Answer :
Given:
The equation of parallel line is [tex]y=-\dfrac{1}{4}x+5[/tex].
Required line passes through (2,-3).
To find:
The equation of line.
Solution:
We have,
[tex]y=-\dfrac{1}{4}x+5[/tex]
On comparing this equation with [tex]y=mx+b[/tex], where m is slope, we get
[tex]m=-\dfrac{1}{4}[/tex]
Slope of two parallel lines are always same. So, slope of required line is [tex]m=-\dfrac{1}{4}[/tex].
The required line passes through the point (2,-3) having slope [tex]m=-\dfrac{1}{4}[/tex], so the equation of line is
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-(-3)=-\dfrac{1}{4}(x-2)[/tex]
[tex]y+3=-\dfrac{1}{4}(x)-\dfrac{1}{4}(-2)[/tex]
[tex]y+3=-\dfrac{1}{4}(x)+\dfrac{1}{2}[/tex]
Subtracting 3 from both sides, we get
[tex]y=-\dfrac{1}{4}(x)+\dfrac{1}{2}-3[/tex]
[tex]y=-\dfrac{1}{4}(x)+\dfrac{1-6}{2}[/tex]
[tex]y=-\dfrac{1}{4}(x)-\dfrac{5}{2}[/tex]
Therefore, the equation of required line is [tex]y=-\dfrac{1}{4}(x)-\dfrac{5}{2}[/tex].