The equation of the line that passes through (4, -5) and is perpendicular to 3x + 4y + 5 = 0 is: 4x - 3y - 31 = 0 .
The equation of the perpendicular line:
Step 1: Find the slope of the given line.
The given line is 3x + 4y + 5 = 0.
To find its slope, we need to rearrange it to slope-intercept form (y = mx + b):
Move the x term to the other side: 4y = -3x - 5
Divide both sides by 4: y = (-3/4)x - 5/4
Therefore, the slope (m) of the given line is -3/4.
Step 2: Find the negative reciprocal of the slope.
Since the line we want is perpendicular to the given line, its slope (m') will be the negative reciprocal of m.
Negative reciprocal of -3/4 is 4/3.
Step 3: Use the point-slope form to write the equation of the perpendicular line.
We know the point-slope form of a line is:
y - y₁ = m'(x - x₁)
where (x₁, y₁) is the point the line passes through, and m' is the slope.
In our case, (x₁, y₁) = (4, -5) and m' = 4/3.
Plug these values into the formula:
y - (-5) = 4/3(x - 4)
Simplify:
y + 5 = 4/3 x - 16/3
Move the constant term to the other side:
y = 4/3 x - 31/3
Therefore, the equation of the line that passes through (4, -5) and is perpendicular to 3x + 4y + 5 = 0 is: 4x - 3y - 31 = 0 .
Question
What is the equation of the line which passes through (4,−5) and is perpendicular to 3x+4y+5=0 ? 4x-3y -31=0 3x-4y -41 =0 4x+ 3y -1=0 3x+4y+8=0