i. Gradient-intercept form
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
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
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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