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