Find the Line Perpendicular to 3y=x+12 Through Point (7,2)

To solve this problem, we need to determine the equation of a line passing through the point $(7, 2)$ and perpendicular to the given line $3y = x + 12$. We can achieve this by following a systematic approach:

• Step 1: Identify the slope of the given line

We start by rewriting the equation of the given line $3y = x + 12$ in slope-intercept form, $y = mx + b$, where $m$ is the slope. To do this, we divide each side of the equation by 3:

$y = \frac{1}{3}x + 4$

Thus, the slope $m_1$ of the given line is $\frac{1}{3}$.

• Step 2: Determine the slope of the perpendicular line

Two lines are perpendicular if the product of their slopes is $-1$. Therefore, if the slope of the line we are looking for is $m_2$, then:

$m_1 \times m_2 = -1$

Substituting the value of $m_1$:

$\frac{1}{3} \times m_2 = -1$

Solving for $m_2$, we find:

$m_2 = -3$

• Step 3: Use the point-slope form to find the equation of the perpendicular line

Now that we know the slope of the perpendicular line is $-3$ and it passes through the point $(7, 2)$, we can use the point-slope form of a line:

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

Here, $m = -3$ and the point is $(x_1, y_1) = (7, 2)$, so:

$y - 2 = -3(x - 7)$

Expanding this equation to solve for $y$, we get:

$y - 2 = -3x + 21$

Adding 2 to both sides to isolate $y$, we have:

$y = -3x + 23$

Thus, the equation of the line that passes through the point $(7, 2)$ and is perpendicular to the line $3y = x + 12$ is $y = -3x + 23$.

The correct answer to this problem is therefore $y = -3x + 23$.