Question

Write an equation of the line satisfying the given conditions.

Through (1,−5); parallel to 5x=6y+7

Answer 1

Explanation:

`equation \: of \: a \: line \: is \: y = mx + b \\`

`5x = 6y + 7 \\ 5x - 7 = 6y \\ ((5x - 7))/(6) = (6y)/(6 )`

Answer 2

The line equation of required line is 5x – 6y – 35 = 0.

SOLUTION:

Given, line equation is 5x = 6y + 7 and the point is p (1, -5).

We have to find the line equation of a line that is parallel to given line and passing through point p.

First, let us find slope of given line.

5x = 6y + 7

5x – 6y – 7 = 0

`\text { Slope of a line }=\frac{-x \text { coefficient }}{y \text { coefficient }}`

`=(-5)/(-6)=(5)/(6)`

We know that, parallel lines will have same slope, so slope of required line is
`(5)/(6)`

Now, we have slope and a point through it.

So, let us find the point slope form of the line i.e
`y-y_(1)=m\left(x-x_(1)\right)`

`\text { Here, } \mathrm{y}_(1)=-5, \mathrm{x}_(1)=1 \text { and } \mathrm{m}=(5)/(6)`

Line equation →
`y-(-5)=(5)/(6)(x-1)`

`y+5=(5)/(6)(x-1)`

6y + 30 = 5x – 5

5x – 6y – 5 – 30 = 0

5x – 6y – 35 = 0

Hence, the line equation of required line is 5x – 6y – 35 = 0.