Question

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

A)
y=-x+10
y = -5x -
y = 5x + 10
yax =2
11)

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 = (-1)/(5)x + (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 = (-1)/(5)x + (10)/(5)\\\\y = (-1)/(5)x + 2`

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

`m = (-1)/(5)`

We know that slopes of parallel lines are equal

So the slope of line parallel to given line is also
`m = (-1)/(5)`

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

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

`3 = (-1)/(5) * 1 + c\\\\15 = -1 + 5c\\\\16 = 5c\\\\c = (16)/(5)`

Thus the required equation is:

`\text {substitute } m=(-1)/(5) \text { and } c=(16)/(5) \text { in eqn } 1`

`y = (-1)/(5)x + (16)/(5)`

Thus the required equation of line is found