Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
(a)5x-y+20=0
(b)6x-4y=9
(c)3x-5y-9=0
(d)None of these

Answer

To find the equation of the line passing through the point (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6), we can follow these steps:

### Step 1: Find the slope of the line through the points (2, 5) and (-3, 6).
The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, let \( (x_1, y_1) = (2, 5) \) and \( (x_2, y_2) = (-3, 6) \).

Calculating the slope:
\[ m = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5} \]

### Step 2: Determine the slope of the perpendicular line.
If two lines are perpendicular, the product of their slopes \( m_1 \) and \( m_2 \) is -1:
\[ m_1 \cdot m_2 = -1 \]
Let \( m_1 = -\frac{1}{5} \) (the slope of the line through (2, 5) and (-3, 6)). Then:
\[ -\frac{1}{5} \cdot m_2 = -1 \implies m_2 = 5 \]

### Step 3: Use the point-slope form to find the equation of the line.
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Using the point (-3, 5) and the slope \( m_2 = 5 \):
\[ y - 5 = 5(x + 3) \]

### Step 4: Simplify the equation.
Expanding the equation:
\[ y - 5 = 5x + 15 \]
Adding 5 to both sides:
\[ y = 5x + 20 \]

### Step 5: Rearranging to standard form.
To express this in standard form \( Ax + By + C = 0 \):
\[ 5x - y + 20 = 0 \]

### Conclusion
The equation of the line is:
\[ 5x - y + 20 = 0 \]
Thus, the correct answer is option (a) \( 5x - y + 20 = 0 \).

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