Question

What are the intercepts of \[3x + 2y = 12\]?

Answer 1

Hint: 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.