AS LEVEL COORDINATE GEOMETRY EXPLAINED

Mathematic pure 1(coordinate geometry) email:racsostudenthelp@gmail.com
3. COODINATE GEOMETRY
 Length- is the distance between two points
 Gradient – is a numerical expression of how steep or gentle a slope is or rate of change.
 Mid- point – is the mid- point between two points.
-Consider the following graph, B( x2 , y2)
Y-axis
(x2-x1)
A(x1 , y1) (x1 , y2)
X-axis
(y2-y1) – is the distance
RACSO PRODUCTS Page 1

Do you remember the Pythagoras theorem? If you want to find distance AB ( the hypotenuse )? The same idea is
used here, it is good for you to know but not necessary.
Getting the distance of the two point, dab= √(푥2 − 푥1 )2 + (푦2 − 푦1 )2
Which can also be written as dab =√Δ푥2 + Δ푦2
Δ =means change and dab= distance AB
Given the gradient of the line,
(a) (b) (c)
-Looking at line (a), is steeper than line (b), that means line “a” has a numerical greater gradient
than line (b), for line (c) it has zero gradient because it is horizontal.
In mathematics, gradient is denote “m”, referring to the graph from previous page
M =
y2 – y1
x2- x1
-When gradient is zero means line is horizontal
-When gradient is undefined means line is vertical (when change in X = 0, or X2 and X1 are equal)

(x1 , y1) ( x2 , y2)
Mid-point
Mid-point = (x1 + x2
2
,
y1 + y2
2
)
To find the equation of a line, you at least need
The gradient and one point two points
OR

-IF GIVEN THE GRADIENT AND A POINT
Recall the following equation, y – y1 = m(x – x1);
Let us solve one question together, example you are given gradient of line as 2 and a point (6, 7).
Therefore y1= 7, x1 = 6 and m = 2, you simply have to plug in the number to the equation and
that is it.
You will then have the equation y – 7 = 2(x – 6) and that is your answer.
-You can write the equation in form of y = mx + c, by opening the brackets and making y the
subject, and equation will be y = 2x – 5. (This will be discuss in detail later)
-IF GIVEN TWO POINTS
-If two points (co-ordinates) are given, you first find the gradient, (please go back learn how to
find gradient if forgotten), when you find gradient, pick any point from the two points given.
Plug in the x and y coordinates in this equation y – y1 = m(x – x1) as I explained above.

 If two lines are parallel, means they have the same gradient, and when two lines are
perpendicular to each other means, if I take the gradient of the 1st line multiply to the
gradient of the 2nd line it is equal to -1. Taking gradient one as M1 and line two as M2 this
means ( M1 x M2 = -1 )
 Looking at the diagram ( from CIE May/June 2007, Question 6)
Line AD is parallel to line BC, but perpendicular to line AB and line DC.
-below is the questions they asked from this diagram.

 How to find equation BC, I know I can get equation AB, but how is that going to
help me? Remember when two lines a perpendicular, the product of their gradient
is (-1). If you calculate gradient of line AB it should come to 1.5 or 3/ 2. From
formula (M1 x M2 = -1) , taking line AB be M1 = 1.5, then gradient of line BC(M2 )
= ( -2/3). You now have the gradient and one point on the line BC which is ( -2, 8).
Then find equation, you can go back and see how to find gradient If given gradient
and a point. Answer should
 How can we find the co-ordinate C and D? I know coordinate C is on the X-axis,
meaning, the value of y=0, so C( x , 0 ). But we have the equation of line BC from
previous part. If you have the equation of any line, if I plug in the value y of any
point on that line, I should get the value of the corresponding value of X of that
specific point and vice versa. I know y= 0, plug in y= 0. You must get X= 10.
Meaning co-ordinate C(10 , 0 ). Then how can we find co-ordinate D? there are two
options, one is getting the equation of line DC and AD and solving them
simultaneously, but that is going to be a long work for only 3 points, I will
introduce something very simple called the vector move. Since I know distance
AD and BC are equal (since it is a rectangle). Then I know moving from
B (-2, 8) C(10,0 )
12 is added to the x value of point B to get the x value of point C, which is 10 .
(-2 +12 =10) and 8 is subtracted from the y value of point B to get the y value of point C,
(8 – 8 =0). We can do the same for point A to get point D, add 12 to the x value and
subtract 8 from the y value.
A (2, 14) D(x, y)
Therefore point D (14, 6), this is what we call vector move.

-We discussed this from previous sections, you can write the equation of a line in form of
y – y1 = m(x – x1) or y = mx + c. I advise you to use y – y1 = m(x – x1) unless specified in
the question that should be written in form of y = mx + c.

4
Y-intercept.
Getting the x-intercept Vertical line is y-axis
From the equation, put y=0 find y, Horizontal line is x-axies
You should get X = -3/2 X -intercept= -3/2
-Something to remember,
(a) (b)

- Whenever you have a line leaning to the right like (a) gradient is always positive and line leaning to
the left like (b) gradient is negative. I think it is now very easy to simply look at a line and know if
gradient is positive or negative, and looking at the value of “C” to determine what the y-intercept is.
 Let us move on to intersection, any idea what intersection is? In coordinate geometry, it is a
point where two or more lines/ curves meet.
 Intersection can occur between a line and line, curve and line, or curve and curve,( just like the
figures showed below) in any case, getting the point of intersection; we need the equations of
the two lines/ curves.
 Something else to remember there may be only one point of intersection or more that one.
( for this level I will only discuss up to 2 points of intersection)
 Look at the following graphs,
(a) (b) (C)
 You will come across the word tangent, which is a line that intersects a curve only once
and does not cross the it! Just like graph (b). In other words we can say the line in graph
(b) is a tangent to the curve.

9709/12/O/N/2009
 How would you solve this? Remember to find the point of intersection you first sove the
two equations simultaniously ( you can either use substitution or elimination method).
The line has equation 2y= x + 5 and curve has equation
y= x2 – 4x + 7. If I make y the subject from equation of line, I should have y= (x + 5)/2.
I can now equate (x + 5)/2 = x2 – 4x + 7 and end up with, with a quadratic equation
y= 2x2 – 9x + 9. Hope you can now solve this. You should get 2 values of x.
x=3 or x =1.5. That is the answer since the questions asks you to find the “X” values, if
asked to find the cordinates of intersection. You would have to take any of the 2
equations you were given, either 2y= x + 5 or y= x2 – 4x + 7, plug in the value of x and
get the corresponding value of y which is y= 4 or y= 3.25. that means the coordinates of
intersection are (3,4) and (1.5 , 3.25).
 let me take you back to quadratics, do you remember how discriminat works? it helps
us determine how many solutions we would obtain from a quadratic equation. There
are 3 options, either obtain 2 solutions, one solution or no solution. From the CIE
question, the line intersects the curve twice,but assuming we didn’t have the graph but
only had the two equations, 2y= x + 5 and y= 2x2 – 9x + 9 and asked to state whether
the curve and line intersect or not, and if yes how many times. I will first substitute the

one equation into the other, make it look like this, y= 2x2 – 9x + 9, and this is the point
where discriminants comes in handy!
- If b2 - 4ac < 0 no solution, if b2 - 4ac = 0, only one solution ( sometimes
means the line only intersects a curve once, meaning the line is a tangent) and if
b2 - 4ac > 0 , there are two solutions( the lines don’t intersect)
-From the expression, b2 - 4ac and looking at equation y= 2x2 – 9x + 9, b= -9 , a=2
and c=9. If solved, 92- 4(2)9= 9. 9 is greater than zero, so there are two
solutions as we have seen from graph, there two values of x.
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AS LEVEL COORDINATE GEOMETRY EXPLAINED

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