Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

1. Analyze the question

We need to find the equation of a line that is perpendicular to a given line ($4y = 7x + 3$) and passes through a specific point $(-3, 1)$.

2. Explain the concept

• Slope-Intercept Form: A line equation in the form $y = mx + c$ has a slope of $m$.
• Perpendicular Slopes: If two lines are perpendicular, the product of their slopes is $-1$ ($m_1 \times m_2 = -1$). This means the slope of the new line will be the negative reciprocal of the original slope ($m_2 = -1 / m_1$).
• Point-Slope Form: Once we have the new slope ($m$) and a point $(x_1, y_1)$, we use the formula: $y - y_1 = m(x - x_1)$.

3. Show the solution

Step 1: Find the slope of the given line. The equation is $4y = 7x + 3$. Divide by 4 to get it into $y = mx + c$ form: $y = \frac{7}{4}x + \frac{3}{4}$ The slope ($m_1$) is $\frac{7}{4}$.

Step 2: Find the slope of the perpendicular line ($m_2$). Since $m_1 \times m_2 = -1$: $\frac{7}{4} \times m_2 = -1$ $m_2 = -\frac{4}{7}$

Step 3: Use the point-slope formula with point $(-3, 1)$. $y - 1 = -\frac{4}{7}(x - (-3))$ $y - 1 = -\frac{4}{7}(x + 3)$

Step 4: Simplify the equation. Multiply both sides by 7 to clear the fraction: $7(y - 1) = -4(x + 3)$ $7y - 7 = -4x - 12$

Rearrange to the form $Ay + Bx + C = 0$: $7y + 4x - 7 + 12 = 0$ $7y + 4x + 5 = 0$

4. Identify the answer

Comparing our result ($7y + 4x + 5 = 0$) to the given options, it matches option A.

5. Tip

When dealing with perpendicular lines, remember the "flip and switch" rule: flip the fraction of the slope and switch the sign (e.g., $\frac{7}{4}$ becomes $-\frac{4}{7}$). This saves you time during your JAMB exam!