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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
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    1 Quiz
  7. Integration | Week 8
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i. Gradient-intercept form

equation of a straight line

In the diagram, the gradient of AB is m and the y-intercept is C. Let P(x, y) be any point on the line, then \( \frac {y \; – \; c}{x} \scriptsize = m \\ \scriptsize or\; y = mx + c \)

ii. Gradient and one point form

straight line e1607613497661

We may however be given the gradient of a line and the coordinates of some other fixed point on the line and not its y-intercept.

Given the gradient of the line AB = m and the coordinates (x, y1) of a point C on the line as shown in the diagram. Let P(x, y) be any other point on the line

Then \( \frac{y \; – \; y_1}{x \; – \; x_1} = \scriptsize m\)

or  y – y1=m (x – x1)

Equation of line through (x, y1) with gradient m is y – y1 = m(x – x1)

Example:

A straight line has a gradient of \(– \frac{3}{2}\) and passes through the point (1,4). 

Find its equation and its intercept on the y-axis.

Solution

(x1, y1) = (1, 4)

m = \( – \frac{3}{2}\)   

y – y1= m(x – x1)

y – 4 = \(– \frac{3}{2} \scriptsize (x – 1) \) 

2y-8 =  -3(x – 1)

2y – 8 = -3x + 3 

∴ 2y+3x = 11

dividing through by 2,

y = \( – \frac {3}{2}\scriptsize x + \frac{11}{2} \)

comparing with y=mx + c, The intercept on y axis is 

:- \( \frac {11}{2} = \scriptsize 5 \frac{1}{2} \)

iii. Two point form

two point form e1607614619285

The points (x1, y1) and (x2, y2) are given on a line AB in the diagram

Then its gradient = \( \frac{y_2 \; – \; y_1}{x_2 \; – \; x_1} = \scriptsize m \)

But y – y1= m (x – x1)

Substituting m we have

y – y1= \( \frac{y_2 \; – \; y_1}{x_2 \; – \; x_1} \scriptsize (x \; – \; x_1) \)

which can be written more clearly in the form

:- \( \frac{y \; – \; y_1}{y_2 \; – \; y_1} = \frac{x \; – \; x_1}{x_2 \; – \; x_1} \)

The equation of a line through the two points (x1,y1),(x2,y2) is 

\( \frac{y \; – \; y_1}{y_2 \; – \; y_1} = \frac{x \; – \; x_1}{x_2 \; – \; x_1} \)

Example:

Find the equation of a line AB which passes through the points (1,-1) and (-2,-13)

Solution

x1= 1, y1 = -1

x2= -2, y2 = – 13

:- \( \frac{y \; – \; y_1}{y_2 \; – \; y_1} = \frac{x \; – \; x_1}{x_2 \; – \; x_1} \)

:- \( \frac{y \; – \; (-1)}{-13 \; – \; (-1)} = \frac{x \; – \; 1}{-2 \; – \; 1} \)

:- \( \frac{y \; +\; 1}{-13 \; + \; 1} = \frac{x \; – \; 1}{-3} \)

-3(y + 1) = -12(x – 1)

-3y – 3 = -12x + 12

-3y = -12x + 15

dividing through with -3

y = 4x – 5

∴ equation of line AB is 

y = 4x – 5 and the gradient is 4

 

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