Perpendicular Lines 2

Question 1 of 7

Find the equation of a line that passes through

(4, 2)

$(4,2)$ and is perpendicular to

y = - \frac{1}{2} x + 9

$y=-1/2x+9$

1. $y = 2 x - 6$ $y=2x-6$
2. $y = 5 x - 1$ $y=5x-1$
3. $y = - 2 x + 3$ $y=-2x+3$
4. $3 x + 2 y = 1$ $3x+2y=1$

Incorrect

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

$m$ $m$ is the gradient of the line
$(x_{1}, y_{1})$ $(x_1,y_1)$ is a point that lies on the line

Remember

The gradients of perpendicular lines are negative reciprocals of each other.

First, identify the gradient of the given equation.

In gradient-intercept form

(y =

$(y=$ $m$ $m$

x + b)

$x+b)$ , $m$ $m$ is the gradient.

$y$ $y$	$=$ $=$	$- \frac{1}{2}$ $-1/2$ $x + 9$ $x+9$

$m_{1}$ $m_1$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

$m_{1}$ $m_1$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

	$=$ $=$	$- 2$ $-2$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$2$ $2$	Change the sign

Use the Point Gradient Formula to find the equation.

Point:

(x_{1}, y_{1}) = (4, 2)

$(x_1,y_1)=(4,2)$

Gradient:

m_{2} = 2

$m_2=2$

$y -$ $y-$ $y_{1}$ $y_1$	$=$ $=$	$m$ $m$ $(x -$ $(x-$ $x_{1}$ $x_1$ $)$ $)$	Point Gradient Formula
$y -$ $y-$ $2$ $2$	$=$ $=$	$2$ $2$ $(x -$ $(x-$ $4$ $4$ $)$ $)$	Substitute values
$y - 2$ $y-2$	$=$ $=$	$2 x - 8$ $2x-8$
$y - 2$ $y-2$ $+ 2$ $+2$	$=$ $=$	$2 x - 8$ $2x-8$ $+ 2$ $+2$	Add $2$ $2$ to both sides
$y$ $y$	$=$ $=$	$2 x - 6$ $2x-6$	Simplify

y = 2 x - 6

$y=2x-6$

Question 2 of 7

Check if the line

y = 2 x + 5

$y=2x+5$ is perpendicular to the line

x + 2 y - 6 = 0

$x+2y-6=0$ .

1. Perpendicular
2. Not Perpendicular

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

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

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

$y$ $y$	$=$ $=$	$2$ $2$ $x + 5$ $x+5$
$m_{1}$ $m_1$	$=$ $=$	$2$ $2$

Line 2

$x + 2 y - 6$ $x+2y-6$	$=$ $=$	$0$ $0$
$x + 2 y - 6$ $x+2y-6$ $- x + 6$ $-x+6$	$=$ $=$	$0$ $0$ $- x + 6$ $-x+6$	Add $- x + 6$ $-x+6$ to both sides
$2 y$ $2y$	$=$ $=$	$- x + 6$ $-x+6$	Simplify
$2 y$ $2y$ $\div 2$ $-:2$	$=$ $=$	$(- x + 6)$ $(-x+6)$ $\div 2$ $-:2$	Divide both sides by $2$ $2$
$y$ $y$	$=$ $=$	$- \frac{1}{2}$ $-1/2$ $x + 3$ $x + 3$

$m_{2}$ $m_2$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

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

- 1

$-1$ .

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

Therefore, Line

1

$1$ and Line

2

$2$ are perpendicular.

Perpendicular

Question 3 of 7

>Find the equation of a line that passes through $(- 8, 5)$ $(-8,5)$ and is perpendicular to $y = - 4 x + 2$ $y=-4x+2$

1. $y = 7 x + 4$ $y=7x+4$
2. $y = x + 7$ $y=x+7$
3. $y = 4 x + 3$ $y=4x+3$
4. $y = \frac{1}{4} x + 7$ $y=1/4x+7$

Incorrect

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

$m$ $color(tomato)(m)$ is the slope of the line
$(x_{1}, y_{1})$ $(\color(royalblue)(x_1,y_1) )$ is a point that lies on the line

Remember

The slope of perpendicular lines are negative reciprocals of each other.

First, identify the slope of the given equation.

In slope intercept form

(y = m x + b)

$(y= color(tomato)(m)x+b)$ ,

m

$color(tomato)(m)$ is the slope.

$y$ $y$	$=$ $=$	$- 4 x + 2$ $color(tomato)(-4)x+2$

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

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

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

	$=$ $=$	$- \frac{1}{4}$ $-1/4$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$\frac{1}{4}$ $1/4$	Change the sign

Use the point slope formula to find the equation.

Point:

(x_{1}, y_{1}) = (- 8, 5)

$(x_1,y_1)=(-8,5)$

Slope:

m_{2} = \frac{1}{4}

$m_2=1/4$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 5$ $y- color(royalblue)(5)$	$=$ $=$	$\frac{1}{4} (x - - 8)$ $color(tomato)(1/4)(x- color(royalblue)(-8))$	Substitute values
$y - 5$ $y-5$	$=$ $=$	$\frac{1}{4} x + 2$ $1/4x+2$
$y - 5 + 5$ $y-5 color(crimson)(+5)$	$=$ $=$	$\frac{1}{4} x + 2 + 5$ $1/4x+2 color(crimson)(+5)$	Add $5$ $5$ to both sides
$y$ $y$	$=$ $=$	$\frac{1}{4} x + 7$ $1/4x+7$	Simplify

y = \frac{1}{4} x + 7

$y=1/4x+7$

Question 4 of 7

>Find the equation of a line that passes through $(- 8, 10)$ $(-8,10)$ and is perpendicular to $y = 4 x + 6$ $y=4x+6$

1. $y = - \frac{1}{4} x + 12$ $y=-1/4x+12$
2. $y = 4 x + 8$ $y=4x+8$
3. $y = 4 x + 12$ $y=4x+12$
4. $y = - x + 4$ $y=-x+4$

Incorrect

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

$m$ $color(tomato)(m)$ is the slope of the line
$(x_{1}, y_{1})$ $(\color(royalblue)(x_1,y_1) )$ is a point that lies on the line

Remember

The slope of perpendicular lines are negative reciprocals of each other.

First, identify the slope of the given equation.

In slope intercept form

(y = m x + b)

$(y= color(tomato)(m)x+b)$ ,

m

$color(tomato)(m)$ is the slope.

$y$ $y$	$=$ $=$	$4 x + 6$ $color(tomato)(4)x+6$

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

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

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

	$=$ $=$	$\frac{1}{4}$ $1/4$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$- \frac{1}{4}$ $-1/4$	Change the sign

Use the point slope formula to find the equation.

Point:

(x_{1}, y_{1}) = (- 8, 10)

$(x_1,y_1)=(-8,10)$

Slope:

m_{2} = - \frac{1}{4}

$m_2=-1/4$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 10$ $y- color(royalblue)(10)$	$=$ $=$	$- \frac{1}{4} (x - - 8)$ $color(tomato)(-1/4)(x- color(royalblue)(-8))$	Substitute values
$y - 10$ $y-10$	$=$ $=$	$- \frac{1}{4} x + 2$ $-1/4x+2$
$y - 10 + 10$ $y-10 color(crimson)(+10)$	$=$ $=$	$- \frac{1}{4} x + 2 + 10$ $-1/4x+2 color(crimson)(+10)$	Add $10$ $10$ to both sides
$y$ $y$	$=$ $=$	$- \frac{1}{4} x + 12$ $-1/4x+12$	Simplify

y = - \frac{1}{4} x + 12

$y=-1/4x+12$

Question 5 of 7

>Find the equation of a line that passes through $(- 3, 5)$ $(-3,5)$ and is perpendicular to $y = 5 x + 6$ $y=5x+6$

1. $y = - \frac{1}{5} x + \frac{28}{5}$ $y=-1/5x+28/5$
2. $y = - \frac{1}{5} x + 28$ $y=-1/5x+28$
3. $y = - 15 x + 285$ $y=-15x+285$
4. $y = 15 x + 28$ $y=15x+28$

Incorrect

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

$m$ $color(tomato)(m)$ is the slope of the line
$(x_{1}, y_{1})$ $(\color(royalblue)(x_1,y_1) )$ is a point that lies on the line

Remember

The slope of perpendicular lines are negative reciprocals of each other.

First, identify the slope of the given equation.

In slope intercept form

(y = m x + b)

$(y= color(tomato)(m)x+b)$ ,

m

$color(tomato)(m)$ is the slope.

$y$ $y$	$=$ $=$	$5 x + 6$ $color(tomato)(5)x+6$

$m_{1}$ $m_1$	$=$ $=$	$5$ $5$

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

$m_{1}$ $m_1$	$=$ $=$	$5$ $5$

	$=$ $=$	$\frac{1}{5}$ $1/5$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$- \frac{1}{5}$ $-1/5$	Change the sign

Use the point slope formula to find the equation.

Point:

(x_{1}, y_{1}) = (- 3, 5)

$(x_1,y_1)=(-3,5)$

Slope:

m_{2} = - \frac{1}{5}

$m_2=-1/5$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - 5$ $y- color(royalblue)(5)$	$=$ $=$	$- \frac{1}{5} (x - - 3)$ $color(tomato)(-1/5)(x- color(royalblue)(-3))$	Substitute values
$y - 5$ $y-5$	$=$ $=$	$- \frac{1}{5} x + \frac{3}{5}$ $-1/5x+3/5$
$y - 5 + 5$ $y-5 color(crimson)(+5)$	$=$ $=$	$- \frac{1}{5} x + \frac{3}{5} + 5$ $-1/5x+3/5 color(crimson)(+5)$	Add $5$ $5$ to both sides
$y$ $y$	$=$ $=$	$- \frac{1}{5} x + \frac{28}{5}$ $-1/5x+28/5$	Simplify

y = - \frac{1}{5} x + \frac{28}{5}

$y=-1/5x+28/5$

Question 6 of 7

>Find the equation of a line that passes through $(4, - 1)$ $(4,-1)$ and is perpendicular to $y = 4 x + 8$ $y=4x+8$

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

Incorrect

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

$m$ $color(tomato)(m)$ is the slope of the line
$(x_{1}, y_{1})$ $(\color(royalblue)(x_1,y_1) )$ is a point that lies on the line

Remember

The slope of perpendicular lines are negative reciprocals of each other.

First, identify the slope of the given equation.

In slope intercept form

(y = m x + b)

$(y= color(tomato)(m)x+b)$ ,

m

$color(tomato)(m)$ is the slope.

$y$ $y$	$=$ $=$	$4 x + 8$ $color(tomato)(4)x+8$

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

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

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

	$=$ $=$	$\frac{1}{4}$ $1/4$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$- \frac{1}{4}$ $-1/4$	Change the sign

Use the point slope formula to find the equation.

Point:

(x_{1}, y_{1}) = (4, - 1)

$(x_1,y_1)=(4,-1)$

Slope:

m_{2} = - \frac{1}{4}

$m_2=-1/4$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - - 1$ $y- color(royalblue)(-1)$	$=$ $=$	$- \frac{1}{4} (x - 4)$ $color(tomato)(-1/4)(x- color(royalblue)(4))$	Substitute values
$y + 1$ $y+1$	$=$ $=$	$- \frac{1}{4} x + 1$ $-1/4x+1$
$y + 1 - 1$ $y+1 color(crimson)(-1)$	$=$ $=$	$- \frac{1}{4} x + 1 - 1$ $-1/4x+1 color(crimson)(-1)$	Subtract $1$ $1$ to both sides
$y$ $y$	$=$ $=$	$- \frac{1}{4} x$ $-1/4x$	Simplify

y = - \frac{1}{4} x

$y=-1/4x$

Question 7 of 7

>Find the equation of a line that passes through $(- 1, - 4)$ $(-1,-4)$ and is perpendicular to $9 x + 3 y = 8$ $9x+3y=8$

1. $y = x - 3$ $y=x-3$
2. $y = 3 x - 33$ $y=3x-33$
3. $y = x - 11$ $y=x-11$
4. $y = \frac{1}{3} x - \frac{11}{3}$ $y=1/3x-11/3$

Incorrect

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

$m$ $color(tomato)(m)$ is the slope of the line
$(x_{1}, y_{1})$ $(\color(royalblue)(x_1,y_1) )$ is a point that lies on the line

Remember

The slope of perpendicular lines are negative reciprocals of each other.

First, convert the equation into gradient-intercept form and identify the gradient.

In gradient-intercept form

(y =

$(y=$ $m$ $m$

x + b)

$x+b)$ , $m$ $m$ is the gradient.

$9 x + 3 y$ $9x+3y$	$=$ $=$	$8$ $8$
$9 x + 3 y$ $9x+3y$ $- 9 x$ $-9x$	$=$ $=$	$8$ $8$ $- 9 x$ $-9x$	Subtract $9 x$ $9x$ on both sides
$3 y$ $3y$	$=$ $=$	$8 - 9 x$ $8-9x$

$\frac{3}{3} y$ $3/3y$	$=$ $=$	$\frac{8}{3} - \frac{9}{3} x$ $8/3-9/3x$	Divide all terms by $3$ $3$

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

Then, identify the slope of the given equation.

In slope intercept form

(y = m x + b)

$(y= color(tomato)(m)x+b)$ ,

m

$color(tomato)(m)$ is the slope.


$y$ $y$	$=$ $=$	$- 3 x + \frac{8}{3}$ $color(tomato)(-3)x+8/3$

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

Get the negative reciprocal of the

m_{1}

$m_1$ by flipping it upside down and changing the sign.

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

	$=$ $=$	$- \frac{1}{3}$ $-1/3$	Flip the number upside down
$m_{2}$ $m_2$	$=$ $=$	$\frac{1}{3}$ $1/3$	Change the sign

Use the point slope formula to find the equation.

Point:

(x_{1}, y_{1}) = (- 1, - 4)

$(x_1,y_1)=(-1,-4)$

Slope:

m_{2} = \frac{1}{3}

$m_2=1/3$

$y - y_{1}$ $y- color(royalblue)(y_1)$	$=$ $=$	$m (x - x_{1})$ $color(tomato)(m)(x- color(royalblue)(x_1))$	Point Slope Formula
$y - - 4$ $y- color(royalblue)(-4)$	$=$ $=$	$\frac{1}{3} (x - - 1)$ $color(tomato)(1/3)(x- color(royalblue)(-1))$	Substitute values
$y + 4$ $y+4$	$=$ $=$	$\frac{1}{3} x + \frac{1}{3}$ $1/3x+1/3$
$y + 4 - 4$ $y+4 color(crimson)(-4)$	$=$ $=$	$\frac{1}{3} x + \frac{1}{3} - 4$ $1/3x+1/3 color(crimson)(-4)$	Subtract $4$ $4$ to both sides
$y$ $y$	$=$ $=$	$\frac{1}{3} x - \frac{11}{3}$ $1/3x-11/3$	Simplify

y = \frac{1}{3} x - \frac{11}{3}

$y=1/3x-11/3$

Perpendicular Lines 2

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Quiz summary

Information

1. Question

Hint

Keep Going!

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

Remember

2. Question

Hint

Well Done!

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

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

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

5. 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

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

Quizzes

Perpendicular Lines 2

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Quiz summary

Information

1. Question

Hint

Keep Going!

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

Remember

2. Question

Hint

Well Done!

Gradient Intercept Form: y=y=y=mmmx+x+x+bbb

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

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

5. 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

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

Quizzes

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

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

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