Question

4.3 extra practice
in exercises 1 and 2, determine which of the lines, if any, are parallel. explain.

1. line a passes through (-1, 4) and (1, 5).

line b passes through (-2, 7) and (0, 4).
line c passes through (0, 4) and (2, 5)

1. line a: 6y = -x + 12

line b: x = 6y + 5
line c: -6y + x = 5
in exercises 3 and 4, write an equation of the line that passes through the given point and is parallel to the given line.

1. (14, 3); 2y - x = 8
2. (3, -8); 3y = 2x - 1

in exercises 5 and 6, determine which of the lines, if any, are parallel or perpendicular. explain.

1. line a passes through (-5, -2) and (1, -1).

line b passes through (-3, 5) and (3, 6).
line c passes through (0, 7) and (1, 1).

1. line a: -x + 2y = 3

line b: -6x = 3y - 1
line c: 4x - 8y = 5
in exercises 7 and 8, write an equation of the line that passes through the given point and is perpendicular to the given line.

1. (-3, 1); y = -5x + 2
2. (8, -5); y = 2x + 3
3. a telecommunications company will be laying new fiber optic cables underground. one cable will be perpendicular to the road shown, passing through the point (7, 2). at what point will the cable pass through the road?

graph of a road with points (-3, 0) and (0, -2) plotted
in exercises 10–12, determine whether the statement is always, sometimes, or never true. explain your reasoning.

1. a line with a positive slope and a line with a negative slope are perpendicular.
2. a vertical line and a horizontal line are perpendicular.
3. two horizontal lines are perpendicular.

Explanation:

Step1: Analyze the road's slope

The road passes through \((-3, 0)\) and \((0, -2)\). The slope \(m_{road}\) is calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 0}{0 - (-3)}=\frac{-2}{3}=-\frac{2}{3}\).

Step2: Find the slope of the cable

Since the cable is perpendicular to the road, the slope of the cable \(m_{cable}\) is the negative reciprocal of \(m_{road}\). So \(m_{cable}=\frac{3}{2}\) (because the negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\)).

Step3: Write the equation of the cable

The cable passes through \((7, 2)\) with slope \(\frac{3}{2}\). Using the point - slope form \(y - y_1=m(x - x_1)\), we have \(y - 2=\frac{3}{2}(x - 7)\).
Simplify it: \(y-2=\frac{3}{2}x-\frac{21}{2}\), then \(y=\frac{3}{2}x-\frac{21}{2}+2=\frac{3}{2}x-\frac{21 + 4}{2}=\frac{3}{2}x-\frac{17}{2}\).

Step4: Find the equation of the road

Using the two - point form for the road with points \((-3,0)\) and \((0,-2)\), the slope - intercept form \(y = mx + b\), where \(m =-\frac{2}{3}\) and \(b=-2\) (from the point \((0,-2)\)), so the equation of the road is \(y =-\frac{2}{3}x-2\).

Step5: Find the intersection point

Set the equations of the cable and the road equal to each other:
\(\frac{3}{2}x-\frac{17}{2}=-\frac{2}{3}x - 2\)
Multiply through by 6 to clear the fractions:
\(6\times\frac{3}{2}x-6\times\frac{17}{2}=6\times(-\frac{2}{3}x)-6\times2\)
\(9x - 51=-4x-12\)
Add \(4x\) to both sides: \(9x + 4x-51=-4x + 4x-12\), \(13x-51=-12\)
Add 51 to both sides: \(13x-51 + 51=-12 + 51\), \(13x = 39\)
Divide both sides by 13: \(x = 3\)
Substitute \(x = 3\) into the equation of the road \(y=-\frac{2}{3}x-2\): \(y=-\frac{2}{3}\times3-2=-2 - 2=-4\)

Answer:

The cable will pass through the road at the point \((3,-4)\)