Question

what is an equation of the line that passes through the point (-4, 4) and is perpendicular to the line 4x - 5y = 10?

Explanation:

Step1: Find slope of given line

Rewrite \(4x - 5y = 10\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(4x\) from both sides: \(-5y=-4x + 10\).
Divide by \(-5\): \(y=\frac{4}{5}x-2\). The slope of this line, \(m_1=\frac{4}{5}\).

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. Then \(\frac{4}{5}\times m_2=-1\), so \(m_2 =-\frac{5}{4}\).

Step3: Use point - slope form

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(-4,4)\) and \(m =-\frac{5}{4}\).
Substitute the values: \(y - 4=-\frac{5}{4}(x + 4)\).
Expand: \(y - 4=-\frac{5}{4}x-5\).
Add 4 to both sides: \(y=-\frac{5}{4}x-1\).
(Or in standard form: \(5x + 4y=-4\))

Answer:

\(y =-\frac{5}{4}x - 1\) (or \(5x+4y=-4\))