Question

1. find the equation of the line that passes through the point (5, -2) and is perpendicular to the line that passes through the points (-5, 7) and (-8, -2).

Explanation:

Step1: Find slope of given line

The slope \( m \) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). For points \((-5,7)\) and \((-8,-2)\), we have \( x_1=-5,y_1 = 7,x_2=-8,y_2=-2 \). So slope \( m_1=\frac{-2 - 7}{-8-(-5)}=\frac{-9}{-3}=3 \).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the required line be \( m_2 \). Then \( m_1\times m_2=-1 \). Since \( m_1 = 3 \), we get \( 3\times m_2=-1\), so \( m_2=-\frac{1}{3} \).

Step3: Use point - slope form to find equation

The point - slope form of a line is \( y - y_0=m(x - x_0) \), where \((x_0,y_0)=(5,-2)\) and \( m =-\frac{1}{3} \). Substituting these values, we get \( y-(-2)=-\frac{1}{3}(x - 5) \).
Simplify the equation:
\( y + 2=-\frac{1}{3}x+\frac{5}{3} \)
\( y=-\frac{1}{3}x+\frac{5}{3}-2 \)
\( y=-\frac{1}{3}x+\frac{5 - 6}{3} \)
\( y=-\frac{1}{3}x-\frac{1}{3} \)
Multiply through by 3 to get the standard form (or keep in slope - intercept form): \( x + 3y=- 1 \) (or \( y=-\frac{1}{3}x-\frac{1}{3} \))

Answer:

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