How can I find the equation of a line that goes through the point (2, 4) and is perpendicular to the line defined by the equation 8x - 7y + 4 = 0?

To find the equation of a line that is perpendicular to another line and passing through a specific point, we can follow these steps:

The given equation is 8x – 7y + 4 = 0. We can rewrite this equation in slope-intercept form (y = mx + b) to easily identify its slope.

Start by isolating y:

8x - 7y + 4 = 0
-7y = -8x - 4
y = (8/7)x + 4/7

From this equation, we see that the slope (m) of the original line is 8/7.

Since we need the slope of the line that is perpendicular to the original line, we will take the negative reciprocal of 8/7.

The negative reciprocal is:

m_{perpendicular} = -7/8

Now that we know the slope of the perpendicular line and the point through which it passes, we can use the point-slope form of the line equation:

y - y_1 = m(x - x_1)

In this case, (x₁, y₁) = (2, 4) and m = -7/8.

Substituting these values into the point-slope form gives:

y - 4 = -7/8(x - 2)

Distributing the slope on the right side:

y - 4 = -7/8x + 7/4

Now, add 4 to both sides to solve for y:

y = -7/8x + 7/4 + 16/4
y = -7/8x + 23/4

The equation of the line that passes through the point (2, 4) and is perpendicular to the line 8x – 7y + 4 = 0 is:

y = -7/8x + 23/4

This is the final equation in slope-intercept form.

How can I find the equation of a line that goes through the point (2, 4) and is perpendicular to the line defined by the equation 8x – 7y + 4 = 0?