We know that the general form of the equation of a straight line is \(y=mx+c\) and that the point with \(x\!=\!2\) and \(y\!=\!-\!3\) fits this equation. So we substitute as follows:

Similar to the previous example, we use the general form of the equation of a straight line and substitute the \(x\)- and \(y\)-coordinates of the point that we know is on the line.

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

Method 1:

\(m=2\) and \((a,b)=(3,-1)\). Use this formula:
$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y+1 &=& 2(x-3)\\[6pt] y+1 &=& 2x-6\\[6pt] y &=& 2x-7 \end{eqnarray} $$

Method 2:

\(m=2\) and \((x,y)=(3,-1)\). Use the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] -1 &=& 2(3)+c\\[6pt] -1 &=& 6+c\\[6pt] c &=& -\!7\\[6pt] y &=& 2x-7 \end{eqnarray} $$

This is similar to the previous example, but the gradient is a fraction. So we should multiply the equation through by its denominator, so that our final equation involves only whole numbers.

Method 1:

We substitute \(m=\frac{3}{4},\) \(a=-2\) and \(b=5\) into the formula:
$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-5 &=& \small\frac{3}{4}\normalsize (x--2)\\[6pt] y-5 &=& \small\frac{3}{4}\normalsize (x+2)\\[6pt] 4(y-5) &=& 3(x+2)\\[6pt] 4y-20 &=& 3x+6\\[6pt] 4y &=& 3x+6+20\\[6pt] 4y &=& 3x+26\\[6pt] -3x+4y &=& 26\\[6pt] \end{eqnarray} $$

Either of the last two lines of working are acceptable as the final form of the equation.

Method 2:

\(m=\frac{3}{4}\) and \((x,y)=(-2,5)\). Use the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] 5 &=& \small\frac{3}{4}\normalsize\left(-2\right)+c\\[6pt] 5 &=& -\small\frac{3}{2}\normalsize+c\\[6pt] \small\frac{10}{2}\normalsize &=& -\small\frac{3}{2}\normalsize+c\\[6pt] c &=& \small\frac{3}{2}\normalsize + \small\frac{10}{2}\normalsize\\[6pt] c &=& \small\frac{13}{2}\normalsize \\[12pt] y &=& \small\frac{3}{4}\normalsize x+\small\frac{13}{2} \\[6pt] y &=& \small\frac{3}{4}\normalsize x+\small\frac{26}{4} \\[6pt] 4y &=& 3x+26 \\[6pt] -3x+4y &=& 26\\[6pt] \end{eqnarray} $$

Method 1 is quicker and easier to understand here. We would always recommend using \(y-b=m(x-a)\) when the gradient is a fraction.

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

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

Method 1:

We could use either point. Let's use \((a,b)=(1,7)\).
$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-7 &=& 2(x-1)\\[6pt] y-7 &=& 2x-2\\[6pt] y &=& 2x+5 \end{eqnarray} $$

Method 2:

Use \((x,y)=(1,7)\) and the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] 7 &=& 2(1)+c\\[6pt] 7 &=& 2+c\\[6pt] c &=& 5\\[6pt] y &=& 2x+5 \end{eqnarray} $$

First we find the gradient. It's going to be a fraction in this example, which will lead us into a slightly different type of working to find the equation of the straight line.

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

Method 1:

We could use either point. Let's use \((a,b)=(-2,5)\).
$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-5 &=& -\small{\frac{3}{2}}\normalsize (x+2)\\[6pt] 2(y-5) &=& -3(x+2)\\[6pt] 2y-10 &=& -3x-6\\[6pt] 2y &=& -\!3x+4\\[6pt] 3x+2y &=& 4\\[6pt] \end{eqnarray} $$

Method 2:

Use \((x,y)=(-2,5)\) and the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] 5 &=& \small\left(\normalsize -\small{\frac{3}{2}}\small\right)\normalsize\left(-2\right)+c\\[6pt] 5 &=& 3+c\\[6pt] c &=& 2\\[6pt] y &=& -\!\small{\frac{3}{2}}\normalsize x+2 \end{eqnarray} $$

You could stop here, if you wish, but we prefer equations of straight lines not to contain fractions. So we would multiply through by 2 and present our final answer in one of the two following forms:

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

$$ \begin{eqnarray} 3x-7y-4\ &=&\ 0\\[6pt] -7y &=&\ -\!3x+4\\[6pt] 7y &=&\ 3x-4\\[6pt] y &=&\ \small\frac{3}{7}\normalsize x-\small\frac{4}{7}\normalsize \end{eqnarray} $$

So the gradient \(m=\frac{3}{7}\).

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

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

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

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

$$ \begin{eqnarray} 4x+3y\ &=&\ 12\\[6pt] 3y &=&\ -\!4x+12\\[6pt] y &=&\ -\!\small\frac{4}{3}\normalsize x+4\\[6pt] \end{eqnarray} $$

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

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

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

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:

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

Method 2:

Use \((x,y)=(3,15)\) and the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] 15 &=& 2(3)+c\\[6pt] 15 &=& 6+c\\[6pt] c &=& 9\\[6pt] y &=& 2x+9 \end{eqnarray} $$

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

$$ \begin{eqnarray} 3x-5y-10\ &=&\ 0\\[6pt] -5y &=&\ -\!3x+10\\[6pt] y &=&\ \small\frac{-3}{-5}\normalsize x+\small\frac{10}{-5}\normalsize\\[6pt] y &=&\ \small\frac{3}{5}\normalsize x-2\\[6pt] \end{eqnarray} $$

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

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

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

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

$$ \begin{eqnarray} 3x+4y-8\ &=&\ 0\\[6pt] 4y &=&\ -\!3x+8\\[6pt] y &=&\ -\!\small\frac{3}{4}\normalsize x+2\\[6pt] \end{eqnarray} $$

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

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

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:

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

Method 2:

Use \((x,y)=(-5,7)\) and the general form of the straight line:
$$ \begin{eqnarray} y &=& mx+c\\[6pt] 7 &=& -\!4(-5)+c\\[6pt] 7 &=& 20+c\\[6pt] c &=& 7-20\\[6pt] c &=& -\!13\\[6pt] y &=& -\!4x-13 \end{eqnarray} $$