Basic Mathematics Form Four Notes – Coordinate Geometry

Basic Mathematics Form Four Notes – Coordinate Geometry

These Basic Mathematics Form Four notes cover Coordinate Geometry. The material is arranged in a clear mobile-friendly reading format while preserving the recognized definitions, explanations, examples, activities, calculations and revision material from the source notes.

Formula & Symbol Clarity

  • Gradient: m = (y₂ – y₁) / (x₂ – x₁)
  • Point-gradient form: y – y₁ = m(x – x₁)
  • Distance: d = √[(x₂ – x₁)² + (y₂ – y₁)²]
  • Midpoint: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

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.

1. horizontal axis known as axis and

2. vertical axis known as 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).

Definition

* Gradient or slope of a line — is defined as the measure of steepness of the line * When using coordinates, gradient is defined as change in y to the change in x

change in y

Gradient = change inx

© Consider two points A(x;,,) and B(x2, y2), the slope between the two points is given by:-

Gradient = 2374 xg — X4

Yi — Y2 Xy—X2

Gradient =
Find the gradient of the lines joining (a) (5,1) and (2,-2) (b) (4,—2)and (—1,0) (c) (—2,—3) and (—4,—7) Solution (a) (5,1) and (2,—2) gradient (b) (4,-2) and (—1,0) gradient

(c) (—2,—3) and (—4,—7)

Y2—-1_—-7–3_-7+3_ -4 Xy—-X, —-4–2 -44+2 -2

gradient = =2
Example 2 (a) The line joining (2, -3) and (k, 5) has a gradient -2. Find k

(b) Find the value of m if the line joining the points (—5, —3) and (6, m) has a slope of i

Solution

(a) Given (2,—3) and (k,5) gradient =2et

5- -3

= —2(k-—2) =5+3
—2k+4=8
—2k=8-4
—2k=4
=2
k=

+. The value of k is —2

(b) Given (—5,—3) and (6,m) Yara

gradient =
2(m + 3) = 11
2m+6=11
2m = 11-6 2m=5
= m²

+. The value of k is

Exercise 1

1. Find the gradientof the line which passes through the following points ;

(3,6) and (-2,8)

b. (0,6) and (99,-12)

c. (4,5)and (5,4)

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:-

e¢ The gradient and the y — intercept (at x =0) or x — intercept ( at y=0) © The gradient and a point on the line * Since only one point is given, then

* Two points on the line

Example 3

Find the equation of the line with the following a. Gradient 2 and intercept b. Gradient and passing through the point

c. Passing through the points and

Solution

(a) Given m=2 y=mxt+ce
y=2x-4 (b) Recall y2—)1
'adient = al X2-X
y- x=2
5=
—2(x — 2) = 3(y- 4) —2x+4=3y-—12 —2x+4-3y+12=0 —2x —3y+16=0 Divide by the negative sign, (—), throughout the equation «The equation of the line is 2x + 3y —-16=0 (c) Recall y2— i
'adient = ———— & x2 —%y

y-N x—X y-4

gradient =
x-3=y-4 x-3-y+4=0 x-y+1=0
«The equation of the line is x -y+1=0

Equation Of A Straight Line In Different Forms

The equation of a line can be expressed in two forms

(i) ax+by+c=0 and (ii) y=mxte Consider the equation of the form y = mx +c
m = Gradient of the line

Example 4

Find the gradient of the following lines
(a) 2y=Sx+1 (b) 2x+3y=5 (c) xt+y= Solution
(a) Express in the form of y = mx +c Divide by both sides
+. Gradient = s (b) Express in the form of y = mx +c Divide by both sides 2x+3y=5 3y =S—2 × 3y = —2x+5
« Gradient = (c) x+y=3 Express in the form of y = mx +c
". Gradient = —1

Intercepts

e Theline ofthe form y= mx +c, crosses the y— axis when x = 0 and alsocrosses x — axis when y =0 e See the figure below
Therefore (i) to get x — intercept, let y = 0 and (ii) to get y — intercept, let x = 0 From the line, y = mx +c y — intercept, let x = 0 y=m(0)+c=0+c=c y — intercept
¢ Therefore, in the equation of the form y = mx +c, mis the gradient and c is the y — intercept

Example 5

Find the y-intercept of the following lines

(a) y=3 × 4+5 9 (b) y=—tx42

Solution

(a) y=3x+5 Compare with y = mx +c y —intercept=c=5
" y — intercept is 5 1 2 (b) y=—Zx45
y — intercept = (c) 3y =2 × 4+1 Express in the form of y = mx +c Divide by 3 both sides
y — intercept =

Example 6

Find the x and y-intercept of the following lines

(a) 2x-—3y-—2=0 (b) 2y-—4x+5=0 Solution
(a) x — intercept, let y = 0 2x —3(0)-2=0 2x-0-2=0 2x-2=0 2x=2
x=5=
y — intercept, let x = 0 2(0) -—3y-2=
" x—intercept=1, y-— intercept = — (b) x — intercept, let y =0
2y —4x+5=0 2(0)-4x+5=0 0-4x+5=0 —4x+5=0 —4x=-5 a= tah | y — intercept, letx = 0 2y-4x+5=0 2y -4(0)+5=0 2y-0+5=0 2y+5=0 2y =-5
yo=–
* x — intercept = ., y — intercept = -f

Exercise 2

Attempt the following Questions.

Find the y-intercept of the line 3x+2y = 18 .

What is the x-intercept of the line passing through (3,3) and (-4,9)?

Calculate the slope of the line given by the equation x-3y= 9

Find the equation of the straight line with a slope -4 and passing through the point (0,0).

Find the equation of the straight line with y-intercept 5 and passing through the point (- 4,8).

Graphs Of Straight Lines

The graph of straight line can be drawn by using the following methods; By using intercepts By using the table of values

Example 7

Sketch the graph of Y = 2X – 1 Solution
(i) By using intercepts y — intercept, let x = 0
x — intercept, let y =

The coordinates are (5.0) and (0, -1)

Then show the straight line through the point (5 ' 0) and (0, -1) on the xy —plane.

By using the table of values

-1} 0 / 1 / 2{3 —3}-1/]1|3{5

Solving Simultaneous Equation By Graphical Method

. Use the intercepts to plot the straight lines of the simultaneous equations . The point where the two lines cross each other is the solution to the simultaneous

equations

Example 8

Solve the following simultaneous equations by graphical method

Palla 3x —-2y=11
Solution Consider 4x +5y=8
ff x=0, 0+Sy=8 y= -=16
ify=0, 4x+0=8 xmtan2

Draw a straight line of the point (2, 1.6) on the xy — plane

Consider 3x—2y=11 if x=0, 0-2y=11 y===-55 i fy =0, 3x—-O=11 x= 37

Draw a straight line of the point (—5.5, 3.7) on the xy — plane

Exercise 3 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; y-x = 3 and 2x+y =9 3x- 4y=-1 and x+y = 2

x =8 and 2x-3y= 10

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 (x;,y:), T with coordinates (x2,y2) and M with coordinates (x,y)

where M is the mid-point of ST. Consider the figure below:

Tx2/¥2)

y2-y ¢ Changeiny M(x,y)

X2-X ion y-% S(x,y)

come x

T

Considering the angles of the triangles SMC and TMD, the triangles SMC and TMD are similar

since their equiangular

x-X, X2.-x

But since M is the midpoint of S and T, then = 21
Therefore + = 1

X2-X

Or x — x4 = Xq-x X2 +X, z
: . SM MC Again, using —— = — ng MT TD

SM __ y-Yi

MT y2-y = y-Y¥s Ya-¥ Or y-y1=yo-y
2y=yory1 = Yat Va

Thus, the coordinates of M(x,y) are eS bi ws ans),

Generally, the coordinates of the mid-point of any line segment is given by ae a $i Li ws.

Example 9

Find the coordinates of the mid-point joining the points (-2,8) and (-4,-2)

Solution

X2 +X V2 = wt

The midpoint is given by ( a

Let x1 be -2, Xp = -4 yy be 8, yp = -2

The midpoint will be:

+ + -2+(-4) 8+(-2) = Xi Y2 Yay = ( —— )

zz 2

= (3)
= (-3, 3).

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(x;,y:) and B(x2,y2) as shown in the figure below:

B(x2,¥2)

The distance between A and B in terms of x), yi,X2, and y2can be found as follows:Join AB and

draw doted lines as shown in the figure above.

Then, AC = x2- x,and BC = yo yi

Since the triangle ABC is a right angled, then by applying Pythagoras theorem to the triangle ABC we obtain

(Ab)? = (Ac)? + (Bc)?

(AB) ? = (xp – x1? + Ya-yi)?
(AB) = {e- x)? + (92 — ys?

Generally the distance between two points is given by:

d= {@: — x,)?+(9, — y)," » whereby d is the distance between two points.
Example Find the distance between the points (-1,7) and (4,-5) Solution The distance between two points is given by: d = |- x)? +(y%2—y)" Let x; be -1 and x» be 4

y; be 7 and yp be -5 thus,

d= /(4—(-1))? + (-5—7)?
= FF Ci = ¥25+ 144
= 7169 =13

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 Miis the slope for Ljand Mois the slope for LathenM)= 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.

'P3(X3,¥3)

mr Qxs,¥2)

P,(x2,Y2)

Consider the points Pj(xi,yi), P2(x2,y2), P3(xs,y3), R(xi,y2) and Q(x3,y2) and the anglesa,p,y(alpha, beta and gamma respectively).

at+B = 90 (complementary angles) a+y= 90 (complementary angles) B = y (alternate interior angles)

Therefore the triangle P2QP3is similar to triangle P;RP2

Ya Y2 × 3- y2~

But the slope of Ly = Mj =
And the slope of Lz = Mj =
ap = From 2— X2 = ¥2~Y1 Y2— Y2 × 27 × 4

ne i ee L

va 92 × 3— X2 _ ya7 Ya _ — _

X2— X4 1— X2

Therefore, —— -m² or MyM>2 = -1.
Generally two perpendicular lines L;and Lzwith slopes Mjand Morespectively the product of their slopes is equal to negative one. That is M;m²= -1.

Example 10

Show that A(-3,1), B(1,2), C(0,-1) and D(-4,-2) are vertices of a parallelogram. Solution

Let us find the slope of the lines AB, DC, AD and BC

change iny

The slope of the line = change inx

Thus, 2-1 1-(-3) 1—(-2)

Slope of the line AB = rr
Slope of the line AD = 3

2-(-1) _4 1-0

-1-(-2)_ 1 o-(-4) 4

Slope of the line BC =
Slope of the line CD =

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 Example 11 Show that A(-3,2), B(5,6) and C(7,2) are vertices of a right angled triangle.

Solution

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

Now,

Slope of the line AB = =
Slope of BC = =

Since the slope of AB and BC are negative reciprocals, then the triangle ABC is a right angled

triangle at B.

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