# PERPENDICULAR LINES IN THE COORDINATE PLANE WORKSHEET

## About "Perpendicular lines in the coordinate plane worksheet"

Perpendicular lines in the coordinate plane worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on perpendicular lines in the coordinate plane.

## Perpendicular lines in the coordinate plane worksheet - Problems

Problem 1 :

In the diagram given below, find the slope of each line. Determine whether the lines jand jare perpendicular.

Problem 2 :

In the diagram given below, find the slope of each line. Determine whether the lines are perpendicular.

Problem 3 :

Decide whether the lines are perpendicular.

Line 1 : y = 3x/4 + 2

Line 2 : y = -4x/3 - 3

Problem 4 :

Decide whether the lines are perpendicular.

Line 1 : 4x + 5y = 2

Line 2 : 5x + 4y = 3

Problem 5 :

In the diagram given below, the equation y = 3x/2 + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point ?

## Perpendicular lines in the coordinate plane worksheet - Solution

Problem 1 :

In the diagram given below, find the slope of each line. Determine whether the lines j1 and j2 are perpendicular.

Solution :

Part 1 :

Find the slope of the line j1. Line j1 is passing through the points (0, 3) and (3, 1).

Let (x1, y1) =  (0, 3) and (x2, y2)  =  (3, 1)

Slope (j1)  =  (y2 - y1)  / (x2 - x1)

Slope (j1)  =  (1 - 3)  / (3 - 0)

Slope (j1)  =  - 2 / 3

Part 2 :

Find the slope of the line j2. Line j2 is passing through the points (0, 3) and (-4, -3).

Let (x1, y1)  =  (0, 3) and (x2, y2)  =  (-4, -3)

Slope (j2)  =  (-3 - 3)  / (-4 - 0)

Slope (j2)  =  (-6)  / (-4)

Slope (j2)  =  3 / 2

Multiply the slopes :

The product is

= (-2/3) x (3/2)

= - 1

Since the product of slopes of the lines j1 and j2 is -1, the lines  j1 and j2 are perpendicular.

Problem 2 :

In the diagram given below, find the slope of each line. Determine whether the lines are perpendicular.

Solution :

Part 1 :

Find the slope of the line AC. Line AC is passing through the points (1, -4) and (4, 2).

Let (x1, y1) =  (1, -4) and (x2, y2)  =  (4, 2)

Slope (AC)  =  (y2 - y1)  / (x2 - x1)

Slope (AC)  =  [(2 - (-4)]  / (4 - 1)

Slope (AC)  =  (2 + 4) / 3

Slope (AC)  =  6 / 3

Slope (AC)  =  2

Part 2 :

Find the slope of the line BD. Line BD is passing through the points (-1, 2) and (5, -1).

Let (x1, y1)  =  (-1, 2) and (x2, y2)  =  (5, -1)

Slope (BD)  =  (-1 - 2)  / [(5 - (-1)]

Slope (BD)  =  (-3)  / 6

Slope (BD)  =  -1 / 2

Multiply the slopes :

The product is

= (2) x (-1/2)

= - 1

Since the product of slopes of the lines is -1, the lines  AC and BD are perpendicular.

Problem 3 :

Decide whether the lines are perpendicular.

Line 1 : y = 3x/4 + 2

Line 2 : y = -4x/3 - 3

Solution :

When we compare the given equations to slope intercept equation of a line y = mx + b, we get

slope of line 1  =  3/4

slope of line 2  =  -4/3

Multiply the slopes :

The product is

= (3/4) x (-4/3)

= - 1

Since the product of slopes of the lines is -1, the given lines are perpendicular.

Problem 4 :

Decide whether the lines are perpendicular.

Line 1 : 4x + 5y = 2

Line 2 : 5x + 4y = 3

Solution :

Rewrite each equation in slope-intercept form to find the slope.

 Line 14x + 5y = 25y = -4x + 2y = -4x/5 + 2/5Slope = -4/5 Line 25x + 4y = 34y = -5x + 3y = -5x/4 + 3/4Slope = -5/4

Multiply the slopes :

The product is

= (-4/5) x (-5/4)

= 1

Since the product of slopes of the lines is not -1, the given lines  are not perpendicular.

Problem 5 :

In the diagram given below, the equation y = 3x/2 + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point ?

Solution :

The slope of the mirror is 3/2. So, the slope of the line p is -2/3.

Let y = mx + b be the equation of the line p.

Substitute (x, y)  =  (-2, 0) and m  =  -2/3 to find the value of b.

0  =  (-2/3)(-2) + b

0  =  4/3 + b

Subtract 4/3 from both sides.

-4/3  =  b

So, the equation of the line p is

y  =  -2x/3 - 4/3

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