Parallel and Perpendicular Lines

We need to write all 3 equations in the form y=mx+c and see which ones have the same gradient.

 

1. a) Add 1 to both sides to get

 

4y = 2x + 1

 

Then, divide both sides by 4 to get

 

y=\dfrac{1}{2}x + \dfrac{1}{4}

 

1. b) Subtract 4x from both sides to get

 

2y = -4x+5

 

Then, divide both sides by 2 to get

 

y = -2x + \dfrac{5}{2}

 

1. c) Add \dfrac{1}{2}x to both sides to get

 

y=\dfrac{1}{2}x+45

 

With all 3 equations written in the desired form, we can see that whilst b) has gradient -2, both a) and c) have gradient \dfrac{1}{2}, therefore a) and c) are parallel.

We need to find the gradient of the line given in the question by writing it in the form y=mx+c. Dividing both sides by 5, we get

 

y = -2x - \dfrac{3}{5}

 

Now, the line we need to draw is parallel to this one so must have the same gradient: -2. We’re not asked to work out the equation (although you’re welcome to do that if it helps) and knowing that the line has gradient -2 and passes through (1, 6) is enough to draw it. The correct plotting is shown below.

 

The line given in the question has gradient \dfrac{3}{7}. The negative reciprocal of this (and therefore the gradient of the perpendicular line) is

 

-\dfrac{7}{3}.

 

Now , since we have the gradient, and we know the lines passes through (5, -4), we can substitute these values into y=mx+c in order to find c. So, we get

 

-4 = \left(-\dfrac{7}{3}\right)\times 5+c = -\dfrac{35}{3}+c

 

The most important thing here is to be careful with your fraction operations. Adding \dfrac{35}{3} to both sides, we get

 

\begin{aligned}c=-4+\dfrac{35}{3}&=-\dfrac{4}{1}+\dfrac{35}{3}\\ \\&=-\dfrac{12}{3}+\dfrac{35}{3}=\dfrac{23}{3}\end{aligned}

 

Therefore, the equation of the line is

 

y=-\dfrac{7}{3}x+\dfrac{23}{3}

The gradient of the given line is 5, which means the gradient of the parallel line must also be 5. We’re given that it passes through (1, 3), and we now know the gradient to be 5, so we can substitute these values into y=mx+c in order to find c. Doing so, we get

 

3 = 1\times 5 +c = 5+c, therefore c =3-5=-2.

 

So, the equation of this line is

 

y=5x-2.

 

We now have plenty of information to plot the graph. The result looks like the figure below.

The remainder of this topic is only relevant for the higher course.

If two lines are perpendicular, then the product of their two gradients (i.e., the result of multiplying them together) is -1. Another way of putting this is: if you have a straight line with gradient m, then a line which is perpendicular to it will have gradient -\dfrac{1}{m}. This is often referred to as the negative reciprocal of m.

 

The gradient of the given line is -3, so the gradient of the perpendicular line must be

 

-\left(\dfrac{1}{-3}\right) = \dfrac{1}{3}

The two minus signs cancel, so the result is a positive number. The gradient of one line and another line perpendicular to it will always have opposite signs – if one is negative the other will always be positive and vice versa.

 

We’re given that the line passes through (-9, -2), and we now know the gradient is \dfrac{1}{3}, so we can substitute these values into y=mx+c in order to find c. Doing so, we get

 

-2=(-9)\times\dfrac{1}{3}+c=-3+c, therefore c=-2+3=1.

 

So, the equation of this line is

 

y=\dfrac{1}{3}x+1.

 

We now have plenty of information to plot the graph. The result looks like the figure below.