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 Worksheet on this topic
 parallel and perpendicular lines
Students are often asked to find the equation of a line that is perpendicular to another line and that passes through a point. Watch the video tutorial below to understand how to do these problems and, if you want, download this free worksheet if you want some extra practice.
Video Tutorial
Practice Problems
Find negative reciprocal of the slope :
Slope = 1 or $$\frac{1}{1} $$
Negative reciprocal = $$\frac{1}{1} =1$$
Plug the x and y given in the equation into the slope interecept formula .
$$0 = 1\left(0\right)+b$$
Solve for b
$$0 = 0 + b$$
$$b = 0$$
Substitute b = 0 into slopeintercept equation.
$$y = x$$
Find negative reciprocal of the slope :
Slope = 5 or $$\frac{5}{1} $$
Negative reciprocal = $$\frac{1}{5} $$
Plug the x and y given in the equation into the slope interecept formula .
$$ 1=\frac{1}{5} \left(0\right)+b$$
Solve for b.
$$ 1 = 0 + b$$
$$ b = 1$$
Substitute b = 1 into slopeintercept equation.
$$y=\frac{1}{5} x+1$$
Find the negative reciprocal of the slope .
Slope = $$\frac{1}{4} $$
Negative reciprocal = $$\frac{4}{1} =4$$
Plug the x and y given in the question into the slope interecept formula .
$$ 4 = 4(4) + b$$
Solve for b.
$$ 4 = 16 + b$$
$$ 16 16$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$b = 20$$
Substitute b into slopeintercept formula.
$$y = 4x  20$$
Line $$y = 8$$ is parallel to the xaxis. So, the perpendicular line to this line must be parallel to the yaxis.
The equation of the perpendicular line is in the form $$x = n$$.
This line is perpendicular to the first line at $$y = 8$$; therefore, the second line must be perpendicular at a point with the xcoordinate of 8. That is, $$x = 8$$.
Find the negative reciprocal of the slope .
Slope = 3 or $$\frac{3}{1} $$
Negative reciprocal = $$\frac{3}{1} =3$$
Plug the x and y given in the question into the slope interecept formula .
$$ 5 = 3(0) + b$$
Solve for b.
$$5 = 0 + b$$
$$b = 5$$
Substitute b into slopeintercept formula.
$$y = 3x  5$$
Line $$x = 32$$ is a vertical line. Any line to be perpendicular to this line must be horizontal.
We can draw infinite number of perpendicular lines to this line. Some of them are, for example,
$$y = 21$$
$$y = 22$$
$$y = 23$$
$$y = 24$$
$$\dots \dots$$
$$\dots \dots$$
$$\dots \dots$$
$$y = 32$$
Out of all the infinite number of perpendicular lines only one passes through (32, 32).
Therefore, $$y = 32$$ is the only perpendicular line the problem asked for.
Multiply both sides of the equation by 3.
$$3\left(\frac{x+y}{3} \right)=3 $$
Simplify the product on the left.
$$x + y = 3$$
Subtract x from each side.
$$x + y  x = 3  x$$
Cancel out the x' on the left.
$$y = x + 3$$
Find the negative reciprocal of the slope .
Slope = 1 or $$\frac{1}{1} $$
Negative reciprocal = $$\frac{1}{1} =1$$
Plug the x and y given in the question into the slope interecept formula .
$$ 6 = 1(3) + b$$
Solve for b.
$$6 = 3 + b$$
$$ +3 +3$$
$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$
$$9 = b$$
Substitute b into slopeintercept formula.
$$y = x + 9$$
Find the slope of the given line first. The line is not in slopeIntercept Form. Add 3x + 5 to both sides.
$$3x+y53x+5=3x+5$$
Cancel out 3x and 3x, and 5 and 5 on the left.
$$y=3x+5$$
Find the negative reciprocal of the slope .
Slope = 3 or $$\frac{3}{1} $$
Negative reciprocal = $$\frac{3}{1} =3$$
Plug the x and y given in the question into the slope interecept formula .
$$5 = 3(0) + b$$
Solve for b.
$$5 = 0 + b$$
$$5 = b$$
Substitute b into slopeintercept formula.
$$y = 3x + 5$$
The slopes of the lines are as follows:
A: 7
B: $$\frac{1}{7} $$
C: $$\frac{1}{7} $$
D: $$\frac{1}{7} $$
Only the products of the slopes of the following pairs give 1.
$$(Slope of A)(Slope of B) = 1$$
$$(Slope of A)(Slope of D) = 1$$
Check the given point to see which lines it fits. Replace (2, 4) in A.
$$4 = 7(2)  10$$
$$4 = 14  10$$
$$4 = 4$$
So, A is acceptable as one of the perpendicular lines.
Replace (2, 4) in B.
$$4=\frac{1}{7} (2)+12$$
$$4=\frac{2}{7} +12$$
$$4=\frac{2+84}{7}$$
$$4=\frac{82}{7}$$
Both sides are not equal. Thus, line B is not accepted.
Replace (4, 2) in D.
$$4=\frac{1}{7} \left(2\right)+\frac{30}{7}$$
$$4=\frac{2}{7} +\frac{30}{7}$$
$$4=\frac{2+30}{7}$$
$$4=\frac{28}{7}$$
$$4=4$$
Both sides are equal. So, the given point is on D.
Point (4, 2) fits two perpendicular lines A and D. Therefore, A and D only are perpendicular lines at (2, 4).
All the equations are not in the form of $$y = mx + b$$. So, we must change their forms to y = mx + b first.
Equation K: Add 2x to both sides.
$$2x + 8y + 2x = 11 + 2x$$
Cancel out the similar terms on the left side.
$$8y = 2x + 11$$
Divide both sides by 8.
$$\frac{8y}{8} =\frac{2x}{8} +\frac{11}{8}$$
Simplify the fractions.
$$y=\frac{1}{4} x+\frac{11}{8}$$
Thus, the slope of K is $$\frac{1}{4} $$.
L: $$y = 5x  6$$
This line is already in SlopeIntercept form. The coefficient of x is 5.
L: slope is 5.
M: $$x = 4y + 12$$
This line is not in slopeintercept form. Subtract 12 from each side.
$$x  12 = 4y + 12  12$$
Cancel out numbers on the right side.
$$x  12 = 4y$$
Divide both sides by 4.
$$\frac{x}{4} \frac{12}{4} =\frac{4y}{4}$$
Simplify all the fractions.
$$y=\frac{1}{4} x3$$
That is, the slope of M is $$\frac{1}{4} $$.
N: $$y = 4x + 9$$ is in slope intercept form.
The coefficient of x is the slope of the line. Thus, 4 is the slope of N.
Slope of $$K = \frac{1}{4} $$
Slope of $$L = 5$$
Slope of $$M = \frac{1}{4} $$
Slope of $$N = 4$$
The product of the slopes of K and N is $$\left(\frac{1}{4} \right)\left(4\right)=1$$. So K and N are perpendicular.