Understanding Slope of Parallel and Perpendicular Lines in Geometry

Equation	Slope	$y -$ Intercept
$y = - 5 x + (- 3)$	$- 5$	$- 3$
$y = - 5 x + (- 1)$	$- 5$	$- 1$

Equation

Slope

y -

Intercept

y = - 5 x + (- 3)

- 5

- 3

y = - 5 x + (- 1)

- 5

- 1

Condition	Conclusion	Example
The lines of the system have the same slope and the same $y -$ intercept.	The lines are coincidental. This means that there are infinitely many points of intersection. Therefore, the system has infinitely many solutions.	${y = 3 x + 1 y = 3 x + 1$
The lines of the system have the same slope but different $y -$ intercept.	The lines are parallel. This means that there is not a point of intersection. Therefore, the system has no solution.	${y = 3 x + 1 y = 3 x + 5$
The lines of the system have different slopes.	The lines are neither parallel nor coincidental. This means that there is one point of intersection. Therefore, the system has one solution.	${y = 3 x + 1 y = - 2 x + 5$

Condition

Conclusion

Example

The lines of the system have the same slope and the same

y -

intercept.

The lines are coincidental. This means that there are infinitely many points of intersection. Therefore, the system has infinitely many solutions.

{y = 3 x + 1 y = 3 x + 1

The lines of the system have the same slope but different

y -

intercept.

The lines are parallel. This means that there is not a point of intersection. Therefore, the system has no solution.

{y = 3 x + 1 y = 3 x + 5

The lines of the system have different slopes.

The lines are neither parallel nor coincidental. This means that there is one point of intersection. Therefore, the system has one solution.

{y = 3 x + 1 y = - 2 x + 5

Side	Points	$Distance Formula (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A O}$	$A (1, m_{1})$ $&$ $O (0, 0)$	$(0 - 1)^{2} + (0 - m_{1})^{2}$	$1 + m_{1}^{2}$
$\overline{C O}$	$C (1, m_{2})$ $&$ $O (0, 0)$	$(0 - 0)^{2} + (0 - m_{2})^{2}$	$1 + m_{2}^{2}$
$\overline{C A}$	$C (1, m_{2})$ $&$ $A (1, m_{1})$	$(1 - 1)^{2} + (m_{1} - m_{2})^{2}$	$m_{1} - m_{2}$

Side

Points

Distance Formula (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Length

\overline{A O}

A (1, m_{1})

&

O (0, 0)

(0 - 1)^{2} + (0 - m_{1})^{2}

1 + m_{1}^{2}

\overline{C O}

C (1, m_{2})

&

O (0, 0)

(0 - 0)^{2} + (0 - m_{2})^{2}

1 + m_{2}^{2}

\overline{C A}

C (1, m_{2})

&

A (1, m_{1})

(1 - 1)^{2} + (m_{1} - m_{2})^{2}

m_{1} - m_{2}

Is the following statement

Always true,
Sometimes true, or
Never true ?

The $x$ -axis is parallel to any other horizontal line.

Two lines with a slope of $2$ are parallel.

Parallel lines have the same slope. The x-axis is a horizontal line. Therefore, it has a slope of 0, like any other horizontal line. Let's examine the graphs of some horizontal lines.

We see that each line is both horizontal and parallel with the x-axis. But let's check another line and draw the graph of y=0.

This line is not parallel with the x-axis as it is coincidental with it. Therefore, the statement is sometimes true.

We know that both lines have a slope of 2. Therefore, let's write their equations in slope-intercept form to investigate if they are parallel. Equation1 y= 2x+b_1 Equation2 y= 2x+b_2 Recall that there are two conditions for parallel lines. One is that they have the same slope. This is true for the two lines. The other condition is that the y-intercepts should not be equal. b_1 ≠ b_2 In other words, the two lines with a slope of 2 are parallel if they also have different y-intercepts. However, if both lines' y-intercepts are equal to the others, the lines are coincidental. Therefore, the statement is sometimes true.

The following equations describe two lines.

Line 1 : 4 y = 2 - x Line 2 : 6 y = p x - 5

What value of

p

makes the lines parallel?

What value of

p

makes the lines perpendicular?

Since the lines are parallel, they will have the same slope. To identify their slopes we will start by rewriting both equations into slope-intercept form.

Line 1

Line 2

Now, we can identify the slopes of the given equations. y=- 1/4x + 1/2 &⇒ m_1= -1/4 [1em] y= p/6x-5/6 &⇒ m_2= p/6 By equating the slopes of the lines, we can find the value of p for which the lines are parallel.

The value of p that will make the lines parallel is - 32.

Two lines are perpendicular when their slopes are opposite reciprocals. This means that their product equals -1. m_1 * m_2=-1 Let's substitutes the slopes we found in Part A into this relationship and solve the equation for p.

When p=24, the lines are perpendicular.

The linear function passes through the points

(1, 2)

and

(3, - 2) .

Find the slope of a line perpendicular to it.

We will first find the slope of the linear function. Then we will find the slope of a line perpendicular to it using that the slopes of perpendicular lines are negative reciprocals.

Finding the Slope of the Function

Let's recall the Slope Formula. m = y_2-y_1/x_2-x_1 Since we have been given two points on the linear function we can substitute these into the formula to find the slope.

The linear function has a slope of - 2.

Finding the Slope of the Perpendicular Line

The slopes of perpendicular lines are negative reciprocals. This means that the product of their slopes equals - 1. m_1 * m_2 = - 1 Let's substitute the slope of the linear function into this relationship to find the slope of the line that is perpendicular to it.

The line perpendicular to the linear function has a slope of 12.

Write the equation of the line that passes through $(4, 1)$ and is perpendicular to the graph of the linear function in the diagram.

When two lines are perpendicular, their slopes multiply to - 1. m_1* m_2 =- 1 First, we need to find the slope of the graph we have been given. This is defined as the ratio of rise over run, or Δ y over Δ x.

As we can see, the rise is -2 and the run is 1. With this information, we can determine the slope.

We will now use this slope to find the slope of the line that passes through (4,1) and is perpendicular to the given line.

We can now start writing the equation of the line. y= mx+ b ⇓ y= 1/2x+ b To complete this equation, we must find its y-intercept. This can be found by substituting the known point (4,1) into the equation and solving for b.

Now that we know the y-intercept, we can write the equation of the line. y=1/2x - 1

	14 Theory slides
	12 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Slope of Parallel and Perpendicular Lines

Catch-Up and Review

Proof

Part 1

Part 2

Hint

Solution

Answer

Hint

Solution

Step 1

Step 2

Answer

Hint

Solution

Proof

Part 1

Part 2

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Line 1

Line 2

Finding the Slope of the Function

Finding the Slope of the Perpendicular Line

Slope of Parallel and Perpendicular Lines

Recommended exercises

Part $1$

Part $2$

Step $1$

Step $2$

Part $1$

Part $2$