How do you find the x and y intercept of $4 x + 4 y = 12$ $4x + 4y = 12$ ?

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How do you find the x and y intercept of $4 x + 4 y = 12$ $4x + 4y = 12$ ?

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Answer 1 · 2016-07-02T07:11:40

Intercept on $x$ $x$ -axis is $3$ $3$ and intercept on $y$ $y$ -axis is also $3$ $3$ .

Explanation:

In $4 x + 4 y = 12$ $4x+4y=12$ , we can find intercept on $x$ $x$ -axis by putting $y = 0$ $y=0$ and intercept on $y$ $y$ -axis by putting $x = 0$ $x=0$ .

Hence intercept on $x$ $x$ -axis is $4 x + 4 \times 0 = 12$ $4x+4xx0=12$ or $4 x = 12$ $4x=12$ or $x = 3$ $x=3$ .

and intercept on $y$ $y$ -axis is $4 \times 0 + 4 y = 12$ $4xx0+4y=12$ or $4 y = 12$ $4y=12$ or $y = 3$ $y=3$ .

Alternatively if intercepts on $x$ $x$ -axis and $y$ $y$ -axis are $a$ $a$ and $b$ $b$ respectively, then equation of line is $\frac{x}{a} + \frac{y}{b} = 1$ $x/a+y/b=1$ . Hence, we can also find intercepts on $x$ $x$ -axis and $y$ $y$ -axis by converting the equation in this form.

Now $4 x + 4 y = 12 \Leftrightarrow \frac{4 x}{12} + \frac{4 y}{12} = 1$ $4x+4y=12hArr(4x)/12+(4y)/12=1$ or

$\frac{x}{3} + \frac{y}{3} = 1$ $x/3+y/3=1$ and hence again we get that intercept on $x$ $x$ -axis is $3$ $3$ and intercept on $y$ $y$ -axis is also $3$ $3$ .

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How do you find the x and y intercept of $4 x + 4 y = 12$ $4x + 4y = 12$ ?

1 Answer

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How do you find the x and y intercept of $4 x + 4 y = 12$ $4x + 4y = 12$ ?