Question

A perpendicular line is drawn from A(2,4)to the line 2y+3y-1=0. Find the equation of the perpendicular line in the form ax+by+c=0

Answer 1

To find the equation of the line that is perpendicular to the line given by the equation \( 2y + 3x - 1 = 0 \) and passes through the point \( A(2, 4) \), we can follow these steps:

1. **Rearrange the given line equation to find its slope:**
The equation \( 2y + 3x - 1 = 0 \) can be rearranged to the slope-intercept form \( y = mx + b \).

\[
2y = -3x + 1
\]
\[
y = -\frac{3}{2}x + \frac{1}{2}
\]

Here, the slope (\( m \)) of the line is \( -\frac{3}{2} \).

2. **Find the slope of the perpendicular line:**
The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope. Thus, the slope of the perpendicular line is:

\[
m_{\text{perpendicular}} = \frac{2}{3}
\]

3. **Use the point-slope form to find the equation of the perpendicular line:**
We can now use the point-slope form of the equation of a line, which is given by:

\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is the point through which the line passes (in this case, \( (2, 4) \)), and \( m \) is the slope.

Substituting in the values:

\[
y - 4 = \frac{2}{3}(x - 2)
\]

Simplifying this:

\[
y - 4 = \frac{2}{3}x - \frac{4}{3}
\]
\[
y = \frac{2}{3}x - \frac{4}{3} + 4
\]
\[
y = \frac{2}{3}x - \frac{4}{3} + \frac{12}{3}
\]
\[
y = \frac{2}{3}x + \frac{8}{3}
\]

4. **Convert to standard form \( ax + by + c = 0 \):**
To convert this equation into the required form \( ax + by + c = 0 \), we can multiply through by 3 to eliminate the fractions:

\[
3y = 2x + 8
\]

Rearranging gives:

\[
-2x + 3y - 8 = 0
\]

Alternatively, we can write it as:

\[
2x - 3y + 8 = 0
\]

Thus, the equation of the perpendicular line in the form \( ax + by + c = 0 \) is:

\[
\boxed{2x - 3y + 8 = 0}
\]