Answer
Point-slope form $ y+4=-\frac{1}{2}(x−6)$.
Slope-intercept form $y=-\frac{1}{2}x-1$.
Work Step by Step
If the line passes through a point $(x_1,y_1)$ and slope is m then point-slope form of the perpendicular line is.
$\Rightarrow y−y_1=m(x−x_1)$
From the question we have
$\Rightarrow (x_1,y_1)=(6,-4)$
Equation of the parallel line.
$\Rightarrow x+2y=5$
Isolate $y$.
$\Rightarrow y=-\frac{1}{2}x+\frac{1}{2}5$
It is in the form of slope-intercept form $y=mx+c$.
The slope of the equation is $m=-\frac{1}{2}$
Two parallel lines have same slopes.
The slope of the required line is
$\Rightarrow m=−\frac{1}{2}$
Substitute all values into the point-slope equation.
$\Rightarrow y-(-4)=(-\frac{1}{2})(x−(6))$
Simplify.
$\Rightarrow y+4=-\frac{1}{2}(x−6)$
The above equation is the point-slope form.
Now isolate y
$\Rightarrow y+4=-\frac{1}{2}(x−6)$
Use distributive property.
$\Rightarrow y+4=-\frac{1}{2}x+\frac{1}{2}\cdot6$
Subtract 4 from both sides.
$\Rightarrow y+4-4=-\frac{1}{2}x+3-4$
Simplify.
$\Rightarrow y=-\frac{1}{2}x-1$
The above equation is the slope-intercept form.