Question

Find the equation of a line perpendicular to the line x-7y+5=0 and having X intercept 3

Answer 1

To find the equation of a line that is perpendicular to the line given by the equation \( x - 7y + 5 = 0 \) and has an x-intercept of 3, we can follow these steps:

1. **Find the slope of the given line.**

The equation \( x - 7y + 5 = 0 \) can be rewritten in slope-intercept form (i.e., \( y = mx + b \)):

\[
7y = x + 5
\]
\[
y = \frac{1}{7}x + \frac{5}{7}
\]

From this, we can see that the slope \( m \) of the given line is \( \frac{1}{7} \).

2. **Determine 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. Therefore, the slope of the line we want to find is:

\[
m_{\text{perpendicular}} = -\frac{1}{\left(\frac{1}{7}\right)} = -7
\]

3. **Use the x-intercept to find the y-intercept.**

Since the x-intercept is given as 3, we know that the point (3, 0) lies on the line. We can use this point to find the equation of the line with slope -7.

We use the point-slope form of the equation of a line:

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

where \( (x_1, y_1) = (3, 0) \) and \( m = -7 \):

\[
y - 0 = -7(x - 3)
\]

Simplifying this, we get:

\[
y = -7x + 21
\]

4. **Rearranging into standard form if needed.**

If you prefer to write it in standard form \( Ax + By + C = 0 \):

\[
7x + y - 21 = 0
\]

So, the equation of the line that is perpendicular to the line \( x - 7y + 5 = 0 \) and has an x-intercept of 3 is:

\[
\boxed{y = -7x + 21}
\]

or in standard form:

\[
\boxed{7x + y - 21 = 0}
\]