Question

Find the gradient of the graph of 3x+2y=6

Answer 1

Solution

$3x+2y=6$

$\Leftrightarrow 2y=-3x+6$

$\Leftrightarrow y=-\dfrac{3}{2} x+3$

$y=-\dfrac{3}{2} x+3$ is a line with the coefficient $-\dfrac{3}{2}$, which is the gradient.

The gradient of the graph is $-\dfrac{3}{2}$.

More information

An equation for a line is $y=ax+b$. Why?

(The 1st Attachment)

Let's move the x-axis to $y=b$.

Then the graph represents a proportional graph $y=ax$. Now we can make infinitely many similar triangles. So the graph is a hypotenuse, which is a line.

So $y=ax+b$ is a line.

This also proves why gradient is constant in a linear equation, with which we can make another formula.

Why is point-slope form $y-y_1=m(x-x_1)$?

(The 2nd attachment)

Let one point on a line be $(x_1,y_1)$.

The gradient is constant in a line, so choose any different points of a line, then $m=\dfrac{y-y_1}{x-x_1}$.

$\therefore y-y_1=m(x-x_1)$

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

• There are three methods to find the gradient or slope of line 3x + 2y = 6

Method 1 :-

• Concept Used :-

Let us consider a line ax + by + c = 0, then slope (m) is given by

$\rm :\implies\:\:\boxed{ \green{ \bf \: slope \: (m) = - \dfrac{coefficient \: of \: x}{coefficient \: of \: y} }}$

So,

• Gradient of line 3x + 2y = 6 is given by

$\tt \longmapsto\: \red{gradient \: = \: - \: \dfrac{3}{2} }$

Method 2 :-

• Reduce the given equation of line in to slope intercept form :-

Concept Used

Let us consider a line in slope intercept form y = mx + c,

where

• m represents slope or gradient of line

• c represents the intercept on y - axis.

Now,

• Equation of line is

$\tt \longmapsto\:3x + 2y = 6$

$\tt \longmapsto\:2y = 6 - 3x$

$\tt \longmapsto\:y = - \dfrac{3}{2} x + \dfrac{6}{2}$

$\rm :\implies\:y = - \dfrac{3}{2} x + 3$

So,

• On comparing with y = mx + c,

we get

$\tt \longmapsto\: \pink{gradient \: = \: - \: \dfrac{3}{2} }$

Method 3 :-

• Method of differentiation

Let us consider a line ax + by + c = 0,

then slope or gradient is evaluated by differentiating the given line w. r. t. x.

Now,

• Given equation of line is

$\tt \longmapsto\:3x + 2y = 6$

On differentiating w. r. t. x, we get

$\tt \longmapsto\:\dfrac{d}{dx}(3x + 2y) = \dfrac{d}{dx}6$

$\tt \longmapsto\:3 + 2\dfrac{dy}{dx} = 0$

$\rm :\implies\:\dfrac{dy}{dx} = - \: \dfrac{3}{2}$

$\tt \longmapsto\: \blue{gradient \: = \: - \: \dfrac{3}{2} }$

Additional Information

Additional Information Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

• Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.
• Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.

2. Point-slope form equation of line

• Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line

• Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line

• Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

• Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, equation of the line is given by x cosβ + y sinβ = p.