Exam Questions - The Equation Of A Straight Line
Introduction
In this article we are going to consider The Equation Of A Straight Line and in particular we are going to be looking at a number of exam questions and explaining how the question needs to be approached.
Where possible, it is highly suggested that a clear diagram is drawn. Geometry is a very visual subject, and if you can see the information then you have a better chance of being able to answer the question.
The Equation Of A Straight Line - Example 1
You are being asked to find the equation of a line given two points. Always take a note of the requirements of a straight line. One thing you need is the gradient and the second is the y-intercept.
First, let us determine the gradient of the line as follows:
\begin{aligned} & (2,1) \quad(4,-5) \\ & x_1 y_1 \quad x_2 y_2 \\ & m=\frac{y_2-y_1}{x_2-x_1} \\ & m=\frac{-5-1}{4-2} \\ & =\frac{-6}{2} \\ & =-3 \end{aligned}Now we have the gradient it is a case of using the equation of a line y = mx + c and either A or B in order to determine the value of “c”: This can be done as follows:
\begin{aligned} & y=-3 x+c \\ & 1=-3(2)+c \\ & 1=-6+c \\ & c=7 \\ & y=-3 x+7 \end{aligned}Once the value of “c” has been obtained it is important to then actually state what the equation of the line is.
The Equation Of A Straight Line - Example 2
The question here is asking if the two lines are parallel, perpendicular or neither. In order to answer this question, the gradient of both lines are needed.
For the line l_1 it is required to perform a rearrangement so as to make “y” the subject. This can be done as follows:
\begin{aligned} 2 x+3 y+5 & =0 \\ 3 y & =-2 x-5 \\ y & =-2 / 3 x-5 / 3 \end{aligned}For the line l_2, the gradient of the line needs to be determined and then picking one of the coordinates to substitute this into the general equation of a straight line. This is done as follows:
\begin{aligned} & (1,7) \quad(5,1) \\ & x_1 y_1 \quad x_2 y_2 \\ & =\frac{y_2-y_1}{x_2-x_1} \\ & =\frac{1-7}{5-1} \\ & =\frac{-6}{4} \\ & =\frac{-3}{2} \end{aligned}Now that both gradients have been obtained it is clear that they are not the same so they are not parallel. Also, they are not the negative reciprocal of one another, so they are not perpendicular to each other either.
So the lines are not parallel and not perpendicular.
The Equation Of A Straight Line – Example 3
For part a) of the question, there is nothing that has not been covered already. The only difference is the form of the answer and it is important to be always reading the question carefully to know what is being asked.
Finding the gradient and the equation is done as follows:
m=\frac{y_2-y_1}{x_2-x_1}\frac{-1-5}{5–2}=\frac{-6}{7}\begin{aligned} & y=-\frac{6}{7} x+c \\ & 5=-\frac{6}{7}(-2)+c \\ & s=\frac{12}{7}+c \\ & c=\frac{23}{7} \end{aligned}The equation of the straight line can now be obtained in the form required as follows:
\begin{aligned} & y=-\frac{6}{7} x+\frac{23}{7} \\ & 7 y=-6 x+23 \\ & 6 x+7 y-23=0 \end{aligned}For part b) it is important that a quick stretch is drawn so as to correctly know what lengths are required as well as knowing where any lines cross the axes.
The area of the triangle can now be found as follows:
\begin{aligned} \text { Area } & =\frac{1}{2} \text { base } \times \text { height } \\ & =\frac{1}{2}\left(\frac{23}{6}\right)\left(\frac{23}{7}\right) \\ & =\frac{529}{84} \text { units }^2 \end{aligned}The Equation Of A Straight Line – Example 4
If dealing with a question such as this one where it asks for the points of intersection, it is a case of solving the equations simultaneously. Both equations are linear so the method of elimination or substitution can be used.
For this particular example, we will use the method of substitution as shown:
\begin{aligned} x=(4 y-1) & \\ 5(4 y-1)+2 y-4 & =0 \\ 20 y-5+2 y-4 & =0 \\ 22 y-9 & =0 \\ 22 y & =9 \\ y & =9 / 22 \end{aligned}\begin{aligned} & x=4\left(\frac{9}{22}\right) \\ & x=\frac{7}{11} \\ & \left(\frac{7}{11}, \frac{9}{22}\right) \end{aligned}Given that the question has asked for “coordinates” then do give the final answer as coordinates.
The Equation Of A Straight Line – Example 5
Part a) For this question you are being asked to determine the equation of the second straight line.
Always look for keywords within a question and in this case it is perpendicular. You can determine the gradient of the second line and to use the coordinate P, but you also need to find the value of y when x = 2 and this is calculated as follows:
\begin{aligned} x=4: \quad 5 y-10 & =8 \\ 5 y & =18 \\ y & =18 / 5 \end{aligned}p=(4,18 / 5)Rearranging the first equation of a straight line in order to determine the perpendicular gradient is calculated as follows:
\begin{aligned} & 5 y=2 x+10 \\ & y=\frac{2}{5} x+2 \end{aligned}So the perpendicular gradient is \frac{-5}{2} for the second straight line.
The equation of the second straight line can now be found as follows, remembering to put the answer in the form that is required within the question:
\begin{aligned} & y=-\frac{5}{2} x+c \quad\left(4, \frac{18}{5}\right) \\ & \frac{18}{5}=-\frac{5}{2}(4)+c \\ & c=\frac{68}{5} \\ & y=-\frac{5}{2} x+\frac{68}{5} \\ & 10 y=-25 x+136 \\ & 25 x+10 y-136=0 \end{aligned}Part b) of the question is asking you to determine the area of a triangle and the sketch would be useful to determine what the lengths are. The sketch can be done as follows:
Make sure that you are able to sketch the straight lines. Remember it is just a sketch that is required in order to determine where the straight lines cross the axes. Having got the sketch the area of the triangle can now be calculated as follows:
\begin{aligned} & \text { Area }=\frac{1}{2} \text { base height } \\ & =\frac{1}{2}\left(\frac{261}{25}\right)\left(\frac{18}{5}\right)=\frac{2349}{125} \text { units }^2 \end{aligned}