Respuesta :
Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is [tex]y=\frac{1}{3} x+2[/tex]
Solution:
Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).
Generic slope intercept form of a line is given by y = mx + c
where m = slope of the line.
Let's first find slope intercept form of 3x + y = -8
3x + y = -8
=> y = -3x - 8
On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3
And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.
Let required slope = x
[tex]\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}[/tex]
So we need to find the equation of a line whose slope is [tex]\frac{1}{3}[/tex] and passing through (-3,1)
Equation of line passing through [tex](x_1 , y_1)[/tex] and having lope of m is given by
[tex]\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)[/tex]
[tex]\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}[/tex]
Substituting the values we get,
[tex]\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}[/tex]
Hence the required equation of line is found using slope intercept form
