Question

The equation of line is $- 3y + 4x = 9$. How do you write the equation of a line that is parallel to the line and passes through the point $\left( { - 12,6} \right)$?

Answer 1

Two lines are either intersecting at one point or never intersect at any point (parallel lines). If two lines are parallel then their slopes are equal and intercepts are different in the slope intercept form of a line.

Complete step by step solution:
The given equation of a line is $- 3y + 4x = 9$.

Write the given equation in a slope intercept form $y = mx + c$ where $m$ represent slope of a line and $c$ represent the y-intercept.

Subtract $4x$ from both sides of the given equation as shown below.
$- 3y + 4x - 4x = 9 - 4x$
$\Rightarrow - 3y = 9 - 4x$

Now, divide by $- 3$ to both sides of the above equation as follows:
$\Rightarrow \dfrac{{ - 3y}}{{ - 3}} = \dfrac{{9 - 4x}}{{ - 3}}$

Simplify the equation as shown below.
$\Rightarrow y = - \dfrac{9}{3} + \dfrac{4}{3}x$
$\Rightarrow y = - 3 + \dfrac{4}{3}x$
$\Rightarrow y = \dfrac{4}{3}x - 3$

Comparing the above equation with slope intercept form, it is observed that slope of line is $\dfrac{4}{3}$ and y-intercept is $- 3$.

Now, to form an equation of a line that is parallel to given equation of a line and passes through the point $\left( { - 12,6} \right)$ we have to use point slope form that is the equation of a line with slope $m$ and passes through the point $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$.

As we know parallel lines have the same slope, therefore the slope of a required parallel line equation is $\dfrac{4}{3}$ but it passes through the point $\left( { - 12,6} \right)$.

Substitute $m$ as $\dfrac{4}{3}$ and $\left( {{x_1},{y_1}} \right)$ as $\left( { - 12,6} \right)$ in the equation $y - {y_1} = m\left( {x - {x_1}} \right)$ as shown below.
$y - \left( 6 \right) = \left( {\dfrac{4}{3}} \right)\left( {x - \left( { - 12} \right)} \right)$
Simplify the equation as shown below.
$\Rightarrow y - 6 = \left( {\dfrac{4}{3}} \right)\left( {x + 12} \right)$

Multiply both sides by number $3$ and simplify further.
$\Rightarrow 3\left( {y - 6} \right) = 4\left( {x + 12} \right)$
$\Rightarrow 3y - 18 = 4x + 48$

Rearrange the equation to write it in standard form as shown below.
$\Rightarrow 4x - 3y + 18 + 48 = 0$
$\Rightarrow 4x - 3y + 66 = 0$

Therefore, the required equation of a line that is parallel to the line of an equation$- 3y + 4x = 9$ and passes through the point $\left( { - 12,6} \right)$ is $4x - 3y + 66 = 0$.

Note: The ratio of coefficient of $x$ is equal to the ratio of coefficient of $y$ and not equal to the ratio of constant numbers for two parallel lines. Similarly the ratio of coefficient of $x$ is not equal to the ratio of coefficient of $y$ for two intersecting lines.