Let's break down the problem and solve each part step by step.

Given:

• Line $l$ has the equation $4x - 3y + 8 = 0$ and passes through point $B(7, 12)$.
• Line $m$ is perpendicular to line $l$ and passes through point $B$.

Part (a): Find the equation of line m.

1. Equation of line $l$:
   We know the equation of line $l$ is $4x - 3y + 8 = 0$. The slope of this line can be found by converting the equation into slope-intercept form $y = mx + b$.

   $4x - 3y + 8 = 0 \implies 3y = 4x + 8 \implies y = \frac{4}{3}x + \frac{8}{3}$ So, the slope $m_l$ of line $l$ is $\frac{4}{3}$.

2. Perpendicular slope:
   The slope of a line perpendicular to line $l$ will be the negative reciprocal of $\frac{4}{3}$. Therefore, the slope of line $m$, $m_m$, is: $m_m = -\frac{3}{4}$

3. Equation of line $m$:
   Line $m$ passes through point $B(7, 12)$ and has slope $-\frac{3}{4}$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 12 = -\frac{3}{4}(x - 7)$ Expanding this: $y - 12 = -\frac{3}{4}x + \frac{21}{4}$ Multiply the whole equation by 4 to eliminate the fraction: $4(y - 12) = -3(x - 7)$ $4y - 48 = -3x + 21$ Bringing everything to one side: $3x + 4y - 69 = 0$ So, the equation of line $m$ is: $3x + 4y - 69 = 0$

Part (b): Find the coordinates of points A and C.

• Point A (where line $l$ crosses the x-axis): On the x-axis, $y = 0$. Substitute $y = 0$ in the equation of line $l$: $4x - 3(0) + 8 = 0 \implies 4x + 8 = 0 \implies x = -2$ So, point $A$ is $(-2, 0)$.

• Point C (where line $m$ crosses the x-axis): Similarly, for line $m$, substitute $y = 0$ in the equation $3x + 4y - 69 = 0$: $3x + 4(0) - 69 = 0 \implies 3x = 69 \implies x = 23$ So, point $C$ is $(23, 0)$.

Part (c): Find the area of triangle ABC.

We have the coordinates of points $A(-2, 0)$, $B(7, 12)$, and $C(23, 0)$. The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Substituting the coordinates $(-2, 0)$, $(7, 12)$, and $(23, 0)$: $\text{Area} = \frac{1}{2} \left| (-2)(12 - 0) + 7(0 - 0) + 23(0 - 12) \right|$ Simplifying: $\text{Area} = \frac{1}{2} \left| (-2)(12) + 7(0) + 23(-12) \right|$ $\text{Area} = \frac{1}{2} \left| -24 - 276 \right| = \frac{1}{2} \times 300 = 150$ So, the area of triangle $ABC$ is $\boxed{150}$.

Summary of Solutions:

• (a) The equation of line $m$ is $3x + 4y - 69 = 0$.
• (b) The coordinates of points $A$ and $C$ are $A(-2, 0)$ and $C(23, 0)$.
• (c) The area of triangle $ABC$ is $150$.

Do you want further clarification on any step or additional details?

Relative Questions:

1. How do you derive the slope of a line from its equation?
2. What is the significance of perpendicular slopes in geometry?
3. Can the area of a triangle be negative? Why or why not?
4. How would the problem change if line $m$ wasn't perpendicular to line $l$?
5. How do you interpret the equation of a line when expressed in general form $ax + by + c = 0$?

Tip:

When two lines are perpendicular, their slopes multiply to give $-1$. This fact is very useful for finding equations of perpendicular lines.