Chapter 1, Problem 25RE

Perpendicular to the line $3x-y=-4$; containing the point $\left(-2,4\right)$

To determine

To find: The equation of a line containing the point $\left(-2,4\right)$ and parallel to the line $3x-y=-4$. To also sketch a graph of the equation.

Answer to Problem 25RE

Solution:

The general form of equation of the line is $3x-y=-4$.

Explanation of Solution

Given:

The points $\left(-2,4\right)$ and the line $3x-y=-4$

Formula used:

Point slope form of an equation of a line.

$(y-y₁)=m(x-x₁)$

Calculation:

To find the slope, write the given line $3x-y=-4$, in the form of $y=mx+c$.

$3x-y-3x=-4-3x$  &nbspSubtract $3x$ from both sides.

$-y=-3x-4$

$y=3x+4$  &nbspMultiply both sides by $–1$.

$y=3x+4$

Hence the slope $m=3$

Now using this slope $m=3$ and the point $\text{A}\left(-2,4\right)$, the equation can be found.

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

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

$y-4=3x+6$

$y-4-3x=3x+6-3x$

$-3x-4+y=6$

$-3x+y-4+4=6+4$  &nbspAdd 4 to both sides

$-3x+y=10$

$3x-y=-10$  &nbspMultiply $\text{−1}$ to both sides

$3x-y=-10$

The general form of equation of the line is $3x-y=-10$.