Slope of a straight line — lesson. Mathematics State Board, Class 10.

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The general form of the equation of the straight line is

$ax + by + c = 0$ .

Here, the coefficient of

$x = a$ .

Coefficient of

$y = b$ .

Constant term

$= c$ .

The equation

$ax + by + c = 0$ can be written as:

$y = -\frac{a}{b}x - \frac{c}{b}$ where

$b \neq 0$

Here, the slope is

$m = -\frac{a}{b}$ , and the

$y$ -intercept is

$-\frac{c}{b}$ .

That is, Slope

$m = - \frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ and

$y$ -intercept is

$-\frac{\text{Constant term}}{\text{Coefficient of y}}$

Example:

Find the slope and

$y$ -intercept of the straight line

$3x + 9y - 6 = 0$ .

Solution:

The given equation of the line is

$3x + 9y - 6 = 0$ .

We know that Slope

$m = - \frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ and

$y$ -intercept is

$-\frac{\text{Constant term}}{\text{Coefficient of y}}$

Here, the Coefficient of

$x = 3$ , Coefficient of

$y = 9$ and Constant term

$= - 6$

Thus, Slope

$m = -\frac{3}{9} = - \frac{1}{3}$

$y$ -intercept

$= - \frac{(-6)}{9} = \frac{2}{3}$

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2. Slope of a straight line

Theory: