Exercise 3 Page 387

Let's label the vertices and count squares to find the coordinates of the vertices. The orthocenter is the point of concurrency of the altitudes. Since the line through $A$ and $B$ is horizontal, the corresponding altitude must be vertical.

The orthocenter is on this vertical line, so the first coordinate must be $1 .$ Thus, answer choices A $(- \frac{3}{4}, - 1)$ and B $(- \frac{4}{3}, 1)$ are certainly not correct. To choose between C $(1, \frac{5}{2})$ and D $(1, \frac{9}{4}),$ let's consider another altitude.

If $M$ is the orthocenter, then $\overline{A C}$ is perpendicular to $\overline{B M} .$ We know that the slopes of perpendicular lines multiply to $- 1,$ so let's calculate the slope of $\overline{A C}$ and $\overline{B M}$ for the two possible $M ’ s$ in answer choices C and D.

Points	Expression of the Slope	Slope
$A (- 2, 4)$ and $C (1, - 2)$	$m = \frac{- 2 - 4}{1 - ( - 2 )}$	$m = - 2$
$M_{C} (1, \frac{5}{2})$ and $B (4, 4)$	$m_{C} = \frac{4 - \frac{5}{2}}{4 - 1}$	$m_{C} = \frac{1}{2}$
$M_{D} (1, \frac{9}{4})$ and $B (4, 4)$	$m_{D} = \frac{4 - \frac{9}{4}}{4 - 1}$	$m_{D} = \frac{7}{1 2}$

We can see, that

m m_{D} = - \frac{1 4}{1 2} \neq = - 1,

so

(1, \frac{9}{4})

is not the orthocenter of the triangle. This eliminates answer choice D, so the correct answer is the remaining answer C,

(1, \frac{5}{2}) .

Alternative Solution

Alternative way of thinking

In the solution above we eliminated choices until only one left. Another way of finding the answer is to calculate the equation of two altitudes and work out the intersection point. This method is used for exactly this triangle in Example 4 of the book on page 338.