Answer:
The answer is below
Step-by-step explanation:
a) The equation of a line passing through the point (0,0) and (16,4) is given by:
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-0=\frac{4-0}{16-4}(x-0)\\\\y=\frac{1}{4}x[/tex]
We have to find the equation of the line perpendicular to y = (1/4)x and passing through the point (4,6).
The line perpendicular to y = (1/4)x has a slope of -4 (product of their slope = -1).
Hence:
[tex]y- y_1=m(x-x_1)\\\\y-6=-4(x-4)\\\\y-6=-4x+16\\\\y=-4x+22[/tex]
The slope of a line passing through the point (16,4) and (4,6) is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{6-4}{4-16}=-\frac{1}{6}[/tex]
We have to find the equation of the line perpendicular to the line with slope of -1/6 and passing through the point (0,0).
The line perpendicular to a line with slope -1/6 has a slope of 6
Hence:
[tex]y- y_1=m(x-x_1)\\\\y-0=6(x-0)\\\\y=6x[/tex]
Solving y = 6x and y = -4x + 22, gives:
x = 2.2, y = 13.2
Hence the orthocenter is at (2.2, 13.2)
b)
The slope of a line passing through the point (3,4) and (11,12) is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{12-4}{11-3}=1[/tex]
We have to find the equation of the line perpendicular to the line with slope of 1 and passing through the point (8,15).
The line perpendicular to a line with slope 1 has a slope of -1
Hence:
[tex]y- y_1=m(x-x_1)\\\\y-15=-1(x-8)\\\\y=-x+8+15\\\\y=-x+23[/tex]
The slope of a line passing through the point (11,12) and (8,15) is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{15-12}{8-11}=-1[/tex]
We have to find the equation of the line perpendicular to the line with slope of -1 and passing through the point (3,4).
The line perpendicular to a line with slope -1 has a slope of 1
Hence:
[tex]y- y_1=m(x-x_1)\\\\y-4=1(x-3)\\\\y=x-3+4\\\\y=x+1[/tex]
Solving y = -x + 23 and y = x + 1, gives:
x = 11, y = 12
Hence the orthocenter is at (11, 12)
