National 5 Maths
Straight Line

We have been given the gradient \((m)\) and the coordinates of a point on the line.

There are two good ways of finding the \(y\)-intercept \((c)\small.\)

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

In this method, we use \((a,\,b)\) for the known point. So \((a,\,b)=(2,\,-3)\small.\)

We substitute \(m\!=\!-4\small,\,\) \(a\!=\!2\small,\,\) \(b\!=\!-3\):

$$ \begin{gather} y-b=m(x-a)\\[6pt] y-(-3)=-4(x-2)\\[6pt] y+3=-4x+8\\[6pt] y=-4x+5\\[6pt] \end{gather} $$

We can read the \(y\)-intercept from this equation: \(c=5\small.\)

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

In this method, we use \((x,\,y)=(2,\,-3)\) because we know that this point satisfies the equation of the line.

So we substitute \(m\!=\!-4\small,\,\) \(x\!=\!2\small,\,\) \(y\!=\!-3\):

$$ \begin{gather} y=mx+c\\[6pt] -3=-4\!\times\!2+c\\[6pt] -3=-8+c\\[6pt] c=-3+8\\[6pt] c=5 \end{gather} $$

Video solution by Clelland Maths 

This example differs from the previous example because we have been given the \(y\)-intercept, not the gradient.

So we can substitute \(c\!=\!-11\small,\,\) \(x\!=\!-3\small,\,\) \(y\!=\!4\,\) into the general equation of a straight line:

$$ \begin{gather} y=mx+c\\[6pt] 4=m(-3)+(-11)\\[6pt] 4=-3m-11\\[6pt] 4+11=-3m\\[6pt] -3m=15\\[6pt] m=-5 \end{gather} $$

So the equation of the line is \(y=-5x-11\small.\)

Note: An alternative method is to use the gradient formula. Because \(c\!=\!-11\small,\) the point \((0,\,-11)\) is on the line. So we can use \((x_1,\,y_1)=(0,\,-11)\) and \((x_2,\,y_2)=(-3,\,4)\) to find the gradient. Why not try it this way for yourself?

Video solution by Clelland Maths 

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

Substitute \(m\!=\!2\) and \((a,\,b)=(3,\,-1)\):

$$ \begin{gather} y-b=m(x-a)\\[6pt] y+1=2(x-3)\\[6pt] y+1=2x-6\\[6pt] y=2x-7 \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

Substitute \(m\!=\!2\) and \((x,\,y)=(3,\,-1)\):

$$ \begin{gather} y=mx+c\\[6pt] -1=2\!\times\!3+c\\[6pt] -1=6+c\\[6pt] c=-7\\[6pt] y=2x-7 \end{gather} $$

Video solution by Clelland Maths 

This looks almost identical to the previous example, but the gradient is a fraction, which complicates matters.

Answering in the form \(y\!=\!mx\!+\!c\) is acceptable when \(m\) is a fraction, but we prefer to multiply the equation through by its denominator, so that our final equation involves only integers.

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

Substitute \(m\!=\!\frac{3}{4}\small,\,\) \(a\!=\!-2\small,\,\) \(b\!=\!5\):

$$ \begin{gather} y-b=m(x-a)\\[6pt] y-5=\textstyle\frac34\normalsize(x+2)\\[6pt] 4(y-5)=3(x+2)\\[8pt] 4y-20=3x+6\\[8pt] 4y=3x+26\\[8pt] \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

Substitute \(m\!=\!\frac{3}{4}\small,\,\) \(x\!=\!-2\small,\,\) \(y\!=\!5\):

$$ \begin{gather} y=mx+c\\[6pt] 5=\textstyle\frac{3}{4}\!\times\!-2+c\\[6pt] 5=-\textstyle\frac{3}{2}+c\\[6pt] c=5+\textstyle\frac{3}{2}\\[6pt] c=\textstyle\frac{13}{2} \end{gather} $$

So the equation of the line is \(y=\frac{3}{4}x+\frac{13}{2}\small.\)

As mentioned above, we would prefer to multiply this through by \(4\) and give the equation as \(4y=3x+26\small,\) but that is really just a style preference. The forms \(3x-4y=-26\) or \(3x-4y+26=0\) are also acceptable.

Note: Method 1 is easier here. We always recommend using \(y\!-\!b\!=\!m(x\!-\!a)\) when the gradient is a fraction. It avoids fiddly fraction arithmetic and makes it easier to multiply through by the denominator.

Video solution by Clelland Maths 

The difference between this and the previous examples is that we need to find the gradient first.

It doesn't matter which point we call \((x_1,\,y_1)\) and which we call \((x_2,\,y_2)\small.\)

So let's use \((x_1,\,y_1)=(1,\,7)\) and \((x_2,\,y_2)=(-1,\,3)\):

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{3-7}{-1-1}\\[6pt] &=&\ \frac{-4}{-2}\\[6pt] &=&\ \,2 \end{eqnarray} $$

Now that we know the gradient, we can find the equation of the line using either of the following two methods.

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

We could choose either point as \((a,\,b)\small,\) because they are both on the line. Let's use \((a,\,b)=(1,\,7)\small.\)

$$ \begin{gather} y-b=m(x-a)\\[6pt] y-7=2(x-1)\\[6pt] y-7=2x-2\\[6pt] y=2x+5 \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

Again, we could use either point, but in this method it's \((x,\,y)\small,\) not \((a,\,b)\):

$$ \begin{gather} y=mx+c\\[6pt] 7=2(1)+c\\[6pt] 7=2+c\\[6pt] c=5\\[6pt] y=2x+5 \end{gather} $$

Video solution by Clelland Maths 

First we need to find the gradient of the line.

It doesn't matter which point we call \((x_1,\,y_1)\) and which we call \((x_2,\,y_2)\small.\)

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{-1-5}{2-(-2)}\\[6pt] &=&\ \small\frac{-6}{4}\\[6pt] &=&\ \small-\!\frac32 \end{eqnarray} $$

As you can see, the gradient is a fraction. So we would recommend Method 1 below, but both methods work.

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

We can use either point. Let's use \((a,\,b)=(-2,\,5)\small.\)
$$ \begin{gather} y-b=m(x-a)\\[6pt] y-5=\textstyle{-\frac32}(x+2)\\[6pt] 2(y-5)=-3(x+2)\\[6pt] 2y-10 =-3x-6\\[6pt] 2y=-3x+4\\[6pt] \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

We can use either point. Let's use \((x,\,y)=(2,\,-1)\) this time, so that you can see that it gives the same equation.
$$ \begin{gather} y=mx+c\\[6pt] 5=\textstyle{\left(-\frac32\right)}\!\times\!\left(-2\right)+c\\[6pt] 5=3+c\\[6pt] c=2\\[6pt] y=\textstyle{-\frac32}x+2 \end{gather} $$

You could keep the equation in this form, if you wish, but we prefer straight line equations not to contain fractions. So we would multiply through by \(2\) and give equation in the form \(2y=-3x+4\small,\) the same as Method 1.

Note: Other forms of the equation, such as \(3x+2y=4\) or \(3x+2y-4=0\), will also receive full marks. The important thing is that there is only one constant term.

Video solution by Clelland Maths 

(a)  We need to rearrange the equation into the form \(y=mx+c\) so that we can read off the gradient \(m.\)

$$ \begin{gather} 3x-7y-4=0\\[6pt] -7y=-3x+4\\[6pt] 7y=3x-4\\[6pt] y=\textstyle\frac37 x-\textstyle\frac47 \end{gather} $$

Now we can read off the gradient: \(m\!=\!\frac37\small.\)

(b)  From the equation, the \(y\)-intercept \(c\!=\!-\frac{4}{7}\small,\) so the point of intersection with the \(y\)-axis is \((0,\,-\frac{4}{7})\small.\)

(c)  The equation of the \(x\)-axis is \(y\!=\!0\small,\) so we substitute \(y\!=\!0\) into the equation of the straight line:
$$ \begin{gather} 3x-7y-4=0\\[6pt] 3x-7(0)-4=0\\[6pt] 3x-4=0\\[6pt] 3x=4\\[6pt] x=\textstyle\frac43 \end{gather} $$

So the point of intersection with the \(x\)-axis is \((\frac{4}{3},\,0)\small.\)

Video solution by Clelland Maths 

(a)  We need to rearrange the equation into the form \(y=mx+c\) so that we can read off the gradient \(m.\)

$$ \begin{gather} 4x+3y=12\\[6pt] 3y=-4x+12\\[6pt] y=\textstyle -\frac{4}{3}x+4\\[6pt] \end{gather} $$

So the gradient \(m=-\frac{4}{3}\small.\)

(b)  The equation of the \(x\)-axis is \(y\!=\!0\small,\) so we substitute \(y\!=\!0\) into the equation of the straight line:
$$ \begin{gather} 4x+3y=12\\[6pt] 4x+3(0)=12\\[6pt] 4x=12\\[6pt] x=3\\[6pt] \end{gather} $$

So the point of intersection with the \(x\)-axis is \((3,\,0)\small.\)

Video solution by Clelland Maths 

First, we need to use the two given points to find the gradient.

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{15-5}{3--2}\\[6pt] &=&\ \frac{10}{5}\\[6pt] &=&\ 2 \end{eqnarray} $$

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

We could use either point. Let's use \((a,\,b)=(3,\,15)\) to avoid negatives.
$$ \begin{gather} y-b=m(x-a)\\[6pt] y-15=2(x-3)\\[6pt] y-15=2x-6\\[6pt] y=2x+9 \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

We could use either point. Let's use \((x,\,y)=(-2,\,5)\) this time, so that you can see that the equation is the same:
$$ \begin{gather} y=mx+c\\[6pt] 5=2(-2)+c\\[6pt] 5=-4+c\\[6pt] c=9\\[6pt] y=2x+9 \end{gather} $$

Video solution by Clelland Maths 

We need to make \(y\) the subject, so that the equation is in the form \(y=mx+c\small.\)

This will enable us to read off the gradient \(m\small.\)

$$ \begin{gather} 3x-5y-10=0\\[6pt] -5y=-3x+10\\[6pt] 5y=3x-10\\[6pt] y=\textstyle\frac{3}{5}x-2\\[6pt] \end{gather} $$

So the gradient \(m=\frac{3}{5}\small.\)

Video solution by Clelland Maths 

The equation of the \(y\)-axis is \(x\!=\!0\small,\) so we substitute this into the equation:

So the point of intersection with the \(y\)-axis is \((0,\,-4)\small.\)

Video solution by Clelland Maths 

(a)  We need to rearrange the equation into the form \(y=mx+c\) so that we can read off the gradient \(m.\)

$$ \begin{gather} 3x+4y-8=0\\[6pt] 4y=-3x+8\\[6pt] y=-\textstyle\frac{3}{4}x+2\\[6pt] \end{gather} $$

So the gradient \(m=\textstyle-\frac{3}{4}\small.\)

(b)   From the previous part, we can see that the \(y\)-intercept is \(2.\) So the line crosses the \(y\)-axis at \((0,\,2)\small.\)

First we need to find the gradient:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \frac{7-(-1)}{-5-(-3)}\\[6pt] &=&\ \frac{7+1}{-5+3}\\[6pt] &=&\ \frac{8}{-2}\\[6pt] &=&\ -\!4 \end{eqnarray} $$

Now we need to use one of the points to find the equation. There are two possible methods for this.

Method 1: \(\boldsymbol{y\!-\!b\!=\!m(x\!-\!a)}\)

We could use either point. Let's use \((a,\,b)=(-5,\,7)\small.\)
$$ \begin{gather} y-b=m(x-a)\\[6pt] y-7=-4\left(x-(-5)\right)\\[6pt] y-7=-4(x+5)\\[6pt] y-7=-4x-20\\[6pt] y=-4x-20+7\\[6pt] y=-4x-13 \end{gather} $$

Method 2: \(\boldsymbol{y\!=\!mx\!+\!c}\)

Again, we could use either point. Let's use \((x,\,y)=(-5,\,7)\):
$$ \begin{gather} y=mx+c\\[6pt] 7=-4(-5)+c\\[6pt] 7=20+c\\[6pt] c=7-20\\[6pt] c=-13\\[6pt] y=-4x-13 \end{gather} $$

Video solution by Clelland Maths 

We need to rearrange the equation into the form \(y=mx+c\) so that we can read off the gradient \(m\small.\)

$$ \begin{align} x+4y-24=0\\[6pt] 4y=-x+24\\[6pt] y=\textstyle-\frac14\normalsize x+6\\[6pt] \end{align} $$

So the gradient \(m=-\frac14\small.\)

Video solution by Clelland Maths