Question

Triangle ABC is defined by the points A(3,8), B(7,5), and C(2,3).

Create an equation for a line passing through point A and perpendicular to line segment BC.

Answer 1

To find the equation of a line passing through point A(3, 8) and perpendicular to line segment BC, we need to follow these steps:

1. **Find the slope of line segment BC:**

The coordinates of points B and C are B(7, 5) and C(2, 3), respectively. The slope of a line segment between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} \]

So for BC:
\[ m_{BC} = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = \frac{2}{5} \]

2. **Determine the slope of the perpendicular line:**

The slope of the line perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope \(m\) of the line perpendicular to BC is:
\[ m = -\frac{1}{m_{BC}} = -\frac{1}{\frac{2}{5}} = -\frac{5}{2} \]

3. **Use the point-slope form of the line equation:**

The point-slope form of the equation of a line passing through a point \((x_1, y_1)\) with slope \(m\) is:
\[ y - y_1 = m(x - x_1) \]

Plugging in point A(3, 8) and the slope \(m = -\frac{5}{2}\):
\[ y - 8 = -\frac{5}{2}(x - 3) \]

4. **Simplify the equation:**

Distribute the slope on the right-hand side:
\[ y - 8 = -\frac{5}{2}x + \frac{15}{2} \]

Add 8 to both sides to isolate \(y\):
\[ y = -\frac{5}{2}x + \frac{15}{2} + 8 \]

Convert 8 to a fraction with a common denominator:
\[ 8 = \frac{16}{1} = \frac{16 \cdot 2}{2} = \frac{16}{2} \]

\[ y = -\frac{5}{2}x + \frac{15}{2} + \frac{16}{2} \]

Combine the fractions:
\[ y = -\frac{5}{2}x + \frac{31}{2} \]

Therefore, the equation of the line passing through point A and perpendicular to line segment BC in slope-intercept form is:

\[ y = -\frac{5}{2}x + \frac{31}{2} \]