Question

What are the intercepts of $3x + 2y = 12$?

Answer 1

Let us consider the equation of a line which cuts off intercepts $a$ and $b$ respectively from the $x$ and $y$ axes are $\dfrac{x}{a} + \dfrac{y}{b} = 1$. It represents that the straight-line cuts $x$ and $y$ axes at the points $(a,0)$ and $(0,b)$ respectively.

Complete step-by-step solution:
It is given that; the equation is $3x + 2y = 12$
We have to find the intercepts of the given equation $3x + 2y = 12$.
Now we convert the given equation into intercept form.
We have,
$3x + 2y = 12$
Dividing both side by $12$we get,
$\dfrac{{3x + 2y}}{{12}} = \dfrac{{12}}{{12}}$
Simplifying we get,
$\dfrac{{3x}}{{12}} + \dfrac{{2y}}{{12}} = \dfrac{{12}}{{12}}$
Simplifying again we get,
$\dfrac{x}{4} + \dfrac{y}{6} = 1$, which is in intercept form.
Therefore, the intercepts are $(4,0)$ and $(0,6)$.
Hence, the intercepts of $3x + 4y = 12$ are $(4,0)$ and $(0,6)$.

Note: (i) The straight line $\dfrac{x}{a} + \dfrac{y}{b} = 1$intersects the x-axis at A $(a,0)$ and the y-axis at B $(0,b)$.
(ii) In $\dfrac{x}{a} + \dfrac{y}{b} = 1$, a is x-intercept and b is y- intercept.These intercept a and b may be positive as well as negative.
(iii) If the straight-line AB passes through the origin, then, a = 0 and b = 0. If we put a = 0 and b = 0 in the intercept form, then $\dfrac{x}{0} + \dfrac{y}{0} = 1$, which is undefined. For this reason, the equation of a straight line passing through the origin cannot be expressed in the intercept form.
(iv) A line parallel to the x-axis does not intercept the x-axis at any finite distance and hence, we cannot get any finite x- intercept (i.e., a) of such a line. For this reason, a line parallel to x-axis cannot be expressed in the intercept from. In like manner, we cannot get any finite y- intercept (i.e., b) of a line parallel to y-axis and hence, such a line cannot be expressed in the intercept form.