Find an equation of the line that passes through the point (2, 4) and is perpendicular to the line 8x + 7y - 4 = 0.

Solution:

Given, the equation of the line is 8x + 7y - 4 = 0 --- (1)

The equation of the line in slope-intercept form is y = mx + c --- (2)

Converting (1) in slope-intercept form,

⇒ 7y = -8x + 4

⇒ y = -(8/7)x + (4/7)

Slope of the line is -8/7.

Slope of the perpendicular line = +7/8

Next, find the value of c by using the point (2,4)

⇒ 4 = (7/8)(2) + c

⇒ 4 = 7/4 + c

0n solving,

⇒ 16 = 7 + 4c

⇒ 4c = 16 - 7

⇒ 4c = 9

⇒ c = 9/4

Put the value of c and slope of perpendicular in (2) we get,

⇒ y = (7/8)x + 9/4

Therefore, the equation of the line is y = (7/8)x + 9/4.

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Find an equation of the line that passes through the point (2, 4) and is perpendicular to the line 8x + 7y - 4 = 0.

Summary:

The equation of the line that passes through the point (2, 4) and is perpendicular to the line 8x + 7y - 4 = 0 is y = (7/8)x + 9/4.