Find the equation of the st.line joining the points `(3,-1)` and `(2,3)` . Also find the equation of another st.line perpendicular to this st.line and passing through `(5,2)`.

Answer

To solve the problem, we need to find the equation of the straight line joining the points (3, -1) and (2, 3), and then find the equation of another straight line that is perpendicular to this line and passes through the point (5, 2).

### Step 1: Find the slope of the line joining the points (3, -1) and (2, 3).

The formula for the slope \( m \) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the points (3, -1) as \((x_1, y_1)\) and (2, 3) as \((x_2, y_2)\):

\[ m = \frac{3 - (-1)}{2 - 3} = \frac{3 + 1}{2 - 3} = \frac{4}{-1} = -4 \]

### Step 2: Use the point-slope form to find the equation of the line.

The point-slope form of the equation of a line is given by:

\[ y - y_1 = m(x - x_1) \]

Using point (3, -1) and the slope \( m = -4 \):

\[ y - (-1) = -4(x - 3) \]

This simplifies to:

\[ y + 1 = -4(x - 3) \]

Expanding this:

\[ y + 1 = -4x + 12 \]

Rearranging gives:

\[ 4x + y - 11 = 0 \]

### Step 3: Find the equation of the line perpendicular to the first line.

The slope of the perpendicular line is the negative reciprocal of the original slope. Therefore, if the slope of the original line is \( -4 \), the slope of the perpendicular line \( m' \) is:

\[ m' = \frac{1}{4} \]

### Step 4: Use the point-slope form for the perpendicular line passing through (5, 2).

Using the point (5, 2) and the slope \( m' = \frac{1}{4} \):

\[ y - 2 = \frac{1}{4}(x - 5) \]

Expanding this:

\[ y - 2 = \frac{1}{4}x - \frac{5}{4} \]

Rearranging gives:

\[ y = \frac{1}{4}x - \frac{5}{4} + 2 \]

Converting 2 to a fraction with a denominator of 4:

\[ y = \frac{1}{4}x - \frac{5}{4} + \frac{8}{4} \]

This simplifies to:

\[ y = \frac{1}{4}x + \frac{3}{4} \]

To express this in standard form, we can multiply through by 4:

\[ 4y = x + 3 \]

Rearranging gives:

\[ x - 4y + 3 = 0 \]

### Final Answers:
1. The equation of the line joining the points (3, -1) and (2, 3) is \( 4x + y - 11 = 0 \).
2. The equation of the line perpendicular to the first line and passing through (5, 2) is \( x - 4y + 3 = 0 \).