Question

Write an equation of the line in slope-intercept form and in standard form for the line passing through (-1,4) and parallel to x+3y=7

Answer 1

Consider that the equation of a line with slope 'm' and y-intercept 'c' is given by,

$y=mx+c$

Now, the equation of the given line is,

$\begin{gathered} x+3y=7 \\ 3y=-x+7 \\ y=\frac{-1}{3}x+\frac{7}{3} \end{gathered}$

Comparing the coefficients, it is found that the slope of the given line is,

$m=\frac{-1}{3}$

Let the slope of the line which is parallel to this given line be 'n'.

Theorem: Two lines will be parallel if their slopes are equal,

$\begin{gathered} n=m \\ n=\frac{-1}{3} \end{gathered}$

Then the equation of the line will be,

$\begin{gathered} y=nx+c \\ y=\frac{-1}{3}x+c \end{gathered}$

Given that this parallel line passes through the point (-1,4), so it must satisfy its equation,

$\begin{gathered} 4=\frac{-1}{3}(-1)+c \\ 4=\frac{1}{3}+c \\ c=4-\frac{1}{3} \\ c=\frac{11}{3} \end{gathered}$

Substitute this value in the equation,

$y=\frac{-1}{3}x+\frac{11}{3}$

Thus, the equation of the line passing through (-1,4) and parallel to x+3y=7 is obtained as,

$y=\frac{-1}{3}x+\frac{11}{3}$

the slope