Question

unit 8: algebraic connections to geometry - cop
select all lines that are perpendicular to ( x = - 2 )
( 3 x - y = 8 )
( x - 3 y = 9 )
line ( m ) containing points ( a ( 4, - 2 ) ) and ( b ( 4, 8 ) )
line ( n ) containing points ( c ( 3, - 1 ) ) and ( d ( - 2, - 1 ) )
( x = - 9 )
( y = 0 )

Explanation:

Step1: Analyze the slope of \(x = - 2\)

The line \(x=-2\) is a vertical line. A vertical line has an undefined slope. A line perpendicular to a vertical line is a horizontal line. A horizontal line has a slope \(m = 0\).

Step2: Analyze each option

• For the line \(3x-y=8\), rewrite it in slope - intercept form \(y = mx + b\). We get \(y=3x - 8\), where \(m = 3

eq0\).

• For the line \(x - 3y=0\), rewrite it as \(y=\frac{1}{3}x\), where \(m=\frac{1}{3}

eq0\).

• For the line passing through \(A(4,-2)\) and \(B(4,8)\), use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here \(x_1=x_2 = 4\), so the slope is undefined (vertical line).
• For the line passing through \(C(3,-1)\) and \(D(-2,-1)\), use the slope formula \(m=\frac{-1-(-1)}{-2 - 3}=\frac{0}{-5}=0\).
• The line \(x=-9\) is a vertical line (undefined slope).
• The line \(y = 0\) is a horizontal line with \(m = 0\).

Answer:

Line \(n\) containing points \(C(3,-1)\) and \(D(-2,-1)\), \(y = 0\)