Coordinate geometry
IDO expect that by now you are comfortable with finding the length, gradient and midpoint of the line joining two given points. These will be further illustrated by giving you the solution to the homework presented last week.
HOMEWORK
1. The coordinates of the points M and N are (- 1, 1) and (3, 9), respectively. Determine the value of: a) the gradient of MN. b) the coordinates of the midpoint of the line MN.
c) the gradient of MP, the line perpendicular to MN.
SOLUTION]
a. Gradient, mo fMN=y2-y1/x2-x 1 Substituting m = 9 - 1/ 3 - - 1 = 8/4 = 2 m=2
b. Midpoint, Mo fMN=x2+x1/2,y+y /2 2 1 Substituting
M = 3 + - 1/2 , 9 + 1 /2
M = ( 1, 5)
c. Since the gradient of MN is 2, if the gradient of the perpendicular be m1. m1xm =-1. Substituting, m1 x 2 = - 1 m1 = - 1/2
2. Given the points A(-6, 1) and B(4 , - 3), find the following with respect to the line AB: (i) Gradient, m.
(ii) midpoint, M.
(iii) length of the line AB.
(iv) Gradient of XY which is parallel to AB. (v) the gradient of AC which is perpendicular to AB.
SOLUTION
(i) Gradient, m of AB = y - y1/x2 - x1 2 Substituting m = -3 -1 /4 - -6 = - 4/10 m = - 2/5
(ii) Midpoint, Mo fAB=x+x/2,y+y 1/2 2 1 2 Substituting
M = 4 + - 6 /2, -3 + 1/2 = - 2/2 , - 2 /2
n The coordinates of the midpoint of AB are ( -1 , -1 )
(iii) The length of AB2 = (x2 - x1) 2 + (y2 - y1) 2 Substituting
AB2 = (4 - -6) 2 + (-3 -1 ) 2 = (10) 2 + (-4) 2 AB 2=100+16=116
AB = √116 = 10.8
The length of AB = 10.8
(iv) Since XY is parallel to AB Then the gradient of XY = - 2/5
(v) Since AC is perpendicular to AB, let AC = m1 and gradient of AB = m. m x m1 = -1. m1 x - 2/5 = -1 m1 = 5/2
The gradient of AC = 5/2
We will continue coordinate geometry by considering:
EQUATION OF STRAIGHT LINES
All straight lines have the equation: y = mx + c, where m is the gradient and c is the intercept; m and c are constants.
When the axes cut at the origin (0 , 0), the equation of the x axis is y = 0 and for the y axis it is x = 0.
y = 3x + 5 is the equation of a line if, for each point (x , y) on the line, the y coordinate is equal to three times the x coordinate of the same point plus 5. The points (1 , 8) and (- 2 , - 1) are, therefore, on the line.
This fact about an equation is not usually emphasised, but must be clearly noted.
The point (x , y) is on the line y = mx + c if it satisfies the equation.
You may show that (1 , - 2) is a point on the line y = 3x - 5 by substituting x = 1 and y = - 2 into the equation. (Substitution shows that - 2 = - 2). Or, by substituting x = 1, you will be able to show that y = - 2.
The value of c, the intercept of a line, is found by substituting x = 0 into its equation. Do you know why? If not, please investigate.
METHODS OF FINDING THE EQUATION
The following are the three methods which are commonly used to find the equation of a straight line.
(i) Evaluating the equation, given the gradient m and the intercept c.
EXAMPLE
Find the equation given that m = - 3 and c = 2
The equation is y = mx + c. Substituting y = - 3x+2
Answer is y = - 3x + 2.
This method can be extended to a given line on a graph. In this case, both the gradient and the intercept can be found from the graph and the equation determined.
(ii) A feature of the second method is: Given the coordinates of two points, (x1 , y1) and (x2 , y2), the equation is
y - y1/x - x1 = y2 - y1/x2 - x1
Using the points A(4 , - 1), B(1 ,1) in the above, then y - -1 /x - 4 = 1- -1/1 - 4 = 2/-3 y + 1 / x - 4 = -2/3
3y + 3 = -2x + 8
3y + 2x = 5.
Answer is 3y + 2x = 5
(iii) The formula given in (ii) may be expressed as y - y1/ x - x1 = m, where m is the gradient of the line. I am sure you realise that m = y2 - y1/ x2 - x1
This formula is used, given the coordinates of a point on the line and the gradient of the line.
EXAMPLE
Find the equation of the line if the gradient m = 2/3 and the point (1 , 2) is on the line. y - y1x - x1 = m, that is y - 2/ x - 1 = 2/3 3y - 6 = 2x - 2
3y - 2x = 6 - 2
3y - 2x = 4
Answer is 3y - 2x = 4
(a) The equation of the line above is y = mx +
c.
(i) State the value of c.
(ii) Determine the value of m.
(iii) Determine the coordinates of the midpoint of the line segment AB.
(b) The point (-2, k) lies on the line. Determine the value of k.
SOLUTION
(i) c is the y coordinate of A. From the graph, A has coordinates (0, 7) c = 7.
(ii) Since B has coordinates (2, 0) and A (0, 7) Gradient, m, of AB = y2 - y1/x2 - x1 = 7 - 0 / 0 - 2
Gradient, m = 7 /-2 = - 7/2
(iii) M, the midpoint, = x2 + x1/2, y2 + y1/2 = 0 + 2/2 , 7 + 0/2 = 1, 7/2
(b) From the above, m = -7 /2 and c = 7. n The equation of AB is y = -7/2 x + 7
2y = -7x + 14.
Since (-2, k) lies on the line, substituting -2 for x,
2y = -7 x -2 + 14 = 14 + 14 = 28. y = 28/2 = 14. k = 14.
Now for your homework.
1. The coordinates of the points M and N are (- 1, 1) and (3, 9), respectively. Determine the value of:
The gradient of MN.
The equation of the line MN.
The equation of the line parallel to MN and passing through the origin.
2. A straight line MN cuts the y axis at M(0 , -3). The gradient of MN is 2. Show that the equation of the line MN is y - 2x = - 3.
3. A straight line K passes through the point M (4, 1) and has a gradient of 3/5. Determine the equation of this line. Given that the line segment MP is
perpendicular to K, determine its equation. 4. A straight line is drawn through the points A(- 5 , 3) and B(1 , 2).
(i) Determine the gradient of AB.
(ii) Find the equation of the line AB.
I must emphasise again that the problems based on this topic are fairly routine. It will do you well to practise them so as not to miss out on the opportunity to score full marks for the question if it is presented in the June exam this year. Have a productive week.