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The problem in the image states:

The points \( A \) and \( B \) have coordinates \( A(-2, 15) \) and \( B(3, 5) \) respectively. The perpendicular to the line \( AB \) at the point \( A(-2, 15) \) crosses the \( y \)-axis at the point \( C \). Find the area of triangle \( ABC \).

Let's break down the solution step by step:

### 1. Equation of Line \( AB \):
First, we find the equation of the line \( AB \).

- The slope of \( AB \) is given by:
\[
\text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 15}{3 - (-2)} = \frac{-10}{5} = -2
\]

- Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \) with \( m = -2 \) and point \( A(-2, 15) \):
\[
y - 15 = -2(x + 2)
\]
Simplifying:
\[
y - 15 = -2x - 4 \quad \Rightarrow \quad y = -2x + 11
\]
Thus, the equation of the line \( AB \) is \( y = -2x + 11 \).

### 2. Equation of the Perpendicular Line:
The line perpendicular to \( AB \) will have a slope that is the negative reciprocal of \( -2 \), which is \( \frac{1}{2} \).

Using the point-slope form again with the point \( A(-2, 15) \) and slope \( \frac{1}{2} \):
\[
y - 15 = \frac{1}{2}(x + 2)
\]
Simplifying:
\[
y - 15 = \frac{1}{2}x + 1 \quad \Rightarrow \quad y = \frac{1}{2}x + 16
\]
Thus, the equation of the perpendicular line is \( y = \frac{1}{2}x + 16 \).

### 3. Finding the Coordinates of Point \( C \):
Point \( C \) is where the perpendicular line crosses the \( y \)-axis. On the \( y \)-axis, \( x = 0 \). Substituting \( x = 0 \) into the equation of the perpendicular line:
\[
y = \frac{1}{2}(0) + 16 = 16
\]
Thus, the coordinates of point \( C \) are \( C(0, 16) \).

### 4. Area of Triangle \( ABC \):
We can now find the area of triangle \( ABC \) using the formula for the area of a triangle given three vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates \( A(-2, 15) \), \( B(3, 5) \), and \( C(0, 16) \):
\[
\text{Area} = \frac{1}{2} \left| -2(5 - 16) + 3(16 - 15) + 0(15 - 5) \right|
\]
Simplifying:
\[
\text{Area} = \frac{1}{2} \left| -2(-11) + 3(1) + 0 \right| = \frac{1}{2} \left| 22 + 3 \right| = \frac{1}{2} \times 25 = 12.5
\]

Thus, the area of triangle \( ABC \) is **12.5 square units**.

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Would you like further clarifications or additional details?

Here are five related questions:
1. How do you calculate the slope of a line from two points?
2. What is the formula for the area of a triangle given three vertices?
3. How can you find the equation of a line perpendicular to a given line?
4. What are the conditions for two lines to be perpendicular?
5. How would the area change if one of the triangle's vertices was moved?

**Tip:** When working with slopes, remember that perpendicular lines have slopes that are negative reciprocals of each other!

The points A and B have coordinates (-2, 15) and (3, 5) respectively. The perpendicular to the line AB at the point A(-2, 15) crosses the y-axis at the point C. Find the area of triangle ABC.

Solve for the area of triangle ABC where A(-2, 15), B(3, 5), and C is the point where the perpendicular from A to line AB crosses the y-axis. This solution involves calculating the slope of a line, using point-slope form, and applying the area of a triangle formula.

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