Question 101

Find the intercepts made by the line 3x + 4y - 12 = 0 on the axes:

Solution

The intercept form of the line = $$\ \frac{\ x}{a}+\ \frac{\ y}{b}\ =\ 1$$ where a is x-intercept and b is y-intercept.

3x + 4y - 12 = 0

3x + 4y = 12

$$\ \frac{3\ x}{12}+\ \frac{\ 4y}{12}\ =\ 1$$

$$\ \frac{\ x}{4}+\ \frac{\ y}{3}\ =\ 1$$

So the intercepts are 4 and 3.

B is the correct answer.

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