Length And Area
Length and Area – Introduction
Quite often with straight line geometry questions you need to be able to determine the length and area that exists between a straight line and the axes which is generally both the x and y axis.
To determine the length you can simply perform Pythagoras’s Theorem and to find the area, usually of a triangle, it is important that you sketch the information that is given to you in order to determine the correct lengths of the triangle that you need in order to determine the area.
Also, it is important to give answers as exact answers, rather than as a decimal, unless of course a question asks for the answer in a particular form. Having an exact answer avoids that chance of losing any marks for poor accuracy in subsequent parts of a question.
Finding Length
If you have a straight line then you can find the distance between two points A and B by using the following standard result:
d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}But you need to understand that this is just Pythagoras, so do not try to recall this formula.
If you are asked to determine the length of a length, draw a quick sketch and complete a triangle as shown below:
You can then determine the vertical and horizontal distances and therefore use Pythagoras to find the required length and remember to square root your answer. Remember to leave your answer in exact form wherever appropriate.
Finding Length - Example
First sketch a straight line. It does not need to be to scale and it does not need to be accurate. You just need to be able to calculate the differences in the x and y values.
You can see that the difference in the x and y values are 3 and 4 respectively. It is now a case of using Pythagoras as you have done with GCSE Maths:
\begin{aligned} d^2 & =(5-2)^2+(7-3)^2 \\ d^2 & =3^2+4^2 \\ d & =\sqrt{\left(3^2+4^2\right)} \\ & =\sqrt{25} \\ & =5 \end{aligned}
Length and Area – Example
For part a) to work out the points of intersection you need to do simultaneous equations of the two straight lines.
Probably the easiest approach is to say y = 4x and then substitute this into the second equation as follows:
\begin{aligned} & 2 x+3 y-21=0 \\ & 2 x+3(4 x)-21=0 \\ & 14 x-21=0 \\ & 14 x=21 \\ & x=\frac{3}{2} \end{aligned}A common mistake with simultaneous equation is to forget to find the value of the second variable, in this case the y coordinates and this is done by a simple substitution as shown:
y=4 \times\left(\frac{3}{2}\right)=6So the coordinates of A are \left(\frac{3}{2}, 6\right)
For part b) it is important that you sketch the two straight lines as this will make the understanding of the maths much easier. The sketches are shown below and are both on the same axis.
You will see that in no means is the sketch to scale or accurate by any means but it does provide the key information to help find the area of the triangle.
The area of the triangle can then be calculated as follows:
\text { Area }=\frac{1}{2} \times 6 \times \frac{21}{2}=\frac{63}{2}Length and Area – Further Examples
In this section of this article we will consider some additional questions and discuss how to best approach them as well as providing a detailed solution.
Q1)
Here is an actual instance where the distance formula is needed to help solve this question. Remember the formula is given as follows:
d^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2Using the information in the question we can obtain the following:
\begin{aligned} 65 & =(x-(-1))^2+(9-13)^2 \\ 65 & =(x+1)^2+(-4)^2 \\ 65 & =x^2+2 x+1+16 \\ x^2+2 x-48 & =0 \\ (x+8)(x-6) & =0 \\ x=-8 \text { or } x= & 6 \end{aligned}Q2)
This is very similar to the question above. Make sure you understand what has been done in the previous question before you attempt this question.
Using the distance formula as above, the two y values can be obtained as follows:
\begin{aligned} & 90=(5-2)^2+(7-y)^2 \\ & 90=3^2+49-14 y+y^2 \\ & y^2-14 y-32=0 \\ &(y-16)(y+2)=0 \\ & y=16 \text { or } y=-2 \end{aligned}Q3)
For part a) you need to find the gradient of each line
\begin{aligned} & l_1: y=2 x+4, \text { gradient }=2 \\ & l_2: 6 x-3 y-9=0 \end{aligned}For the second line you need to make y the subject and this is done as follows:
\begin{aligned} 3 y & =6 x-9 \\ y & =2 x-3, \text { gradient }=2 \end{aligned}They both have the same gradients so they are both parallel.
For part b) the third line will have gradient of -\frac{1}{2}
And so the equation can be calculated as follows:
\begin{aligned} y-y_1 & =m\left(x-x_1\right) \\ y & -10=-\frac{1}{2}(x-3) \\ 2 y & -20=-x+3 \\ x+2 y & -23=0 \text { is the equation of } l_3 \end{aligned}For part c) you need to perform simultaneous equations with the two straight lines mentioned. This is done as follows:
\begin{aligned} & l_2: y=2 x-3 \\ & l_3: 2 y=-x+23 \end{aligned}\begin{aligned} 2(2 x-3) & =-x+23 \\ 4 x-6 & =-x+23 \\ 5 x & =29 \\ x & =\frac{29}{5} \end{aligned}\begin{aligned} \therefore y & =2\left(\frac{2 a}{5}\right)-3 \\ & =\frac{43}{5} \end{aligned}\text { ie }\left(\frac{29}{5}, \frac{43}{5}\right)What is needed is to now find the distance between (3,10) \text { and }\left(\frac{29}{5}, \frac{43}{5}\right).
This can be done by Pyhtagoras as follows:
Length and Area – Exam Style Question
This is a typical exam style question and we will go through each part in turn.
Part a)
Here you are simply being asked to determine the equation of a straight line and this is calculated as follows:
\begin{aligned} y-y_1 & =m\left(x-x_1\right) \\ y & -14=-\frac{1}{4}(x-(-4)) \\ 4 y & -56=-x-4 \\ x+4 y & -52=0 \end{aligned}Part b)
A straight line will cross the y-axis when x = 0.
\begin{array}{r} 0+4 y-52=0 \\ y=13 \end{array}Which means that the coordinates of A are (0, 13)
Part c)
The equation of the line l_2 \text { is } y=3 x
To find the coordinates of B it is case of solving a pair of simultaneous equations and this can be done by substitution as follows:
\begin{array}{r} x+4(3 x)-52=0 \\ 13 x-52=0 \\ x=4 \\ y=12 \end{array}B is (4, 12)
Part d)
Again to help to ensure you are using the correct dimensions a quick sketch will be beneficial here and this is shown below:
What is needed is the area of the triangle OAB. The base is OA which is 13 and the height is 4 so the area of the triangle can be calculated as follows:
\frac{1}{2} \times 13 \times 4=26