Answer: 5a. (15, -3) 5b. 5x - 7y - 26 = 0 5c. (15, 7)
Step-by-step explanation:
5a) D is the Midpoint of A and C.
[tex]D_x=\dfrac{A_x+C_x}{2}\qquad \qquad \qquad D_y=\dfrac{A_y+C_y}{2}\\\\\\8=\dfrac{1+p}{2}\qquad \qquad \qquad \qquad \quad 2=\dfrac{7+q}{2}\\[/tex]
16 = 1 + p 4 = 7 + q
15 = p -3 = q
D = (15, -3)
********************************************************************************************
5b) Find the perpendicular slope and use the Point-Slope formula.
[tex]m=\dfrac{y_A-y_D}{x_A-x_D}\quad =\dfrac{1-8}{7-2}\quad =\dfrac{-7}{5}\quad \rightarrow \quad m_{\perp}=\dfrac{5}{7}[/tex]
[tex]y-y_D=m(x-x_D)\\\\y-2=\dfrac{5}{7}(x-8)\\\\\\7(y-2)=5(x-8)\\\\7y-14=5x-40\\\\0=5x-7y-40+14\\\\\bold{5x-7y-26=0}[/tex]
**********************************************************************************************
5c) line AE: y = 7
line DE: 5x - 7y - 26 = 0
Use Substitution method to solve the system for x, given that y = 7:
5x - 7(7) - 26 = 0
5x - 49 - 26 = 0
5x - 75 = 0
5x = 75
x = 15 y = 7
E = (15, 7)
