Perpendicular Lines 1

Topics

Algebra 2

Linear Equations and Graphs

Perpendicular Lines

Perpendicular Lines 1

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Question 1 of 8

1. Question
Are the two lines at right angles with each other?
- 1. Yes
- 2. No
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Perpendicular lines have gradients which are negative reciprocals of one another.

Gradient Intercept Form: $y = m x + b$ $y=mx+b$

$m$ $m$ is the gradient of the line

$b$ $b$ is the y-intercept (where the line cuts the y-axis)

The gradient is given by the coefficient of $x$ $x$ or the value of $m$ $m$ .

$y$ $y$ $=$ $=$ $- \frac{1}{2} x + 4$ $-1/2x+4$

$m$ $m$ $=$ $=$ $- \frac{1}{2}$ $-1/2$

$y$ $y$ $=$ $=$ $2 x - 3$ $2x-3$

$m$ $m$ $=$ $=$ $2$ $2$

The gradients are negative reciprocals of each other, so the lines are perpendicular or are at right angles with each other.

The lines are perpendicular.

Question 2 of 8

>Find the equation of a line that passes through $(3, 2)$ $(3,2)$ and is perpendicular to $y = x + 5$ $y=x+5$

1. $y = - 3 x + 5$ $y=-3x+5$
2. $y = - x + 5$ $y=-x+5$
3. $y = 2 x + 3$ $y=2x+3$
4. $y = x + 15$ $y=x+15$

Question 3 of 8

>Find the equation of a line that passes through $(3, 7)$ $(3,7)$ and is perpendicular to $y = 3 x - 2$ $y=3x-2$

1. $y = - \frac{1}{3} x + 8$ $y=-1/3x+8$
2. $y = 3 x + 4$ $y=3x+4$
3. $y = - x + \frac{8}{3}$ $y=-x+8/3$
4. $y = - x + 3$ $y=-x+3$

Question 4 of 8

Find the equation of a line that passes through

(6, 4)

$(6,4)$ and is perpendicular to

3 x - 4 y + 8 = 0

$3x-4y+8=0$

1. $y = - \frac{4}{3} x + 12$ $y=-4/3x+12$
2. $y = \frac{4}{3} x + 12$ $y=4/3x+12$
3. $y = - \frac{4}{3} x - 12$ $y=-4/3x-12$
4. $y = \frac{3}{4} x + 12$ $y=3/4x+12$

Question 5 of 8

Check if the line passing through

R (- 2, 2)

$R(-2,2)$ and

S (1, 5)

$S(1,5)$ is perpendicular to the line passing through

T (1, 2)

$T(1,2)$ and

U (4, - 1)

$U(4,-1)$ .

1. Not Perpendicular
2. Perpendicular

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Gradient Formula

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m=\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$

Remember

To prove the perpendicularity of two lines, the product of their gradients should be equal to $- 1$ $-1$

First, solve for the gradient of each line using the gradient formula

Line 1

$($ $($ $x_{1}$ $x_1$ $,$ $,$ $y_{1}$ $y_1$ $)$ $)$	$=$ $=$	$S ($ $S($ $- 2$ $-2$ $,$ $,$ $2$ $2$ $)$ $)$
$($ $($ $x_{2}$ $x_2$ $,$ $,$ $y_{2}$ $y_2$ $)$ $)$	$=$ $=$	$R ($ $R($ $1$ $1$ $,$ $,$ $5$ $5$ $)$ $)$

$m_{1}$ $m_1$	$=$ $=$	$\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$	Gradient Formula

	$=$ $=$	$\frac{\color{#9a00c7}{5}-\color{#9a00c7}{2}}{\color{#00880a}{1}-(\color{#00880a}{-2})}$	Substitute values

	$=$ $=$	$\frac{3}{3}$ $3/3$	Simplify

	$=$ $=$	$1$ $1$

Line 2

$($ $($ $x_{1}$ $x_1$ $,$ $,$ $y_{1}$ $y_1$ $)$ $)$	$=$ $=$	$T ($ $T($ $1$ $1$ $,$ $,$ $2$ $2$ $)$ $)$
$($ $($ $x_{2}$ $x_2$ $,$ $,$ $y_{2}$ $y_2$ $)$ $)$	$=$ $=$	$U ($ $U($ $4$ $4$ $,$ $,$ $- 1$ $-1$ $)$ $)$

$m_{2}$ $m_2$	$=$ $=$	$\frac{\color{#9a00c7}{y_2}-\color{#9a00c7}{y_1}}{\color{#00880a}{x_2}-\color{#00880a}{x_1}}$	Gradient Formula

	$=$ $=$	$\frac{\color{#9a00c7}{-1}-\color{#9a00c7}{2}}{\color{#00880a}{4}-\color{#00880a}{1}}$	Substitute values

	$=$ $=$	$- \frac{3}{3}$ $-3/3$	Simplify

	$=$ $=$	$- 1$ $-1$

To prove that these two lines are perpendicular, check if the product of the two gradients is equal to

- 1

$-1$ .

$m_{1} \times m_{2}$ $m_1 times m_2$	$=$ $=$	$1 \times - 1$ $1 times -1$
	$=$ $=$	$- 1$ $-1$

Therefore, Line

1

$1$ and Line

2

$2$ are perpendicular.

Perpendicular

Question 6 of 8

>Find the equation of a line that passes through $(2, 2)$ $(2,2)$ and is perpendicular to $y = - 2 x + 6$ $y=-2x+6$

1. $y = x - 1$ $y=x-1$
2. $y = x + 2$ $y=x+2$
3. $y = 2 x + 4$ $y=2x+4$
4. $y = \frac{1}{2} x + 1$ $y=1/2x+1$

Question 7 of 8

>Find the equation of a line that passes through $(- 3, 1)$ $(-3,1)$ and is perpendicular to $y = 4 x + 1$ $y=4x+1$

1. $y = - \frac{1}{4} x + \frac{1}{4}$ $y=-1/4x+1/4$
2. $y = - 4 x + 4$ $y=-4x+4$
3. $y = - x + 1$ $y=-x+1$
4. $y = - x + \frac{1}{4}$ $y=-x+1/4$

Question 8 of 8

>Find the equation of a line that passes through $(-6,4)$ and is perpendicular to $3y=2x-3$

1. $y=-3/2x-5$
2. $y=-3x-5$
3. $y=-3x-10$
4. $y=x-5$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 2$ $y- color(royalblue)(2)$	$=$ $=$	$- 1 (x - 3)$ $color(tomato)(-1)(x- color(royalblue)(3))$	Substitute values
$y - 2$ $y-2$	$=$ $=$	$- x + 3$ $-x+3$
$y - 2 + 2$ $y-2 color(crimson)(+2)$	$=$ $=$	$- x + 3 + 2$ $-x+3 color(crimson)(+2)$	Add $2$ $2$ to both sides
$y$ $y$	$=$ $=$	$- x + 5$ $-x+5$	Simplify

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 7$ $y- color(royalblue)(7)$	$=$ $=$	$- \frac{1}{3} (x - 3)$ $color(tomato)(-1/3)(x- color(royalblue)(3))$	Substitute values
$y - 7$ $y-7$	$=$ $=$	$- \frac{1}{3} x + 1$ $-1/3x+1$
$y - 7 + 7$ $y-7 color(crimson)(+7)$	$=$ $=$	$- \frac{1}{3} x + 1 + 7$ $-1/3x+1 color(crimson)(+7)$	Add $7$ $7$ to both sides
$y$ $y$	$=$ $=$	$- \frac{1}{3} x + 8$ $-1/3x+8$	Simplify

$3 x - 4 y + 8$ $3x-4y+8$	$=$ $=$	$0$ $0$
$3 x - 4 y + 8$ $3x-4y+8$ $+ 4 y$ $+4y$	$=$ $=$	$0$ $0$ $+ 4 y$ $+4y$	Add $4 y$ $4y$ on both sides
$3 x + 8$ $3x+8$	$=$ $=$	$4 y$ $4y$

$\frac{3}{4} x + \frac{8}{4}$ $3/4x+8/4$	$=$ $=$	$\frac{4}{4} y$ $4/4y$	Divide all terms by $4$ $4$

$\frac{3}{4} x + 2$ $3/4 x + 2$	$=$ $=$	$y$ $y$

$y$ $y$	$=$ $=$	$\frac{3}{4}$ $3/4$ $x + 2$ $x+2$	Identify the slope

$m_{1}$ $m_1$	$=$ $=$	$\frac{3}{4}$ $3/4$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point Gradient Formula

$y -$ $y-$ $4$ $4$	$=$ $=$	$- \frac{4}{3}$ $-4/3$ $(x -$ $(x-$ $6$ $6$ $)$ $)$	Substitute values

$y - 4$ $y-4$	$=$ $=$	$- \frac{4}{3} x + 8$ $-4/3x+8$

$y - 4$ $y-4$ $+ 4$ $+4$	$=$ $=$	$- \frac{4}{3} x + 8$ $-4/3x + 8$ $+ 4$ $+4$	Add $4$ $4$ to both sides

$y$ $y$	$=$ $=$	$- \frac{4}{3} x + 12$ $-4/3x +12$	Simplify

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 2$ $y- color(royalblue)(2)$	$=$ $=$	$\frac{1}{2} (x - 2)$ $color(tomato)(1/2)(x- color(royalblue)(2))$	Substitute values
$y - 2$ $y-2$	$=$ $=$	$\frac{1}{2} x - 1$ $1/2x-1$
$y - 2 + 2$ $y-2 color(crimson)(+2)$	$=$ $=$	$\frac{1}{2} x - 1 + 2$ $1/2x-1 color(crimson)(+2)$	Add $2$ $2$ to both sides
$y$ $y$	$=$ $=$	$\frac{1}{2} x + 1$ $1/2x+1$	Simplify

Perpendicular Lines 1

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Quiz summary

Information

1. Question

Hint

Correct!

Gradient Intercept Form: $y = m x + b$ $y=mx+b$

2. Question

Keep Going!

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Remember

3. Question

Keep Going!

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Remember

4. Question

Hint

Excellent!

Point Gradient Form: $y -$ $y-$ $y_{1}$ $y_1$ $=$ $=$ $m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$

Remember

5. Question

Hint

Great Work!

Gradient Formula

Remember

6. Question

Keep Going!

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Remember

7. Question

Keep Going!

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Remember

8. Question

Keep Going!

Point Slope Form: $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Remember

Quizzes

$y$ $y$	$=$ $=$	$- \frac{1}{2} x + 4$ $-1/2x+4$
$m$ $m$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

$y$ $y$	$=$ $=$	$1 x + 5$ $color(tomato)(1)x+5$

$m_{1}$ $m_1$	$=$ $=$	$1$ $1$

$y$ $y$	$=$ $=$	$3 x + 2$ $color(tomato)(3)x+2$

$m_{1}$ $m_1$	$=$ $=$	$3$ $3$

$y$ $y$	$=$ $=$	$- 2 x + 6$ $color(tomato)(-2)x+6$

$m_{1}$ $m_1$	$=$ $=$	$- 2$ $-2$

$y$ $y$	$=$ $=$	$4 x + 1$ $color(tomato)(4)x+1$

$m_{1}$ $m_1$	$=$ $=$	$4$ $4$

$3y$	$=$	$2x-3$
$y$	$=$	$2/3x-3/3$	Divide all terms by $3$

$y$	$=$	$2/3x-1$

$m_1$	$=$	$2/3$

	$=$	$3/2$	Flip the number upside down
$m_2$	$=$	$-3/2$	Change the sign

$y- color(royalblue)(y_1)$	$=$	$color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y- color(royalblue)(4)$	$=$	$color(tomato)(-3/2)(x- color(royalblue)(-6))$	Substitute values
$y-4$	$=$	$-3/2x-9$
$y-4 color(crimson)(+4)$	$=$	$-3/2x-9 color(crimson)(+4)$	Add $4$ to both sides
$y$	$=$	$-3/2x-5$	Simplify

Perpendicular Lines 1

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Quiz summary

Information

1. Question

Hint

Correct!

Gradient Intercept Form: y=mx+by=mx+by=mx+b

2. Question

Keep Going!

Point Slope Form: y-y1=m(x-x1)y−y1=m(x−x1)y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))

Remember

3. Question

Keep Going!

Point Slope Form: y-y1=m(x-x1)y−y1=m(x−x1)y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))

Remember

4. Question

Hint

Excellent!

Point Gradient Form: y-y−y-y1y1y_1===mmm(x-(x−(x-x1x1x_1)))

Remember

5. Question

Hint

Great Work!

Gradient Formula

Remember

6. Question

Keep Going!

Point Slope Form: y-y1=m(x-x1)y−y1=m(x−x1)y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))

Remember

7. Question

Keep Going!

Point Slope Form: y-y1=m(x-x1)y−y1=m(x−x1)y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))

Remember

8. Question

Keep Going!

Point Slope Form: y-y1=m(x-x1)y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))

Remember

Quizzes

Gradient Intercept Form: $y = m x + b$ $y=mx+b$

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Point Gradient Form: $y -$ $y-$ $y_{1}$ $y_1$ $=$ $=$ $m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Point Slope Form: $y - y_{1} = m (x - x_{1})$ $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$

Point Slope Form: $y- color(royalblue)(y_1)= color(tomato)(m)(x- color(royalblue)(x_1))$