Question

Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1, 3).

Answer 1

The equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1, 3) in slope intercept form is $y = \frac{-1}{5}x + \frac{16}{5}$

Solution:

Given that a line is parallel to line x + 5y = 10 and passes through the point (1, 3)

We have to find the equation of line

The slope intercept form is given as:

y = mx + c  -------- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Let us first find the slope of line

Given equation of line is x + 5y = 10

$5y = -x + 10\\\\y = \frac{-1}{5}x + \frac{10}{5}\\\\y = \frac{-1}{5}x + 2$

On comparing the above equation of line with slope intercept form,

$m = \frac{-1}{5}$

We know that slopes of parallel lines are equal

So the slope of line parallel to given line is also $m = \frac{-1}{5}$

Let us find the equation of line with slope m = -1/5 and passes through the point (1, 3)

$\text {substitute } m=\frac{-1}{5} \text { and }(x, y)=(1,3) \text { in eqn } 1$

$3 = \frac{-1}{5} \times 1 + c\\\\15 = -1 + 5c\\\\16 = 5c\\\\c = \frac{16}{5}$

Thus the required equation is:

$\text {substitute } m=\frac{-1}{5} \text { and } c=\frac{16}{5} \text { in eqn } 1$

$y = \frac{-1}{5}x + \frac{16}{5}$

Thus the required equation of line is found