x-4y=4
Step-by-step explanation:
The given line has the equation [tex]y=-4x+3[/tex]. We want to find a line perpendicular to the given line and intersecting ( passing through ) [tex]( 8, 1 )[/tex].
The given line, on comparision with the standard slope-intercept equation of line : [tex]y=mx+c[/tex] , yields [tex]m=-4[/tex]. Slope of given line is [tex]-4[/tex].
If two lines with slopes [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are perpendicular, then [tex]m_{1}\times m_{2}=-1[/tex]
So, [tex]\text{(Slope of perpendicular line)}\times(-4)=-1[/tex]
Slope of line = [tex]\dfrac{1}{4}[/tex].
A line having slope [tex]m[/tex] and passing through [tex](x_{1},y_{1})[/tex] has the equation [tex]\dfrac{y-y_{1}}{x-x_{1}}=m[/tex]
Required line is [tex]\dfrac{y-1}{x-8}=\dfrac{1}{4}\\\\ 4(y-1)=(x-8)\\x-4y=4[/tex].
∴ Required line is [tex]x-4y=4[/tex]
Changing it to the form mentioned, [tex]y=\dfrac{1}{4}x+(-1)[/tex]
∴ ? in the attachment is 1
