Question

which equation represents a line which is perpendicular to the line $8x + 3y = 3$?

answer
$\circ\\ y = \frac{3}{8}x - 4$ \qquad \qquad $\circ\\ y = -\frac{3}{8}x - 7$
$\circ\\ y = -\frac{8}{3}x + 5$ \qquad \qquad $\circ\\ y = \frac{8}{3}x - 5$

Explanation:

Step1: Find slope of given line

Rewrite \(8x + 3y = 3\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(8x\) from both sides: \(3y=-8x + 3\).
Divide by 3: \(y =-\frac{8}{3}x + 1\). So the slope of the given line \(m_1=-\frac{8}{3}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\).
We have \(-\frac{8}{3}\times m_2=-1\).
Solve for \(m_2\): \(m_2=\frac{3}{8}\) (multiply both sides by \(-\frac{3}{8}\)).

Step3: Identify the line with slope \(\frac{3}{8}\)

Check the slopes of the given options:

• For \(y=\frac{3}{8}x - 4\), slope \(m=\frac{3}{8}\).
• For \(y =-\frac{3}{8}x-7\), slope \(m =-\frac{3}{8}\).
• For \(y=-\frac{8}{3}x + 5\), slope \(m=-\frac{8}{3}\).
• For \(y=\frac{8}{3}x-5\), slope \(m=\frac{8}{3}\).

Answer:

\(y=\frac{3}{8}x - 4\)