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Find the equation of a line passing through the point (-3, -7) that is perpendicular to the line 5x-3y=10. Show detailed steps.

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To find the equation of a line passing through a given point that is perpendicular to a given line, we need to follow these steps:

Find the slope of the given line by rearranging the equation into slope-intercept form y = mx + b, where m is the slope.

Find the slope of a line perpendicular to the given line by taking the negative reciprocal of the slope found in step 1.

Use the point-slope form y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope found in step 2, to write the equation of the perpendicular line.

Step 1: Find the slope of the given line 5x - 3y = 10

We can rearrange the equation into slope-intercept form by solving for y:

5x - 3y = 10

-3y = -5x + 10

y = (5/3)x - (10/3)

The slope of this line is 5/3.

Step 2: Find the slope of a line perpendicular to the given line

The slope of a line perpendicular to the given line is the negative reciprocal of 5/3:

m_perp = -3/5

Step 3: Use the point-slope form to write the equation of the perpendicular line passing through (-3, -7).

We plug in the values we found for the slope and the given point into the point-slope form:

y - y1 = m_perp(x - x1)

y - (-7) = (-3/5)(x - (-3))

Simplifying this equation, we get:

y + 7 = (-3/5)x - 9/5

Now we can rearrange this equation into slope-intercept form y = mx + b, where m is the slope and b is the y-intercept:

y = (-3/5)x - 44/5

Therefore, the equation of the line passing through the point (-3, -7) and perpendicular to the line 5x - 3y = 10 is y = (-3/5)x - 44/5.

User Chander Shoor
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