CHAPTER 1: COORDINATE GEOMETRY
By the end of this chapter, you should be able to:
Equation of a Line
The General Equation of a Straight Line
Derive the general equation of a straight line
COORDINATES OF A POINT
•The coordinates of a points - are the values of x and y enclosed by the brackets which are used to describe the position of point in a line in the plane.
The plane is called xy-plane and it has two axis.
Consider the xy-plane below
The coordinates of points A, B, C ,D and E are A(2, 3), B(4, 4), C(-3, -1), D(2, -4) and E(1, 0).
Find the gradient of the lines joining
(a) The line joining (2, -3) and (k, 5) has a gradient -2. Find k
1. Find the gradientof the line which passes through the following points ;
2. A line passes through (3, a) and (4, -2), what is the value of a if the slope of the line is 4?
3. The gradient of the linewhich goes through (4,3) and (-5,k) is 2. Find the value of k.
FINDING THE EQUATION OF A STRAIGHT LINE
The equation of a straight line can be determined if one of the following is given:-
Find the equation of the line with the following
EQUATION OF A STRAIGHT LINE IN DIFFERENT FORMS
The equation of a line can be expressed in two forms
Find the gradient of the following lines
Find the y-intercept of the following lines
Find the x and y-intercept of the following lines
Attempt the following Questions.
GRAPHS OF STRAIGHT LINES
The graph of straight line can be drawn by using the following methods;
Sketch the graph of Y = 2X - 1
SOLVING SIMULTANEOUS EQUATION BY GRAPHICAL METHOD
Solve the following simultaneous equations by graphical method
1. Draw the line 4x-2y=7 and 3x+y=7 on the same axis and hence determine their intersection point
2. Find the solutionfor each pair the following simultaneous equations by graphical method;
Midpoint of a Line Segment
The Coordinates of the Midpoint of a Line Segment
Determine the coordinates of the midpoint of a line segment
Let S be a point with coordinates (x1,y1), T with coordinates (x2,y2) and M with coordinates (x,y) where M is the mid-point of ST. Consider the figure below:
Considering the angles of the triangles SMC and TMD, the triangles SMC and TMD are similar since their equiangular
Find the coordinates of the mid-point joining the points (-2,8) and (-4,-2)
Therefore the coordinates of the midpoint of the line joining the points (-2,8) and (-4, -2) is (-3,3).
Distance Between Two Points on a Plane
The Distance Between Two Points on a Plane
Calculate the distance between two points on a plane
Consider two points, A(x1,y1) and B(x2,y2) as shown in the figure below:
The distance between A and B in terms of x1, y1,x2, and y2can be found as follows:Join AB and draw doted lines as shown in the figure above.
Then, AC = x2– x1and BC = y2– y1
Since the triangle ABC is a right angled, then by applying Pythagoras theorem to the triangle ABC we obtain
Therefore the distance is 13 units.
Parallel and Perpendicular Lines
Gradients in order to Determine the Conditions for any Two Lines to be Parallel
Compute gradients in order to determine the conditions for any two lines to be parallel
The two lines which never meet when produced infinitely are called parallel lines. See figure below:
The two parallel lines must have the same slope. That is, if M1is the slope for L1and M2is the slope for L2thenM1= M2
Gradients in order to Determine the Conditions for any Two Lines to be Perpendicular
Compute gradients in order to determine the conditions for any two lines to be perpendicular
When two straight lines intersect at right angle, we say that the lines are perpendicular lines. See an illustration below.
Consider the points P1(x1,y1), P2(x2,y2), P3(x3,y3), R(x1,y2) and Q(x3,y2) and the anglesα,β,γ(alpha, beta and gamma respectively).
Therefore the triangle P2QP3is similar to triangle P1RP2.
Generally two perpendicular lines L1and L2with slopes M1and M2respectively the product of their slopes is equal to negative one. That is M1M2= -1.
Show that A(-3,1), B(1,2), C(0,-1) and D(-4,-2) are vertices of a parallelogram.
Let us find the slope of the lines AB, DC, AD and BC
We see that each two opposite sides of the parallelogram have equal slope. This means that the two opposite sides are parallel to each other, which is the distinctive feature of the parallelogram. Therefore the given vertices are the vertices of a parallelogram.
Problems on Parallel and Perpendicular Lines
Solve problems on parallel and perpendicular lines
Show that A(-3,2), B(5,6) and C(7,2) are vertices of a right angled triangle.
Right angled triangle has two sides that are perpendicular, they form 90°.We know that the slope of the line is given by: slope = change in y/change in x
Since the slope of AB and BC are negative reciprocals, then the triangle ABC is a right angled triangle at B.
1. Find the distance between point (? 3,? 2) and the point midway between (2,13) and (4,7). Write your answer in the form a?c where a and c are positive real numbers.
Let (x, y) be the midpoint between
(2, 13) and (4,7)
The midpoint is (3, 10)
Let d be distance between (3,10) and (-3,-2)
2. Find the equation of the line through the points (4,6) and the midpoint of (2,4) and (10,4).Soln.
Given points (4,6), (2,4) and (10,4)
Then new points are (4,6) and 6,4)
Equation, y=m(x – x1)+y. Take point P(4,6)
y=m(x – x1) + y1
y=-1(x – 4)+6
The equation of line y=-x+10
3. Find the equation of the line passing through (6,4) and perpendicular to the line whose equation is 12x + 6y = 9.
If line L1 with slope M1 is perpendicular to line L2 of slope M2 then, M1 M2=-1
Let the equation of line L1 be: 12x + 6y=9 (given). Rearranging in the form y=mx + c,
Using the given coordinates,
is the required equation.
4. The distance between (1,5) and (k+5, k+1) is 8. Find K, given that it is positive
Therefore k =4 since k is positive. 0.5 marks.
5. The gradient of line L1is -2. Another line L2 is perpendicular to L1 and passes through point (-3, -2). What is the equation of L2?
let M1be the gradient of L1, M1 = −2let M2be the gradient of L2,
but M1M2 = -1 For perpendicular lines
- 0.5 marks
-2y − 4 = −x − 3
−2y = −x − 3 + 4 0.5 marks
The equation is 0.5 marks
6(a) The coordinate of P, Q and R are (2, m), (-3, 1) and (6, n) respectively. If the length of PQ is units and midpoint of QR is find the possible value of m and n (b)The gradient of line is -2. Another line L2 is perpendicular to L1 and passes through (-3,-2). What is the equation of L2
(a) Length of PQ= …………………….00
50 = 25 +
25 = 1-2m+m2
m-6=0 or m+4=0
Mid point of QR
From m(x,y) = ………………………00.
-1 = ……………………00
-2 = 1+n
(b) Let m be the gradient of L1
Let m2 be the gradient of L2
........................................................ 00 ½
The equation of L2 is .................................00 ½
7. OBCA form a quadrilateral with coordinates A (-5, 0), B(4,3) C(-1,3) and O(0,0)
Soln. The coordinates of the quadrilateral OBCA are O(0,0), A(-5,0), C(-1,3) and B(4,3)
(i) Slopes AO=(-5-0)/(0-0)=0
Slope of CB =
AO and CB are parallel
(ii) Slope of AC and OB
Slope of AC=
Slope of OB =
OB and AC are parallel
All the sides are equal in length
(c)Equation of AB
Slope of AB=
Equation of line AB
Equation of CO
Equation of line
Equation of CO:
The lines AB and CO are perpendicular
Since, all the sides are equal and opposite sides are parallel, the quadrilateral OBCA is a rhombus.
8 An engineer is in the process of constructing two straight roads, RI and R, which will meet at the right angles. If RI will be represented by the equation 2x—3y—4=0 and R, will pass through the point , find the equation representing R2 in the form ax+by+c=0
Consider the figure below
The equation of R1?2x – 3y – 4=0
2x – 3y – 4=0
3y=2x – 4
hence slope (m)=2/3
Since the two line are in perpendicular
m1m2 = -1
But R2 pass through point (4, -2)
9 Find the length of a line segment joining the points (3,-2) and (15,3) .
L=13 unit length.
10 Let P and Q be two points at (2,5) and (4, -1) respectively. Find
(i)Find equation of the line that passes through the midpoint of PQ and is perpendicular to it in form of ax + by +c=0
(ii)The distance between P and Q
Given P(2,5) and Q(4, -1)
Slope of PQ
Form perpendicular line, M1M2 = -1
-3 x M2 =-1
M2 = 1/3
Now, midpoint of line PQ
M.P=(3 , 2)
x - 3 = 3y - 6
x - 3y + 3 =0
11 Find the equation of the line passing at point (6, 2) and it is perpendicular to the line that crosses the x-axis at 3 and the y-axis at 4.
Given that; point (6, -2) is perpendicular to line that passes at point (3,0) and (0,-4)
Now, lets find slope(M1) of line athe passes at the two points
But for perpendicular lines, M1M2=-1
(4/3 ) M2=-1
-3(x - 6)=4(y + 2)
-3x +18= 4y +8
3x +4y +8 -18=0
Therefore, Equation Of A Line Is 3x +4y - 10=0
12 Find the equation of the line passing passing through the midpoint of the points A(? 3 2, ) and B(1,? )4 and which is perpendicular to line AB .
Lets find the midpoint of A(? 3 2, ) and B(1,? 4)
From midpoint formula
Now lets find the slope of line AB
But for perpendicula lines
M2=-1/ M1 = -1 ÷ -3/2
Now Eqn. of line
Therefore, Eqn. of a line is 2x -3y=1
13 . Find the point of intersection of the lines x ? 2y = ?5 and 2x + 7y ? 34 = 0
Point of intersection btn line x ? 2y = ?5 and 2x + 7y ? 34 = 0 can be obtained by solve simultaneous equation
2x+7y=34 ...(i) Eqn 1
Substitute x in the eqn1 that 2x+7y=34
4y-10 + 7y=34
Therefore, point of intersection is (3, 4)
14 The points P, Q and R are (5, —3), (-6, I) and (1, 8) respectively. Show that these points form an isosceles triangle.
Given point P, Q and R are (5,-3) (-6,1) and (1,8) respectivelly
Required to show these points form an is isosceles triangle.(Two sides are equal) Consider the triangle below:
So, if is an isosceles triangle; PQ=PR
Recall distance formula
Now, find Length of PR
Since, PQ=PR, Hence triangle formed by points P, Q, R is an Isosceles triangle.
15 Find the equation of a line which passes through the point A(-3, 4) and which is parallel to the line 3x + 4y - 15=0.
Condition for parallel line M1=M2
from eqn y= mx +c then, M1=-3/4
Therefore, M2= -3/4
Equation of line passes though point A(-3,4)
Now, cross multiplication
4(y-4) = -3(x+3)
4y-16 = -3x-9
.:The equation of the line is 4y+ 3x-7=0
16 Show that triangle ABC is right-angled where A = (-2,-1), B = (2,1) and C = (1,3).
Given, the vertices Of Triangle ABC are A(-2, -1), B(2, 1) and C(1, 3)
For triangle ABC to be right angled triangle, two lines must meet perpendicularly
For perpendicular lines: M1M2=-1
Slope of AB(M1)
M1= 1/ 2
Slope Of BC(M2)
M1 x M2= 1/ 2 x -2 =-1;
Therefore, triangle ABC is right-angled triangle.