Given:
The equation of line is
[tex]y=\dfrac{1}{3}x-1[/tex]
To find:
The equation of line in slope-intercept form parallel to given line and passes through (6,10).
Solution:
The slope intercept form of a line is
[tex]y=mx+b[/tex] ...(i)
where, m is slope and b is y-intercept.
The equation of given line is
[tex]y=\dfrac{1}{3}x-1[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]m=\dfrac{1}{3}[/tex]
Slope of given line is [tex]\dfrac{1}{3}[/tex].
We know that slope of parallel lines are same. So, the slope of required line is [tex]\dfrac{1}{3}[/tex].
Since slope of required line is [tex]\dfrac{1}{3}[/tex] and it passes through (6,10), therefore the equation of line is
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-10=\dfrac{1}{3}(x-6)[/tex]
[tex]y-10=\dfrac{1}{3}(x)-\dfrac{1}{3}(6)[/tex]
[tex]y-10=\dfrac{1}{3}(x)-2[/tex]
Add 10 on both sides.
[tex]y-10+10=\dfrac{1}{3}(x)-2+10[/tex]
[tex]y=\dfrac{1}{3}(x)+8[/tex]
Therefore, the required equation of line is [tex]y=\dfrac{1}{3}(x)+8[/tex].
