I want to find the equation of the straight line EB.
My attempt:
$$D\equiv\left(\frac{2+6}{2}, \frac{1+3}{2}\right)$$
$$\equiv(4,2)$$
Now, the slope of AD,
$$\frac{2-5}{4-3}$$
$$=-3$$
Now, let $k$ is a constant. The slope of AD and EB are the same as they are parallel. Now, the equation of EB,
$$y=-3x+k\tag{1}$$
Now, $B(2,1)$ is a point of the straight line EB. So, inputting $x=2$ & $y=1$ in $(1)$, we get
$$1=-6+k$$
$$k=7$$
Now, inputting the value of $k$ in $(1)$,
$$y=-3x+7\tag{i}$$
This is the equation of EB.
My book's attempt:
Slope of BC,
$$\frac{3-1}{6-2}$$
$$=\frac{2}{4}$$
$$=\frac{1}{2}$$
Now, let $k$ is a constant. The equation of EB is
$$y=-2x+k\tag{2}$$
Now, $B(2,1)$ is a point of the straight line EB. So, inputting $x=2$ & $y=1$ in $(1)$, we get,
$$1=-4+k$$
$$k=5$$
Inputting the value of $k$ in $(2)$,
$$y=-2x+5\tag{ii}$$
Comments:
$(i)$ & $(ii)$ don't match. I think this is because the figure isn't drawn to scale and AD isn't parallel to EB. Am I correct?
AB and AC do not have the same length hence AD is not the height. You are right.