Answer:
A. (5,-8)
Step-by-step explanation:
Assume your list of points is
A. (5, -8); B. (4, -8); C. (-5, -8); D. (5, 8); E. (4, 5); F. (5, 0)
One way to solve this problem is to insert the values into the expression to see what works
A. (5, -8)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\-8 + 8 & = & 4(5 - 5)\\0 & = & 4(0)\\0 & = & 0\\\end{array}\\\textbf{TRUE}[/tex]
B. (4, -8)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\-8 + 8 & = & 4(4 - 5)\\0 &=&4(-1)\\0 & = & -4\\\end{array}\\\textbf{False}[/tex]
C. (-5, -8)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\-8 + 8 & = & 4(-5 - 5)\\0& = & 4(-10)\\0 & = & -40\\\end{array}\\\textbf{False}[/tex]
D. (5, 8)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\8 + 8 & = & 4(5 - 5)\\16 & = & 4(0)\\16 & = & 0\\\end{array}\\\textbf{False}[/tex]
E. (4, 5)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\5 + 8 & = & 4(4 - 5)\\13 & = & 4(-1)\\13 & = & -4\\\end{array}\\\textbf{False}[/tex]
F. (5, 0)
[tex]\begin{array}{rcl}y + 8 & = & 4(x - 5)\\0 + 8 & = & 4(5 - 5)\\8 & = & 4(0)\\8 & = & 0\\\end{array}\\\textbf{False}[/tex]
Only Point A satisfies the equation.
The graph below shows that only Point A is on the line.
