Answer:
[tex]4y-x = 15[/tex]
Step-by-step explanation:
We are given the following information in the question:
We have to find the equation of line passing through the point (5,5) anfd parallel to the line with slope [tex]\frac{1}{4}[/tex].
Since the line is parallel, they will have the same slope.
Thus, slope of line = [tex]\displaystyle\frac{1}{4}[/tex]
Point-slope form:
[tex](y-y_1) = m(x-x_1)[/tex],
where m is the slope of line and [tex](x_1,y_1)[/tex] is a point on the line.
Putting all the values, we have the equation of line:
[tex](y-y_1) = m(x-x_1)\\(y-5) = \displaystyle\frac{1}{4}(x -5)\\\\4(y-5) = (x-5)\\4y-20 = x -5\\4y-x = -5+20\\4y-x = 15[/tex]
The above equation is the required equation of the line.