Given:
Slope of a straight line is [tex]\sqrt{3}[/tex].
The line passing through the point (-1,5).
To find:
The equation of the straight line.
Solution:
The point slope form of a line is:
[tex]y-y_1=m(x-x_1)[/tex]
Where, m is the slope and [tex](x_1,y_1)[/tex] is the point.
The line passing through the point (-1,5) with slope [tex]\sqrt{3}[/tex]. So, the equation of the line is:
[tex]y-5=\sqrt{3}(x-(-1))[/tex]
[tex]y-5=\sqrt{3}(x+1)[/tex]
[tex]y=\sqrt{3}x+\sqrt{3}+5[/tex]
Therefore, the equation of the straight line is [tex]y=\sqrt{3}x+\sqrt{3}+5[/tex].
