y=4x-28
Explanationwhen two lines are perpendicular the product o their slopes equals, -1
[tex]\begin{gathered} if \\ line_1\perp\text{ line}_2 \\ then \\ slope_1*slope_2=-1 \end{gathered}[/tex]so
Step 1
find the slope of the line AC
a)the function of the line DB is given in the form
[tex]\begin{gathered} y=mx+b \\ where \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \end{gathered}[/tex]so
[tex]\frac{1}{2}x+2y=12[/tex]b)isolate y to get the slope-intercept form, and check the m value
[tex]\begin{gathered} \frac{1}{2}x+2y=12 \\ subtract\text{ }\frac{1}{2}x\text{ in both sides} \\ \frac{1}{2}x+2y-\frac{1}{2}x=12-\frac{1}{2}x \\ 2y=-\frac{1}{2}x+12 \\ divide\text{ both sides by 2} \\ \frac{2y}{2}=\frac{-\frac{1}{2}x+12}{2} \\ y=-\frac{1}{4}x+6\Rightarrow line\text{ 1} \\ so \\ slope_1=-\frac{1}{4} \end{gathered}[/tex]c) now, as the lines AC and DB are perpendicular replace to find slope 2
[tex]\begin{gathered} slope_1*slope_2=-1 \\ -\frac{1}{4}*slope_2=-1 \\ Multiply\text{ both sides by -4} \\ -\frac{1}{4}slope_2*-4=-1*-4 \\ slope_2=4 \end{gathered}[/tex]therefore
the slope of the line AC is 4
Step 2
now, we can use the slope-point formula to find the equaiton of the line,it says
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ where\text{ m is the slope} \\ (x_1,y_1)\text{ is a point from the line} \end{gathered}[/tex]so
a)let
[tex]\begin{gathered} slope=4 \\ point=(8,4)\text{ this is a common point for both lines} \end{gathered}[/tex]b) finally, replace and solve for y
[tex]\begin{gathered} y-y_{1}=m(x-x_{1}) \\ y-4=4(x-8) \\ y-4=4x-32 \\ add\text{ 4 in both sides} \\ y-4+4=4x-32+4 \\ y=4x-28 \end{gathered}[/tex]so, the equation of the line AC ia
y=4x-28
I hope this helps you
