Part (i)
We can use the slope formula on the points A(0,4) and C(8,0)
[tex]m = \frac{y_2-y_1}{x_2-x_1}\\\\
m = \frac{0-4}{8-0}\\\\
m = \frac{-4}{8}\\\\
m = -\frac{1}{2}[/tex]
The slope of line AC is -1/2.
Line BM is perpendicular to AC. This indicates we'll apply the negative reciprocal to -1/2 to end up with 2.
The slope of line BM is 2.
Note that the perpendicular slopes multiply to -1. This is true of any perpendicular pair of slopes as long as neither line is vertical nor horizontal.
Answers:
- Slope of line AC = -1/2
- Slope of line BM = 2
=============================================================
Part (ii)
The slope of [tex]\text{line }\text{B}\text{M}
[/tex] was m = 2 from the previous part above.
We'll use the coordinates of point B(9,7) along with this slope value to find the equation of [tex]\text{line }\text{B}\text{M}[/tex].
[tex]y - y_1 = m(x - x_1)\\\\
y - 7 = 2(x - 9)\\\\
y - 7 = 2x - 18\\\\
y = 2x - 18+7\\\\
y = 2x - 11\\\\[/tex]
Answer: y = 2x-11
=============================================================
Part (iii)
The equations for lines [tex]\text{A}\text{L}[/tex] and [tex]\text{B}\text{M}[/tex] are [tex]7y+x = 28[/tex] and [tex]y = 2x-11[/tex] in that order. Solving this system of equations will produce the location of point P, where the lines intersect.
Let's apply substitution to solve for x.
[tex]7y+x = 28\\\\
7( y ) + x = 28\\\\
7( 2x-11 ) + x = 28 \ \text{ ... y replaced with 2x-11}\\\\
14x-77+x = 28\\\\
15x = 28+77\\\\
15x = 105 \\\\
x = 105/15\\\\
x = 7\\\\[/tex]
Then use this to find the y coordinate.
[tex]y = 2x-11\\\\
y = 2(7)-11\\\\
y = 14-11\\\\
y = 3[/tex]
The location of point P is (7,3). This is the orthocenter of the triangle where the altitudes intersect.
Answer: (7,3)
