Finding the Altitude Slope in Triangle ABC

The problem asks for the slope of the altitude AD in triangle ABC, given the coordinates of points A(1, 5), B(2, 1), and C(6, 3).

An altitude from a vertex to the opposite side is perpendicular to that side. Therefore, the altitude AD is perpendicular to the side BC.

Calculating the Slope of Side BC

The slope ($m$) of a line segment between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Using the coordinates of B(2, 1) and C(6, 3):

• $x_1 = 2$, $y_1 = 1$
• $x_2 = 6$, $y_2 = 3$

The slope of BC ($m_{BC}$) is:

$m_{BC} = \frac{3 - 1}{6 - 2} = \frac{2}{4} = \frac{1}{2}$

Determining the Slope of Altitude AD

Since the altitude AD is perpendicular to the side BC, the product of their slopes must be -1.

Let $m_{AD}$ be the slope of the altitude AD.

The condition for perpendicular lines is:

$m_{AD} \times m_{BC} = -1$

Substitute the calculated slope of BC:

$m_{AD} \times \frac{1}{2} = -1$

Solve for $m_{AD}$:

$m_{AD} = -1 \times 2$

$m_{AD} = -2$

The slope of the altitude AD is -2.