Question

1. write an equation of a line perpendicular to the line 4x - 3y = 15 and passes through the point (8, -5).

Explanation:

Step1: Find slope of given line

Rewrite \(4x - 3y = 15\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(4x\) from both sides: \(-3y=-4x + 15\).
Divide by \(-3\): \(y=\frac{4}{3}x-5\). So the slope of the given line \(m_1=\frac{4}{3}\).

Step2: Find slope of perpendicular line

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

Step3: Use point - slope form to find equation

The point - slope form is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(8,-5)\) and \(m =-\frac{3}{4}\).
Substitute the values: \(y-(-5)=-\frac{3}{4}(x - 8)\).
Simplify: \(y + 5=-\frac{3}{4}x+6\).
Subtract 5 from both sides: \(y=-\frac{3}{4}x + 1\).
(We can also write it in standard form: \(3x+4y = 4\) by multiplying through by 4: \(4y=-3x + 4\), then \(3x + 4y=4\))

Answer:

The equation of the line is \(y =-\frac{3}{4}x + 1\) (or \(3x + 4y=4\))