The work is shown in the attached images. Each equation is labeled for slope intercept and standard form
y=-1/5x +18/5
5y+x=18
The equation of line passing through point (-2,4) and parallel to x+5y=7 in
standard form is [tex]x+5y =18[/tex]
slope intercept form is [tex]y = \frac{-1}{5}x + \frac{18}{5}[/tex]
The equation of line is a linear equation with degree of one and an algebraic form of representing the set of points, which together form a line in a coordinate system.
[tex](y-y_{1} ) = m(x-x_{1} )[/tex]
where,
[tex](x_{1}, y_{1} )[/tex] are the points through which the line passes
m is the slope of line
(slopes of parallel lines are equals)
According to the given question
we have
Equation of line x+5y=7
a point (-2,4)
slope of the line x+5y = 7 by placing it into slope intercept form:
x+5y = 7
⇒ 5y = -x+7
⇒ y = [tex]\frac{-x}{5} +\frac{7}{5}[/tex]
the slope of the line, m=[tex]\frac{-1}{5}[/tex]
here, the point (-2,4) and slope is m= [tex]\frac{-1}{5}[/tex]
therefore the equation of line which is parallel to x+5y =7 and passing through (-2,4) is:
[tex](y-4)=\frac{-1}{5}(x-(-2))[/tex]
⇒[tex](y-4) = \frac{-1}{5} (x+2)[/tex]
⇒ [tex]5(y-4)= -x-2[/tex]
⇒[tex]5y-20 = -x-2[/tex]
⇒[tex]x+5y= 20-2[/tex]
⇒ [tex]x+5y = 18[/tex]
so,
(a). the equation of line in slope intercept form is given by
[tex]5y= -x+18[/tex]
[tex]y = \frac{-1}{5}x+\frac{18}{5}[/tex]
(b). the standard form of line is x+5y = 18.
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