Certainly! Let's find the equations of the three altitudes of triangle \(\( \triangle ABC \)\) with vertices \(\( A(-3, 0) \)\), \(\( B(0, 6) \)\), and \(\( C(4, 6) \)\).
1. Finding the slopes of the sides:
- Slope of \(\( BC \)\):
Points \(\( B(0, 6) \)\) and \(\( C(4, 6) \)\) have the same y-coordinate, so the line is horizontal.
\(\[
\text{slope of } BC = 0
\]\)
- Slope of \(\( AC \)\):
Points \(\( A(-3, 0) \)\) and \(\( C(4, 6) \)\):
\(\[
\text{slope of } AC = \frac{6 - 0}{4 - (-3)} = \frac{6}{7}
\]\)
- Slope of \(\( AB \)\):
Points \(\( A(-3, 0) \)\) and \(\( B(0, 6) \)\):
\(\[
\text{slope of } AB = \frac{6 - 0}{0 - (-3)} = \frac{6}{3} = 2
\]\)
2. Finding the perpendicular slopes (slopes of the altitudes):
- The perpendicular slope to \(\( BC \)\) (horizontal line) is undefined (vertical line). This altitude must pass through point \(\( A(-3, 0) \)\), so the line is:
\(\[
x = -3
\]\)
- The perpendicular slope to \(\( AC \)\):
\(\[
\text{perpendicular slope to } AC = -\frac{1}{\frac{6}{7}} = -\frac{7}{6}
\]\)
Passing through point \(\( B(0, 6) \)\), the equation is:
\(\[
y - 6 = -\frac{7}{6}(x - 0) \implies y = -\frac{7}{6}x + 6
\]\)
- The perpendicular slope to \(\( AB \)\):
\(\[
\text{perpendicular slope to } AB = -\frac{1}{2}
\]\)
Passing through point \(\( C(4, 6) \)\), the equation is:
\(\[
y - 6 = -\frac{1}{2}(x - 4) \implies y = -\frac{1}{2}x + 2 + 6 \implies y = -\frac{1}{2}x + 8
\]\)
3. Final equations of the altitudes:
- Altitude from \(\( A \)\):
\(\[
x = -3
\]\)
- Altitude from \(\( B \)\):
\(\[
y = -\frac{7}{6}x + 6
\]\)
- Altitude from \(\( C \)\):
\(\[
y = -\frac{1}{2}x + 8
\]\)
These are the equations of the three altitudes of \(\( \triangle ABC \)\).
