Chapter 1, Problem 33RE

To determine

To find:The Equation of the line

Answer to Problem 33RE

The equation of the line in general form is $2x-3y=-19$

Explanation of Solution

Given:

Parallel to the line $2x-3y=-4$

  $\left(-5,3\right)$

Formula Used:

The slope-intercept form $y=mx+b$

Calculation:

Since the two lines are parallel, their slopes will be equal.

For finding the slope of the line $2x-3y=-4$ , write the equation in the slope-intercept form $y=mx+b$

Subtract $2x$ from both sides of the equation.

  $2x-3y-2x=-4-2x$

  $-3y=-4-2x$

Multiply by $-1$ on both sides of the equation.

  $\left(-1\right)\left(-3y\right)=\left(-1\right)\left(-4\right)+\left(-1\right)\left(-2x\right)$

  $3y=2x+4$

Divide by 3 on both sides of the equation.

  $\frac{3y}{3}=\frac{2}{3}x+\frac{4}{3}$

  $y=\frac{2}{3}x+\frac{4}{3}$

So, the slope of the line is $\frac{2}{3}$ .

Since the line parallel to $2x-3y=-4$ also has the slope $\frac{2}{3}$ and contains the point $\left(-5,3\right)$ , use the point-slope form the obtain its equation.

  $y-y_1=m\left(x-x_1\right)$

  $y-3=\frac{2}{3}\left(x-\left(-5\right)\right)$

Remove the parentheses using the distributive property of multiplication.

  $y-3=\frac{2}{3}x+\frac{10}{3}$

Add 3 to both sides of the equation.

  $y-3+3=\frac{2}{3}x+\frac{10}{3}+3$

  $y=\frac{2}{3}x+\frac{19}{3}$

Therefore, the equation of the lines is $y=\frac{2}{3}x+\frac{19}{3}$ .

In order to get the equation of the line in the general form, simply $y=\frac{2}{3}x+\frac{19}{3}$ .

Multiply by 3 on both sides of the equation.

  $3y=3\cdot \frac{2}{3}x+3\cdot \frac{19}{3}$

  $3y=2x+19$

Subtract $2x$ from both sides of the equation.

  $3y-2x=2x+19-2x$

  $-2x+3y=19$

Multiply by $-1$ on both sides of the equation.

  $\left(-1\right)\left(-2x\right)+\left(-1\right)3y=\left(-1\right)19$

  $2x-3y=-19$

Conclusion:

Therefore, the equation of the line in general form is $2x-3y=-19$