Parellel lines have equal gradients, so rearrange the given equation to \(y=-2x+5\) and we can see that the gradient of both lines is \(-2.\)

Now we can substitute \(a=4, b=-1\) and \(m=-2\) into the equation below:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y+1 &=& -2(x-4)\\[6pt] y+1 &=& -2x+8\\[6pt] y &=& -2x+7 \end{eqnarray} $$

Alternative method:
Substitute \(x=4, y=-1\) and \(m=-2\) into the general equation of a straight line. This method is usually easier when the gradient is an integer.

$$ \begin{eqnarray} y &=& mx+c\\[6pt] -1 &=& -2(4)+c\\[6pt] -1 &=& -8+c \\[6pt] c &=& 7\\[6pt] y &=& -2x+7 \end{eqnarray} $$

We start by rearranging the equation to determine the gradient of both lines:

$$ \begin{eqnarray} 5y-x &=& 9 \\[6pt] 5y &=& x+9 \\[6pt] y &=& \small\frac{1}{5}\normalsize x+\small\frac{9}{5}\normalsize \end{eqnarray} $$

So \(m=\frac{1}{5}.\) Now we use the definition of the gradient of a line between two points:

$$ \begin{eqnarray} m\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] \frac{1}{5} &=&\ \frac{t-7}{2+8}\\[8pt] \frac{1}{5} &=&\ \frac{t-7}{10}\\[8pt] 5(t-7) &=& 10\\[6pt] t-7 &=& 2 \\[6pt] t &=& 9 \end{eqnarray} $$

We start by rearranging the equation to determine the gradient of \(l_1.\)

$$ \begin{eqnarray} 3x-4y &=& 1 \\[6pt] -4y &=& -3x+1 \\[6pt] 4y &=& 3x-1 \\[6pt] y &=& \small\frac{3}{4}\normalsize x-\small\frac{1}{4}\normalsize \end{eqnarray} $$

So \(m_{l_1}\!=\!\large\frac{3}{4}\small.\) Now we use the fact that the gradients of two perpendicular lines multiply to \(-1.\)

$$ \begin{eqnarray} m_{l_1}\,m_{l_2} &=& -\!1 \\[6pt] \small\frac{3}{4}\normalsize \,m_{l_2} &=& -\!1 \\[6pt] m_{l_2} &=& -\!\small\frac{4}{3}\normalsize \end{eqnarray} $$

Now that we know the gradient of \(l_2\) and one point on \(l_2,\) finding its equation is easy:

$$ \begin{eqnarray} y-b &=& m(x-a) \\[6pt] y-2 &=& -\small\frac{4}{3}\normalsize (x-3) \\[6pt] 3(y-2) &=& -\!4(x-3) \\[6pt] 3y-6 &=& -\!4x+12 \\[6pt] 3y &=& -\!4x+18 \\[6pt] \end{eqnarray} $$

Note: This equation is equivalent to \(4x+3y=18\small,\) \(4x+3y-18=0\) and \(y=-\large\frac{4}{3}\normalsize x+6.\) Any of these forms are acceptable as your final answer.

We choose any two pairs of points and calculate the gradients between them. Let's compare \(m_{\textsf{AC}}\) and \(m_{\textsf{BC}}.\)

$$ \begin{eqnarray} m_{\textsf{AC}}\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] &=&\ \frac{-1+5}{4-1}\\[8pt] &=&\ \frac{4}{3}\\[14pt] m_{\textsf{BC}}\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] &=&\ \frac{-1-7}{4-10}\\[8pt] &=&\ \frac{-8}{-6}\\[8pt] &=&\ \frac{4}{3}\\[12pt] \end{eqnarray} $$

\(m_{\textsf{AC}}=m_{\textsf{BC}}\) so AC and BC are parallel. C is a common point. So A, B and C are collinear.

Note: The course specification specifically states that candidates should "include the phrases ‘parallel’ and ‘common point’ in their answers to show collinearity," so please be careful to use this wording.

We choose any two pairs of points and calculate the gradients between them. Let's compare \(m_{\textsf{PQ}}\) and \(m_{\textsf{PR}}.\)

$$ \begin{eqnarray} m_{\textsf{PQ}}\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] &=&\ \frac{8-2}{2+1}\\[8pt] &=&\ \frac{6}{3}\\[8pt] &=&\ 2\\[14pt] m_{\textsf{PR}}\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] &=&\ \frac{-7-2}{-4+1}\\[8pt] &=&\ \frac{-9}{-3}\\[8pt] &=&\ 3\\[12pt] \end{eqnarray} $$

\(m_{\textsf{PQ}}\ne m_{\textsf{PR}}\) so PQ and PR are not parallel. P, Q and R are not therefore collinear.

We use the fact that the gradient \(m=tan\ \theta\), where \(\theta\) is the angle with the positive \(x\)-axis.

So \(m_{l_1}=tan\ 30^\circ = \frac{1}{\sqrt{3}}\)

As \(l_2\) is perpendicular, \(m_{l_2}\!=\!\!-\!\sqrt{3}\) (the negative reciprocal).

Now we can substitute \(a\!=\!\!-\!3, b\!=\!2\sqrt{3}\) and \(m\!=\!\!-\!\sqrt{3}\) as follows:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-2\sqrt{3} &=& -\!\sqrt{3}(x+3)\\[6pt] y-2\sqrt{3} &=& -\!\sqrt{3}\,x-3\sqrt{3}\\[6pt] y &=& -\!\sqrt{3}\,x-3\sqrt{3}+2\sqrt{3}\\[6pt] y &=& -\!\sqrt{3}\,x-\sqrt{3} \end{eqnarray} $$

This question is very similar the previous example, except now \(l_1\) is a downhill line and \(30^\circ\) is the angle with the negative side of the \(x\)-axis.

A quick sketch should help you see that \(30^\circ\) with the negative \(x\)-axis is the same as \(180^\circ-30^\circ=150^\circ\) above the positive \(x\)-axis. Note that \(150^\circ\) is a Quadrant II (obtuse) angle, so \(tan\ 150^\circ\) is negative, as you would expect.

\(m_{l_1}=tan\ 150^\circ=-tan\ 30^\circ=-\frac{1}{\sqrt{3}}\)

As \(l_2\) is perpendicular, \(m_{l_2}=\sqrt{3}\)

Now we can substitute \(a=-3, b=2\sqrt{3}\) and \(m=\sqrt{3}\) as follows:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-2\sqrt{3} &=& \sqrt{3}(x+3)\\[6pt] y-2\sqrt{3} &=& \sqrt{3}\,x+3\sqrt{3}\\[6pt] y &=& \sqrt{3}\,x+3\sqrt{3}+2\sqrt{3}\\[6pt] y &=& \sqrt{3}\,x+5\sqrt{3} \end{eqnarray} $$

This example is a little twist on the \(m=tan\ \theta\) theme.

Firstly, let's rearrange \(2x-3y=1\) so that we know whether it has a positive or negative gradient.

$$ \begin{eqnarray} 2x-3y &=& 1\\[6pt] -3y &=& -2x+1\\[6pt] y &=& \small\frac{2}{3}\normalsize x-\small\frac{1}{3}\normalsize \end{eqnarray} $$

So the gradient is positive. It's an "uphill" line.

Now satisfy yourself from the diagram below that if \(a^\circ\) is the angle with the positive direction of the \(x\)-axis then \((90-a)^\circ\) is the acute angle with the \(y\)-axis.

So now we can calculate the required angle:

$$ \begin{eqnarray} m &=& tan\ a^\circ\\[6pt] \small\frac{2}{3}\normalsize &=& tan\ a^\circ\\[6pt] a &=& tan^{-1}\,\small\frac{2}{3}\normalsize\\[6pt] a &\approx& 33.7^\circ \end{eqnarray} $$

So the acute angle with the \(y\)-axis is \((90-33.7)^\circ = 56.3^\circ\) (correct to 1 decimal place).

The median through R is the straight line that connects R and the midpoint of PQ.

Let's use M for the midpoint of PQ:

$$ \begin{eqnarray} \textsf{M} &=& \left(\frac{x_{\small\textsf{P}\normalsize}+x_{\small\textsf{Q}\normalsize}}{2},\frac{y_{\small\textsf{P}\normalsize}+y_{\small\textsf{Q}\normalsize}}{2}\right)\\[6pt] &=& \left(\frac{2+6}{2},\frac{-1+7}{2}\right)\\[6pt] &=& \left(4,3\right) \end{eqnarray} $$

Now we start working towards obtaining the equation of the median RM by finding its gradient:

$$ \begin{eqnarray} m_{\small\textsf{RM}\normalsize} &=& \frac{y_{\small\textsf{R}\normalsize}-y_{\small\textsf{M}\normalsize}}{x_{\small\textsf{R}\normalsize}-x_{\small\textsf{M}\normalsize}}\\[6pt] &=& \frac{-2-3}{3-4} \\[6pt] &=& \frac{-5}{-1}\\[6pt] &=& 5 \end{eqnarray} $$

Finally, we substitute the coordinates of either point M or R to find the equation of the median. Let's use M:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-3 &=& 5(x-4)\\[6pt] y-3 &=& 5x-20 \\[6pt] y &=& 5x-17 \end{eqnarray} $$

First we need to find the midpoint of AB. Let's call it M.

$$ \begin{eqnarray} \textsf{M} &=& \left(\frac{x_{\small\textsf{A}\normalsize}+x_{\small\textsf{B}\normalsize}}{2},\frac{y_{\small\textsf{A}\normalsize}+y_{\small\textsf{B}\normalsize}}{2}\right)\\[6pt] &=& \left(\frac{-3+7}{2},\frac{4+-6}{2}\right)\\[6pt] &=& \left(\frac{4}{2},\frac{-2}{2}\right)\\[6pt] &=& \left(2,-1\right) \end{eqnarray} $$

Next we need to know the gradient of AB:

$$ \begin{eqnarray} m_{\small\textsf{AB}\normalsize} &=& \frac{y_{\small\textsf{A}\normalsize}-y_{\small\textsf{B}\normalsize}}{x_{\small\textsf{A}\normalsize}-x_{\small\textsf{B}\normalsize}}\\[6pt] &=& \frac{4--6}{-3-7} \\[6pt] &=& \frac{10}{-10}\\[6pt] &=& -1 \end{eqnarray} $$

So the gradient of the perpendicular bisector is \(1\), because \((-1)(1)=-1.\)

Now we are ready to find the equation of the perpendicular bisector of AB:

$$ \begin{eqnarray} y+1 &=& 1(x-2)\\[6pt] y+1 &=& x-2\\[6pt] y &=& x-3 \end{eqnarray} $$

(a)  The altitude from K is perpendicular to LM, so first we need to find \(m_{\small\textsf{LM}\normalsize}.\)

$$ \begin{eqnarray} m_{\small\textsf{LM}\normalsize} &=& \frac{y_{\small\textsf{L}\normalsize}-y_{\small\textsf{M}\normalsize}}{x_{\small\textsf{L}\normalsize}-x_{\small\textsf{M}\normalsize}}\\[6pt] &=& \frac{7-3}{1-3} \\[6pt] &=& \frac{4}{-2}\\[6pt] &=& -\!2 \end{eqnarray} $$

\((-2)(\frac{1}{2})=-1,\) so the altitude has gradient \(\frac{1}{2}.\)

We use this gradient and the coordinates of K to find the equation of the altitude:

$$ \begin{eqnarray} y-1 &=& \small\frac{1}{2}\normalsize(x+4) \\[6pt] y-1 &=& \small\frac{1}{2}\normalsize x+2 \\[6pt] y &=& \small\frac{1}{2}\normalsize x+3 \\[6pt] 2y &=& x+6 \\[6pt] x-2y &=& -\!6 \\[6pt] \end{eqnarray} $$

(b)  To find where the altitude intersects with LM (or LM extended) we first need to know the equation of LM.

From part (a), we know that \(m_{\small\textsf{LM}\normalsize}=-2\), so let's use point L to obtain its equation:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-7 &=& -\!\!2(x-1)\\[6pt] y-7 &=& -\!\!2x+2\\[6pt] y &=& -\!\!2x+9 \end{eqnarray} $$

To find the point of intersection we must solve the two equations simultaneously. As they both have \(y\) as their subjects, we can simply equate their right hand sides:

$$ \begin{eqnarray} \small\frac{1}{2}\normalsize x+3 &=& -2x+9 \\[6pt] x+6 &=& -4x+18 \\[6pt] x+4x &=& 18-6 \\[6pt] 5x &=& 12\\[6pt] x &=& \small\frac{12}{5}\normalsize \end{eqnarray} $$

To find the corresponding \(y\)-coordinate, we substitute into one of the equations:

$$ \begin{eqnarray} y &=& -2x+9 \\[6pt] &=& -2\small\left(\frac{12}{5}\right)\normalsize+9 \\[6pt] &=& -\small\frac{24}{5}\normalsize+9 \\[6pt] &=& -\small\frac{24}{5}\normalsize+\small\frac{45}{5}\normalsize \\[6pt] &=& \small\frac{21}{5}\normalsize \end{eqnarray} $$

So the point of intersection is \(\left(\large\frac{12}{5}\normalsize,\large\frac{21}{5}\normalsize\right).\)

The median through C is the straight line that connects C and the midpoint of AB.

Let's use M for the midpoint of AB:

$$ \begin{eqnarray} \textsf{M} &=& \left(\frac{x_{\small\textsf{A}\normalsize}+x_{\small\textsf{B}\normalsize}}{2},\frac{y_{\small\textsf{A}\normalsize}+y_{\small\textsf{B}\normalsize}}{2}\right)\\[6pt] &=& \left(\frac{-3+7}{2},\frac{5+9}{2}\right)\\[6pt] &=& \left(2,7\right) \end{eqnarray} $$

Now we start working towards obtaining the equation of the median CM by finding its gradient:

$$ \begin{eqnarray} m_{\small\textsf{CM}\normalsize} &=& \frac{y_{\small\textsf{C}\normalsize}-y_{\small\textsf{M}\normalsize}}{x_{\small\textsf{C}\normalsize}-x_{\small\textsf{M}\normalsize}}\\[6pt] &=& \frac{11-7}{2-2} \\[6pt] &=& \frac{4}{0} \end{eqnarray} $$

The gradient of the median is undefined. This indicates a vertical line. So the equation of the median is \(x=2\small.\)

The method here is to find the gradient \(m_1\) of the line passing through the given points, the gradient \(m_2\) of the line with the given equation, and to see if they meet the \(m_1m_2=-1\) condition for perpendicular gradients.

Using \((x_1,y_1)=(-4,2)\) and \((x_2,y_2)=(2,-7)\):

$$ \begin{eqnarray} m_1\ &=&\ \frac{y_2-y_1}{x_2-x_1}\\[8pt] &=&\ \frac{-7-2}{2--4}\\[8pt] &=&\ \frac{-9}{6}\\[8pt] &=& -\!\frac{3}{2} \end{eqnarray} $$

Now let's make \(y\) the subject of the equation of the other line:

$$ \begin{eqnarray} 3y &=& 2x+9 \\[6pt] y &=& \small\frac{2}{3}\normalsize x+3 \\[6pt] \end{eqnarray} $$

So \(m_2=\large\frac{2}{3}\normalsize\) and \(m_1m_2=-\large\frac{3}{2}\normalsize\times\large\frac{2}{3}\normalsize= -1,\) so the lines are perpendicular.