Question:
Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.
(a) Find the equation of the line passing through B and perpendicular to the line AC
(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.
Answer:
[tex]y = \frac{1}{5}x + \frac{12}{5}[/tex]
[tex]D = (\frac{43}{26},\frac{71}{26})[/tex]
Step-by-step explanation:
Given
[tex]\triangle ABC[/tex]
[tex]A = (2,1)[/tex]
[tex]B = (3,3)[/tex]
[tex]C = (1,6)[/tex]
Solving (a): Line that passes through B, perpendicular to AC.
First, calculate the slope of AC
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
Where:
[tex]A = (2,1)[/tex] --- [tex](x_1,y_1)[/tex]
[tex]C = (1,6)[/tex] --- [tex](x_2,y_2)[/tex]
The slope is:
[tex]m = \frac{6- 1}{1 - 2}[/tex]
[tex]m = \frac{5}{-1}[/tex]
[tex]m = -5[/tex]
The slope of the line that passes through B is calculated as:
[tex]m_2 = -\frac{1}{m}[/tex] --- because it is perpendicular to AC.
So, we have:
[tex]m_2 = -\frac{1}{-5}[/tex]
[tex]m_2 = \frac{1}{5}[/tex]
The equation of the line is the calculated using:
[tex]m_2 = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
Where:
[tex]m_2 = \frac{1}{5}[/tex]
[tex]B = (3,3)[/tex] --- [tex](x_1,y_1)[/tex]
[tex](x_2,y_2) = (x,y)[/tex]
So, we have:
[tex]\frac{1}{5} = \frac{y - 3}{x - 3}[/tex]
Cross multiply
[tex]5(y-3) = 1(x - 3)[/tex]
[tex]5y - 15 = x - 3[/tex]
[tex]5y = x - 3 + 15[/tex]
[tex]5y = x +12[/tex]
Make y the subject
[tex]y = \frac{1}{5}x + \frac{12}{5}[/tex]
Solving (b): Point of intersection between AC and [tex]y = \frac{1}{5}x + \frac{12}{5}[/tex]
First, calculate the equation of AC using:
[tex]y = m(x - x_1) + y_1[/tex]
Where:
[tex]A = (2,1)[/tex] --- [tex](x_1,y_1)[/tex]
[tex]m = -5[/tex]
So:
[tex]y=-5(x - 2) + 1[/tex]
[tex]y=-5x + 10 + 1[/tex]
[tex]y=-5x + 11[/tex]
So, we have:
[tex]y=-5x + 11[/tex] and [tex]y = \frac{1}{5}x + \frac{12}{5}[/tex]
Equate both to solve for x
i.e.
[tex]y = y[/tex]
[tex]-5x + 11 = \frac{1}{5}x + \frac{12}{5}[/tex]
Collect like terms
[tex]-5x -\frac{1}{5}x = \frac{12}{5} - 11[/tex]
Multiply through by 5
[tex]-25x-x = 12 - 55[/tex]
Collect like terms
[tex]-26x = -43[/tex]
Solve for x
[tex]x = \frac{-43}{-26}[/tex]
[tex]x = \frac{43}{26}[/tex]
Substitute [tex]x = \frac{43}{26}[/tex] in [tex]y=-5x + 11[/tex]
[tex]y = -5 * \frac{43}{26} + 11[/tex]
[tex]y = \frac{-5 *43}{26} + 11[/tex]
Take LCM
[tex]y = \frac{-5 *43+11 * 26}{26}[/tex]
[tex]y = \frac{71}{26}[/tex]
Hence, the coordinates of D is:
[tex]D = (\frac{43}{26},\frac{71}{26})[/tex]
