Question

1. write an equation, in point - slope form, for the line that is perpendicular to (-x + 2y = 8) and passes through the point ((6,2)).

Explanation:

Step1: Find slope of given line

First, rewrite \(-x + 2y = 8\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Starting with \(-x + 2y=8\), add \(x\) to both sides: \(2y=x + 8\). Then divide by 2: \(y=\frac{1}{2}x + 4\). The slope of this line, \(m_1=\frac{1}{2}\).
If two lines are perpendicular, the product of their slopes \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\). So \(\frac{1}{2}\times m_2=-1\), which gives \(m_2=-2\).

Step2: Use point - slope form

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope.
We know that the line passes through the point \((6,2)\) (so \(x_1 = 6\), \(y_1=2\)) and has a slope \(m=-2\).
Substitute these values into the point - slope formula: \(y - 2=-2(x - 6)\).

Answer:

The equation of the line in point - slope form is \(y - 2=-2(x - 6)\)