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}

Which of the lines are parallel?

i : i i : i i i : 4 x + y + 2 = 0 2 y + 4 = 8 x 8 x = 6 - 2 y

If two nonvertical lines are parallel they have the same slope. Let's recall the slope-intercept form of a linear function. y= mx+ b Here, m is the slope and b is the y-intercept of the line. To identify the slopes of the lines, we need to rewrite the equations into slope-intercept form.

Line i

The slope of line i is - 4.

Line ii

Line ii has a slope of 4.

Line iii

This linear function labeled iii has the slope - 4.

Conclusion

Comparing the lines, we see that i and iii both have a slope of - 4. Since the lines also have different y-intercept, b = - 2 and b = 3, they do not coincide. Therefore, line i and line iii, are parallel.

Write the equation of the line that passes through $(- 2, 2)$ and is parallel to the graph of the linear function in the diagram.

When lines are parallel, they have the same slope. Therefore, we first need to find the slope of the given graph which is defined as the ratio of rise to run, or Δ y over Δ x.

As we can see, the rise is 3 and the run is 1. With this information, we can determine the slope.

Since the line we want to find is parallel to the one in the graph it must have the same slope. We can now start writing its equation. y= mx+ b ⇓ y= 3x+ b To complete the equation, we must find the y-intercept b. It can be found by substituting the known point (-2,2) into the equation.

Now we can complete the equation. y=3x+8

Write the equation of the line that passes through

(5, 4)

and is perpendicular to

y = 2 x + 9 .

To write an equation of a line perpendicular to the given equation, we first need to determine its slope. Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of the slopes of perpendicular lines equals -1. m_1* m_2=-1 Notice that the given equation is written in slope-intercept form. y= 2x+9 In the above formula we can see that the slope is 2. By substituting this value for m_1 into our equation, we can solve for the slope m_2 of the perpendicular line.

Any perpendicular line to the given equation will have a slope of m= - 12. With this information, we can write a partial equation in slope-intercept form for a perpendicular line to the given equation. y= -1/2x+b By substituting the given point (5,4) into this equation we can solve for the y-intercept b of the perpendicular line.

Now that we have the y-intercept, we can write the equation of the perpendicular line. y=-1/2x+ 13/2

Lines

i

and

i i

are parallel and have different

y -

intercepts. Describe the transformation from the graph of

i

i i .

Line i : Line i i : y - intercept of 4 y - intercept of - 2

First, let's recall the slope-intercept form of linear functions. y= mx+ b Here m is the slope and b is the y-intercept. We have two lines i and ii that have y-intercepts of 4 and - 2 respectively. Let's write this into the equations of the lines. Linei & Lineii y= m_ix+ 4 & y= m_(ii)x+( - 2) We know that the lines are parallel. Therefore, they have the same slope. Let's call this slope m. Linei & Lineii y= mx+ 4 & y= mx+( - 2) Let's explore these lines for various slopes.

Notice that for any slope the vertical distance between the lines is always 6 units. Therefore, a vertical translation 6 units down transforms the graph of i to ii.

The linear function

f (x)

passes through the origin and the point

(2, 6) .

Find the equation of a line parallel to

f (x)

and has a

y -

intercept of

- 3 .

Parallel lines have the same slope. Therefore, we will start by finding the slope of the line parallel to f(x). Then we will substitute the y-intercept into the equation.

Finding the Slope

We can find the slope of f(x) by substituting the points that its line passes through into the Slope Formula.

Now that we have found the slope, we can start writing the equation of the line. y= mx+b ⇓ y= 3x+b

Substituting the y-intercept

From the exercise, we also know that the line has a y-intercept of - 3. Let's substitute this into our equation.

The equation of the line is y=3x-3.

Find the number of solutions to the system of equations without solving the system.

⎩ ⎪ ⎨ ⎪ ⎧ - 9 x + 12 = 6 y 3 x + 2 y = 4

⎩ ⎪ ⎨ ⎪ ⎧ 4 x + 2 y = - 1 3 y - \frac{3}{2} = - 6 x

We have been given a system of equations. - 9x+12=6y & (I) 3x+2y=4 & (II) We want to find the number of solutions it has without solving it. Let's recall what the possible number of solutions are and the conditions for each one.

Number of Solutions	Condition
No solution	The lines have the same slope but different y-intercepts.
One solution	The lines have different slopes.
Infinitely many solutions	The lines have the same slope and the same y-intercept.

As we can see, we need to compare the slope and the y-intercept of the equations. Recall that if a line is written in slope-intercept form we can identify the slope m and the y-intercept b. y= mx+ b Let's rewrite the equations in our system in slope-intercept form.

The lines have the same slope - 32 and the same y-intercept 2. Therefore, the lines are coincidental and the system has infinitely many solutions.

In Part A we went over the conditions that applies to a system of linear equations with no solution, one solution, and infinitely many solutions. Like in Part A, we have to rewrite the equations in slope-intercept form.

These equations have the same slope but different y-intercepts. Therefore, they are parallel lines which means they never intersect. Therefore, the system has no solution.

	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 i

Line ii

Line iii

Conclusion

Finding the Slope

Substituting the y-intercept

Slope of Parallel and Perpendicular Lines

Recommended exercises

Part $1$

Part $2$

Step $1$

Step $2$

Part $1$

Part $2$