Coordinate geometry
WE WILL continue the review of coordinate geometry with problems in the Cartesian plane. If I may, I will begin with the homework given last week. HOMEWORK
1. 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.
2. A straight line MN cuts the y axis at M(0 , 2). The gradient of MN is – 3. Show that the equation of the line MN is y + 3x = 2 SOLUTION
The line, therefore, has gradient – 3 and intercept 2.
Using the equation y = mx + c and substituting.
The equation is y = – 3 x + 2 y = – 3 x + 2 or y + 3x = 2
3. A straight line is drawn through the points A(– 5 , 3) and B(1 , 2).
(i) Determine the gradient of AB. (ii) Write the equation of the line AB.
It is clear from the above that you need to study the various methods of finding the equation of a line by being competent in identifying the appropriate formula. Having done this, you are only required to select and present the formula and perform effectively the required substitution and clearing.
As we continue to review problems in the Cartesian plane, I wish to remind you that in some instances you are given information which enables you to find a point on the line and the gradient of the line. This information is then used in the appropriate formula.
EXAMPLE
Given the points A (2 , 3) and B (6 , – 1), determine the equation of the perpendicular bisector of AB and state the co-ordinates of the point at which the perpendicular bisector meets the y axis.
The equation of the perpendicular bisector is y – x = – 3.
At the point where this line cuts the y axis, x = 0.
Substituting, y – 0 = –3 y = –3
The coordinates of the point
(0, – 3)
The line cuts the y axis at (0, –3).
Remembering the numerous poor attempts made on this topic in past examinations, I am recommending that you do a conscientious review of this topic.
Let us now attempt the following together.
The coordinates of A and B are (3 , – 2 ) and (7 , 0), respectively. X is the mid point of AB.
(a) Calculate (i) the length of AB (ii) The gradient of AB
(iii) The coordinates of X
(b) Determine the equation of the perpendicular bisector of AB and state the coordinates of the point at which the perpendicular bisector meets the y axis.
Now for your homework.
The coordinates of the points L and N are (5 , 6) and (8 , – 2), respectively.
1. (i) State the coordinates of the midpoint M, of the line LN.
(ii) Calculate the length of the line LN.
(iii) Calculate the gradient of the line LN.
(iv) Determine the equation of the straight line which is perpendicular to LN and which passes through the point M.
2. The line L joining the points (x , 2) and (3 , – 1) has gradient – 3/4. Determine:
(i) The value of x
(ii) The coordinates of the mid point of the line joining the point (5 , 6) to the point (3 , – 1)
(iii) The equation of the line perpendicular to the line represented by y = x + 3 and passing through the point (3 , – 1).
3. A is the point (– 3 , 5) and B the point (3 , – 1).
(i) Find the coordinates M, the midpoint of AB.
(ii) The gradient of AB.
I expect, as usual, that you will find more exercises of this nature in your textbooks and past papers. Please practise as many of them as you can.