Answer:
[tex]y=\frac{1}{4}x+(-1)[/tex]
Step-by-step explanation:
Given equation of line:
[tex]y=-4x+3[/tex]
To find the equation of line perpendicular to the line of the given equation and passes through point (8,1).
Applying slope relationship between perpendicular lines.
[tex]m_1=-\frac{1}{m_2}[/tex]
where [tex]m_1[/tex] and [tex]m_2[/tex] are slopes of perpendicular lines.
For the given equation in the form [tex]y=mx+b[/tex] the slope [tex]m_2[/tex] can be found by comparing [tex]y=-4x+3[/tex] with standard form.
∴ [tex]m_2=-4[/tex]
Thus slope of line perpendicular to this line [tex]m_1[/tex] would be given as:
[tex]m_1=-\frac{1}{-4}[/tex]
∴ [tex]m_1=\frac{1}{4}[/tex]
The line passes through point (8,1)
Using point slope form:
[tex]y-y_1=m(x-x_1)[/tex]
Where [tex](x_1,y_1)\rightarrow (8,1)[/tex] and [tex]m=m_2=\frac{1}{4}[/tex]
So,
[tex]y-1=\frac{1}{4}(x-8)[/tex]
Using distribution.
[tex]y-1=(\frac{1}{4}x)-(\frac{1}{4}\times 8)[/tex]
[tex]y-1=\frac{1}{4}x-2[/tex]
Adding 1 to both sides.
[tex]y-1+1=\frac{1}{4}x-2+1[/tex]
[tex]y=\frac{1}{4}x-1[/tex]
Thus the equation of line in standard form is given by:
[tex]y=\frac{1}{4}x+(-1)[/tex]
