Find the Perpendicular Line: Solving 3y=4x-y through Point (-3,-15)

To solve this problem, we'll follow these steps:

• Step 1: Rearrange the given line equation to standard slope-intercept form.
• Step 2: Determine the slope of the perpendicular line.
• Step 3: Use the point-slope form to find the equation of the desired line.

Now, let's work through each step:

Step 1: The given line equation is $3y = 4x - y$. First, simplify this equation:

$3y = 4x - y$

Add $y$ to both sides to consolidate $y$:

$3y + y = 4x$

$4y = 4x$

Divide both sides by 4:

$y = x$

The slope $m$ of this line is 1.

Step 2: Since the line we want is perpendicular to this line, the slope $m_1$ of the desired line should satisfy:

$m \times m_1 = -1$

Given $m = 1$, we have:

$1 \times m_1 = -1$

So, $m_1 = -1$.

Step 3: Use the point-slope form of a line $y - y_1 = m(x - x_1)$ with point $(-3, -15)$ and slope $-1$:

$y - (-15) = -1(x - (-3))$

Simplify the equation:

$y + 15 = -1(x + 3)$

Expand:

$y + 15 = -x - 3$

Subtract 15 from both sides:

$y = -x - 3 - 15$

$y = -x - 18$

Rearrange to the standard form:

$y + x = -18$

The equation of the line that passes through $(-3, -15)$ and is perpendicular to $3y = 4x - y$ is:

$y + x = -18$