Question

which line is parallel to the line whose equation is 4x+3y=7 and also passes through the point (-5,2)?

Answer 1

If a line is parallel to another they will have the same gradient. We can rearrange this equation into the form y=mx+c to start.

3y=-4x+7
y=-4/3x+7/3

Now we know that the gradient of the line is -4/3 so
y=-4/3x+c

Now we can substitute the points (-5,2) into this to find the y intercept

2=(-4/3 times -5)+c
2=20/3+c
-14/3=c

therefore the line is y=-4/3x-14/3

Answer 2

The equation of the required line is $y=-\dfrac{4}{3}x-\dfrac{14}{3}$.

Given:

The equation of the parallel line is:

$4x+3y=7$

The required line passes through the point $(-5,2)$.

To find:

The equation of the required line.

Explanation:

The slope of line $ax+by+c=0$ is:

$m=\dfrac{-a}{b}$

So, the slope of the line $4x+3y=7$ is:

$m=\dfrac{-4}{3}$

The slopes of two parallel lines are always equal. So, the slope of the required line is $m=\dfrac{-4}{3}$.

Point-slope form of a line is:

$y-y_1=m(x-x_1)$

Where, m is the slope and $(x_1,y_1)$ is the point.

The slope of the required line is $m=\dfrac{-4}{3}$ and it passes through the point $(-5,2)$. So, the equation of the required line is:

$y-2=-\dfrac{4}{3}(x-(-5))$

$y-2=-\dfrac{4}{3}(x+5)$

$y-2=-\dfrac{4}{3}x-\dfrac{20}{3}$

Adding 2 on both sides, we get

$y=-\dfrac{4}{3}x-\dfrac{20}{3}+2$

$y=-\dfrac{4}{3}x+\dfrac{-20+6}{3}$

$y=-\dfrac{4}{3}x-\dfrac{14}{3}$

Therefore, the equation of the required line is $y=-\dfrac{4}{3}x-\dfrac{14}{3}$.

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