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What is the equation in slope-intercept form for the line that passes through the point (–6, –5) and is perpendicular to the graph of the equation 4x + 3y = -6?

a. y = 3x + 17
b. y = -4x - 29
c. y = -3/4x - 19/4
d. y = 4/3x + 23/3

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Final answer:

The equation in slope-intercept form for the line that passes through the point (-6, -5) and is perpendicular to the graph of the equation 4x + 3y = -6 is y = (3/4)x - 11/4.

Step-by-step explanation:

To find the equation in slope-intercept form for a line that is perpendicular to the line represented by the equation 4x + 3y = -6, we need to first determine the slope of the original line.

The equation of the original line can be rearranged to be in the form y = mx + b, where m is the slope.

So, 4x + 3y =-6 becomes 3y = -4x - 6 and then y = (-4/3)x - 2.

Since the line we want is perpendicular to this line, its slope will be the negative reciprocal of (-4/3). The negative reciprocal of a number is found by flipping its fraction and changing its sign.

So, the slope of the new line will be 3/4.

The new line passes through the point (-6, -5). We can use the slope-intercept form y = mx + b and substitute the point and the slope to find the value of b. So, -5 = (3/4)(-6) + b.

Solving for b, we get b = -5 + (3/4)(6) = -5 + 9/4 = -20/4 + 9/4 = -11/4.

Therefore, the equation in slope-intercept form for the line that passes through the point (-6, -5) and is perpendicular to the graph of the equation 4x + 3y = -6 is y = (3/4)x - 11/4.

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