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# How do you graph -3x=6y-2 using the intercepts?

May 8, 2016

#### Explanation:

Intercepts formed by the line $- 3 x = 6 y - 2$ on $x$ axis and $y$ axis can be obtained by putting $y = 0$ and $x = 0$ respectively in the equation $- 3 x = 6 y - 2$.

Doing so, intercept on $x$ axis is given by $- 3 x = 6 \cdot 0 - 2 = - 2$ or $x = \frac{2}{3}$. And intercept on $y$ axis is given by $- 3 \cdot 0 = 6 y - 2$ or $y = \frac{1}{3}$.

Hence, intercept on $x$ axis is $\frac{2}{3}$ and that on $y$ axis is $\frac{1}{3}$. So plotting points $\left(\frac{2}{3} , 0\right)$ and $\left(0 , \frac{1}{3}\right)$ and joining them wil give us the desired graph.

graph{3x+6y-2=0 [-0.2865, 0.9635, -0.1375, 0.4875]}

Additional information - Equation of a line which forms an intercept $a$ on $x$ axis and $b$ on $y$ axis is given by $\frac{x}{a} + \frac{y}{b} = 1$.

May 8, 2016

Mart the points
color(green)(=>y_("intercept")->(x,y)->(0,1/3)
=>color(green)(x_("intercept") ->(x,y)->(2/3,0)
and draw a straight line through them but extend it to the edges of the graph.

#### Explanation:

We could if we so chose manipulate the given equation so that we have the standard form of $y = m x + c$ and then determine the intercepts, but there is no need to.

$\textcolor{g r e e n}{\text{Determine x-intercept}}$
Using first principle method

Knowing the First principle method comes in handy when doing higher level math.

The $x \text{-intercept}$ is when $\textcolor{b l u e}{y = 0}$

So by substitution we have:

$\textcolor{b r o w n}{- 3 x = 6 y - 2 \text{ "->" } - 3 x = 6 \left(\textcolor{b l u e}{0}\right) - 2}$

$\implies - 3 x = - 2$

Multiply both sides by $\left(- 1\right)$

$\implies + 3 x = + 2$

Divide both sides by 3

$\frac{3}{3} x = \frac{2}{3}$

But $\frac{3}{3} = 1$

$\textcolor{g r e e n}{{x}_{\text{intercept}} = \frac{2}{3}} \to \left(x , y\right) \to \left(\frac{2}{3} , 0\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Determine y-intercept}}$
Using shortcuts method

The $y \text{-intercept}$ is when $x = 0$

$- 3 \left(0\right) = 6 y - 2$

$\textcolor{g r e e n}{\implies y = + \frac{2}{6} = \frac{1}{3}} \to \left(x , y\right) \to \left(0 , \frac{1}{3}\right)$

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