Question

find the equation of a straight line which is parallel to the line with equation 5x+7y=14 and passes through a point (-2,-3)​

Answer 1

`7y=-5x-11`

Explanation:

Since it says the lines are parallel both of their gradients will be same.

so gradient of line 1 =
`y=(14-5x)/(7)` =
`y=(14)/(7)-(5x)/(7)`=
`-(5)/(7)`

so equation of line 2 =
`y-y_(1)=m(x-x_(1) )`

=
`y-(-3)_{}=(-5)/(7) (x-(-2)_{} )`

=
`y +3=-(5)/(7)x+(10)/(7)`

=
`y =-(5x+10)/(7)-3`

=
`7y=-5x-11`

Answer 2

Explanation:

basically a line equation typically looks like

y = ax + b

with a being the slope, and b bent the y-intercept (y value when x = 0).

5x + 7y = 14

7y = -5x + 14

y = -5/7 x + 14/7 = -5/7 x + 2

so, wie know the slope of the original line : -5/7 .

any line parallel to it must have the same slope.

our desired line looks like

y = -5/7 x + b

to get b we use the provided point (-2, -3).

-3 = -5/7 × -2 + b = 10/7 + b

-21/7 = 10/7 + b

-31/7 = b

and the full equation is

y = -5/7 x - 31/7

7y = -5x - 31

5x + 7y = -31