Question

find the equation of the line in slope intercept form perpendicular to $4x - 3y = 5$ and goes through the point $(4, -1)$
y =
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Explanation:

Step1: Find slope of given line

Rewrite \(4x - 3y = 5\) in slope - intercept form (\(y=mx + b\)).
Subtract \(4x\) from both sides: \(-3y=-4x + 5\).
Divide by \(-3\): \(y=\frac{4}{3}x-\frac{5}{3}\).
The slope (\(m_1\)) of this line is \(\frac{4}{3}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\) (\(m_1\times m_2=-1\)).
Let \(m_2\) be the slope of the perpendicular line.
\(\frac{4}{3}\times m_2=-1\)
Solve for \(m_2\): \(m_2=-\frac{3}{4}\).

Step3: Use point - slope form to find equation

Point - slope form is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(4,-1)\) and \(m =-\frac{3}{4}\).
Substitute values: \(y-(-1)=-\frac{3}{4}(x - 4)\).
Simplify: \(y + 1=-\frac{3}{4}x+3\).
Subtract \(1\) from both sides: \(y=-\frac{3}{4}x+2\).

Answer:

\(y = -\frac{3}{4}x + 2\)