A Level Maths: Straight Line Geometry
Introduction
In this article we are going to consider a geometry question based on the use of equations of a straight line as well as using knowledge of parallel and perpendicular lines. A Level Maths covers geometry and it is an important area of the course where questions can be quite detailed. It is important with geometry questions that you are showing all your working out very clearly and that you make use of a diagram.
The Question
Make sure you read the question. The very first sentence is telling you that geometry must be used and no other method.
For part (i) you need to know that the midpoint of a line is determined by finding the average of the x and y coordinates.
\begin{aligned} & M=\left(\frac{-1+7}{2}, \frac{4+8}{2}\right) \\ & M=(3,6) \end{aligned}To find the perpendicular bisector of AB it is first necessary to determine the gradient of AB.
Gradient of AB =\frac{8-4}{7–1}=\frac{4}{8}=\frac{1}{2}
The perpendicular gradient is then determined by finding the negative reciprocal which is -2
You can now find the equation of the perpendicular bisector:
\begin{aligned} & y-6=-2(x-3) \\ & y-6=-2 x+6 \\ & y+2 x=12 \end{aligned}Show also that...
And this is the required answer. Note the question says “Show also that….” With questions like this you need to derive the answer but you cannot use the answer that has been provided. The question could have quite easily stated, “determine the equation of the perpendicular bisector”.
Part (ii) of the question is asking us to find the area of the shaded region. Look carefully at the diagram provided and you should note the following:
The triangle has vertices as shown by the 3 circles. The coordinates of M are known so the other two vertices are needed. It is important to note that the triangle does not start at point A. Point A is not on the y axis.
The equation of the perpendicular bisector is known and this cuts the y axis at 0, 12. In order to determine the third vertices of the triangle the equation of AB is needed. The gradient of AB has already been found.
\begin{aligned} & y-4=\frac{1}{2}(x+1) \\ & y-4=\frac{x}{2}+\frac{1}{2} \\ & y=\frac{x}{2}+\frac{9}{2} \end{aligned}The y intercept is \frac{9}{2} so the coordinates of the third vertices is \left(0, \frac{9}{2}\right)
The diagram below shows the length of the perpendicular height and also the dimension of the base:
The area of the triangle can now be calculated.
\text { Area }=\frac{1}{2} \times 7.5 \times 3=\frac{45}{4} \text { units }^2The overall question is worth 12 marks and for that you are going need to do some detailed working out. For A Level Maths all working must be shown and it is important that your working is clear, logical and well presented so that it can be easily understood.
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