frac{ab}{c}\right).$

Slope of line $r$

Given that the slope of $\overline{AC}$ is $\frac{c}{\text{-}a}$ and $\overline{AC}\perp r,$ to find the slope of line $r,$ we will proceed in the same way as we did to find the slope of line $p.$

Equation of line $r$

Since the slope of line $r$ is $\frac{a}{c}$ and it passes through the point $B(b,0),$ we can write its equation in point-slope form.

Point of intersection of lines $r$ and $q$

Let's solve the system formed by the equations of lines $r$ and $q$ to find the point of intersection. Again, the $x\text{-}$coordinate of the point is $0,$ and to find its $y\text{-}$coordinate, we will use the Substitution Method. The coordinates of the point of intersection are $(0,\text{-}\frac{ab}{c}).$

Coordinates of the orthocenter of $△ABC$

Recall that the orthocenter of a triangle is the point where the altitudes of the triangle intersect. We have proven that the altitudes of $△ABC$ intersect the point $(0,\text{-}\frac{ab}{c}),$ so the orthocenter of $△ABC$ is $(0,\text{-}\frac{ab}{c}).$