answer the following question
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Answers
First draw trianle ABC, as BC as base. and AD is perpendicular to BC, likewise AE perpendicular to side BC and BF perpendicular to side AC
I can't draw here. So follow the instruction
Slope of side AB = 1 - ⁻2 = ⁻ 3
1 - 2
The altitude CD is perpendicular toside AB
slope of CD = 1 = 1
slope of AB ⁻ 3
The altitude CD passes through C ( -1, 0) using slope form of the equation of a line, so
equation of CD = Y - 0 = -1/3 (x - ⁻ 1)
= y - 0 = ⁻ x/3 - 1/3
= 3y = -x - 1
= x + 3y + 1 = 0
This is the required equation of the altitude from C to AB.
The slope of BC = 0 - 1 = 1
⁻ 1 - 1 2
The altitude AE is perpendicular to side BC, so
the slope AE = 1 = 1 = 2
Slope of BC 1/2
The altitude CD passes through A ( 2, -2) using slope form of the equation of a line, so
equation of CD = Y - ⁻ 2 = 2 (x - 2)
= y + 2 = 2x - 4
= 2x - y - 2 = 0
This is the required equation of the altitude from A to BC.
The slope of AC = 0 - 2 = 2
⁻ 1 - 2 3
The altitude BF is perpendicular to side AC, so
the slope BF = 1 = 1 = 3
Slope of aC 2/3 2
The altitude CD passes through B ( 1, 1) using slope form of the equation of a line, so
equation of CD = Y - 1 = 3/2 (x - 1)
= 2y -2 = 3x - 3
= 3x - 2y - 1 = 0
This is the required equation of the altitude from B to AC.