Question

On a coordinate plane, line M N goes through points (2, 3) and (negative 3, 2). Point K is at (3, negative 3).
Which point could be on the line that is perpendicular to Line M N and passes through point K?

(0, −12)
(2, 2)
(4, 8)
(5, 13)

Answer 1

To find a point that lies on the line perpendicular to line MN and passes through point K (3, -3), we first need to determine the slope of line MN using the points (2, 3) and (-3, 2).

1. **Calculate the slope of line MN**:

The slope \( m \) is calculated using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substituting the given points:
- \( (x_1, y_1) = (2, 3) \)
- \( (x_2, y_2) = (-3, 2) \)

\[
m = \frac{2 - 3}{-3 - 2} = \frac{-1}{-5} = \frac{1}{5}
\]

2. **Find the slope of the line perpendicular to MN**:

The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. Therefore, the slope \( m_{perpendicular} \) of the line that is perpendicular to MN is:
\[
m_{perpendicular} = -\frac{1}{m} = -5
\]

3. **Use the point-slope form to find the equation of the perpendicular line**:

Using the point \( K(3, -3) \) and the slope \( m_{perpendicular} = -5 \):
\[
y - y_1 = m(x - x_1)
\]
Plugging in the values:
\[
y + 3 = -5(x - 3)
\]
Distributing the -5:
\[
y + 3 = -5x + 15
\]
Subtracting 3 from both sides:
\[
y = -5x + 12
\]

4. **Identify which of the given points lies on this line**:

Now we will check each potential point by substituting the x-values into the line equation \( y = -5x + 12 \):

- For **(0, -12)**:
\[
y = -5(0) + 12 = 12 \quad \text{(not } -12\text{)}
\]

- For **(2, 2)**:
\[
y = -5(2) + 12 = -10 + 12 = 2 \quad \text{(this is a solution)}
\]

- For **(4, 8)**:
\[
y = -5(4) + 12 = -20 + 12 = -8 \quad \text{(not } 8\text{)}
\]

- For **(5, 13)**:
\[
y = -5(5) + 12 = -25 + 12 = -13 \quad \text{(not } 13\text{)}
\]

The only point that lies on the line that is perpendicular to line MN and passes through point K is:

\[
\boxed{(2, 2)}
\]