What is the equation of the line which passes through (4, -5) and is perpendicular to 3x+ 4y+ 5 = 0?a)4x - 3y - 31 = 0b)3x - 4y - 41 = 0c)4x + 3y - 1 = 0d)3x + 4y + 8 = 0Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Equation of the line perpendicular to 3x + 4y + 5 = 0 passing through (4, -5)

The given line has the equation 3x + 4y + 5 = 0. To find the equation of a line perpendicular to this, we need to determine the slope of the given line first.

- Finding the slope of the given line:
The given line is in the form Ax + By + C = 0. To find the slope, we rearrange the equation in slope-intercept form (y = mx + b), where m is the slope.
3x + 4y + 5 = 0
4y = -3x - 5
y = -3/4x - 5/4

The slope of the given line is -3/4.

- Finding the slope of the perpendicular line:
The product of the slopes of two perpendicular lines is -1. Therefore, the slope of the perpendicular line will be the negative reciprocal of -3/4, which is 4/3.

- Using point-slope form to find the equation:
We have the point (4, -5) through which the line passes and the slope 4/3. We can use the point-slope form of the equation of a line to find the equation.
y - y1 = m(x - x1)
y + 5 = 4/3(x - 4)
3y + 15 = 4x - 16
4x - 3y - 31 = 0

Therefore, the equation of the line passing through (4, -5) and perpendicular to 3x + 4y + 5 = 0 is 4x - 3y - 31 = 0, which corresponds to option 'A'.

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