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.
SOLUTION
Given K, the line which passes through the point (4, 1) with gradient (m) 3/5, then the equation is given by: y - y1/ x - x1 = m Substituting, y - 1/ x - 4 = 3 /5 Simplifying,
5y - 5 = 3x - 12
5y - 3x = 5 - 12
5y - 3x = - 7
Answer: 5y - 3x = - 7
Let m1 be the gradient of the perpendicular, m1 x m = - 1 m1 x 3/5 = - 1 m1 = -5/3 and it passes through (4, 1)
The equation is: y - y1 / x -x1 = m y - 1/ x - 4 = -5 /3 Simplifying
3y - 3 = - 5x + 20.
3y+5x = 23.
Answer: 3y + 5x = 23
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.
SOLUTION
Given the points A(- 5 , 3) and B(1 , 2), then (i) Gradient of AB, m = y2 - y1/ x2 - x1
= 2 - 3/1- -5 = - 1/6 m = -1/6
(iii) Since the gradient of AB = -1/6, and AB passes through the point (- 5 3), then the formula for finding the equation of AB is:
y - y1/ x - x1 = m. Substituting the coordinates of the point (- 5 , 3)
y - 3/x - -5 = - 1/6
y - 3/ x + 5 = - 1/6. Clearing, 6(y - 3) = -1(x + 5)
6y - 18 = - x - 5
6y + x = 18 - 5
6y + x = 13
Answer: 6y + x = 13.
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 coordinates of the point at which the perpendicular bisector meets the y axis.
Given the points A (2 , 3) and B (6, - 1), the gradient of AB = y2 - y1 / x2 - x1
= -1 - 3 /6 - 2
= - 4 /4 = - 1
Let the gradient of the line perpendicular to AB be m
m x - 1 = - 1 (product of the gradients of perpendicular lines is - 1)
m=1
The midpoint of AB is M = x2 + x1/2 , y2 + y1/2 Substituting the coordinates of the points, M = 6 + 2/2 , - 1 + 3/2 = 8/2 , 2 /2 = (4 , 1)
The perpendicular bisector of AB has gradient 1 and passes through the point M (4 , 1).
The equation of the perpendicular bisector is found using the formula: y - y1/ x - x1 = m y - 1/x - 4 = 1 Clearing, y-1=x-4y-x=1-4=-3
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 is (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 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 midpoint 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.
SOLUTION
(a)
(i) the length of AB
AB2 = (7 - 3)2 + (0 - - 2) 2 = 42 + 22 = 20 the length of AB = √20
(ii) the gradient of AB Gradient of AB or m = y2 - y1 / x2 - x1 = 0 - - 2/7 - 3 = 2/4 = 1/2 m = 1/2
(iii) the coordinates of X the midpoint of AB or M = x2 + x1/2, y2 + y1/2 M = 7 + 3/2, 0 + -2/2 = (5 , - 1)
M = (5 , - 1)
(b) Since the gradient m of AB is 1/2 The gradient of the perpendicular to AB = - 2.
(N.B. 1/2 x -2 = -1)
Since the bisector of AB passes through (5 , - 1), the midpoint of AB and has gradient - 2 , The equation is found by using the formula: y - y1 / x - x1 = m
By substituting: y - - 1/X - 5 = - 2 Clearing, y + 1 = - 2 x + 10 y = - 2 x - 1 + 10 y=-2x+9
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 midpoint 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)
“Education is the passport to the future, for tomorrow belongs to those who prepare for it today.” – Malcolm X
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.