07.5 Perpendicular Lines - Cardiff Tutor Company

Question 1

LEVEL 6

Find which, if any, of the following lines are perpendicular.

$a$ ) $4y=2x+12$

$b$ ) $\dfrac{1}{2}y=x+\dfrac{5}{2}$

$c$ ) $4y+2x=-16$

$d$ ) $\dfrac{1}{3}y=x+\frac{2}{3}$

Select the correct answer from the list below:

A: $b$ and $c$

B: $b$ , $c$ , and $d$

C: $b$ and $d$

D: $a$ , $b$ , $c$ , and $d$

CORRECT ANSWER: A: $b$ and $c$

WORKED SOLUTION:

To find out which lines are perpendicular we need to change them into the form $y=mx+c$ to find their $m$ values. If the lines are perpendicular then their $m$ values will multiply together to give $-1$ .

Line a
$4y=2x+12$
Divide both sides by $4$
$y=\frac{1}{2}x+3$
$m=\frac{1}{2}$

Line b
$\dfrac{1}{2}y=x+\dfrac{5}{2}$
Multiply both sides by $2$
$y=2x+5$
$m=2$

Line c
$4y+2x=-16$
Subtract $2x$ from both sides
$4y=-2x-16$
Divide both sides by $4$
$y=-\frac{1}{2}x+4$
$m=-\frac{1}{2}$

Line d
$\dfrac{1}{3}y=x+\frac{2}{3}$
Multiply both sides by $3$
$y=3x+2$
$m=3$

We now need to multiply these $m$ values together and chooses the ones that have a product of $-1$ .

From the table we can see that only lines $b$ and $c$ are perpendicular.

Question 2

LEVEL 6

Plot the graph, from $x=-3$ to $x=1$ of the straight line that is perpendicular to the line $3y-x=7$ and passes through the point $(-2,2)$ .

Select the correct graph from the list below:

CORRECT ANSWER: D

WORKED SOLUTION:

We need to change this equation into the form $y=mx+c$ to find its gradient, which we do by first adding $x$ .

$3y-x=7$
$3y=x+7$

And then dividing by $3$

y=\frac{1}{3}x+\frac{7}{3}

To find the gradient of our perpendicular line, we need to divide $-1$ by the gradient of our current line, $\dfrac{1}{3}$ .

-1\div\frac{1}{3}=-1\times3=-3

To draw our perpendicular line from the point we need to find this point, read across $1$ and then down $3$ (because it is negative). Doing this will give us two points on our line.

And now, all we need to do is connect these points with a straight line.

Question 3

LEVEL 6

Find the equation the line that is perpendicular to $4y+5x=2$ and passes through the point $(4,3)$ .

Select the correct answer from the list below:

A: $y=-\frac{5}{4}x-\frac{1}{2}$

B: $y=\frac{4}{5}x-\frac{1}{5}$

C: $y=-\frac{4}{5}x+\frac{3}{5}$

D: $y=5x+2$

CORRECT ANSWER: B: $y=\frac{4}{5}x-\frac{1}{5}$

WORKED SOLUTION:

We need to change this equation into the form $y=mx+c$ to find its gradient, which we do by first subtracting $5x$ from both sides.

$4y+5x=2$
$4y=-5x+2$

And then dividing by $3$

y=-\frac{5}{4}x+\frac{1}{2}

To find the gradient of our perpendicular line, we need to divide $-1$ by the gradient of our current line, $-\dfrac{5}{4}$ .

-1\div-\frac{5}{4}=-1\times-\frac{4}{5}=\frac{4}{5}

Now we have the gradient and the point it passes through, $(4,3)$ , we can substitute these values into $y=mx+c$ to find the value of $c$

$3=\frac{4}{5}\times4+c$

$3=\frac{16}{5}+c$

And now we need to subtract $\dfrac{16}{5}$ from both sides.

$3-\frac{16}{5}=c$

$\frac{15}{5}-\frac{16}{5}=c$

$-\frac{1}{5}=c$

$c=-\frac{1}{5}$

Therefore, the line perpendicular to $4y+5x=2$ and passing through $(4,3)$ is given by

y=\frac{4}{5}x-\frac{1}{5}

Question 4

LEVEL 6

Find the equation the line that is perpendicular to $2=y-x$ and passes through the point $(3,5)$ .

Select the correct answer from the list below:

A: $y=-x+8$

B: $y=x+8$

C: $y=-2x+4$

D: $y=-x$

CORRECT ANSWER: A: $y=-x+8$

WORKED SOLUTION:

We need to change this equation into the form $y=mx+c$ to find its gradient, which we do by adding $x$ to both sides.

$2=y-x$

$y=x+2$

To find the gradient of our perpendicular line, we need to divide $-1$ by the gradient of our current line, $1$ .

-1\div 1=-1

Now we have the gradient and the point it passes through, $(3,5)$ , we can substitute these values into $y=mx+c$ to find the value of $c$

$5=-1\times 3+c$

5=-3+c

$c=8$

Therefore, the line perpendicular to $2=y-x$ and passing through $(3,5)$ is given by

y=-x+8

Question 5

LEVEL 6

Choose the equation that is NOT perpendicular to the line in the following graph.

Select the correct answer from the list below:

A: $y = \frac{1}{3}x + 1$

B: $y = - \frac{1}{3}x + 12$

C: $3y + x = 63$

D: $x = 12 - 3y$

CORRECT ANSWER: A: $y = \frac{1}{3}x + 1$

WORKED SOLUTION:

The equation of the line in the graph is $y = 3x+3$ , which has a gradient of $3$

Therefore any line perpendicular to it has gradient $- \frac{1}{3}$ , since it is the negative reciprocal

$y = \frac{1}{3}x + 1$ is the only equation that does not have a gradient of $\frac{1}{3}$

All other equations can be rearranged into the form $y=mx+c$ , where $m = - \frac{1}{3}$