answer the following question
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.