Question

Reduce the following equations into intercept form and find their intercepts on the axes.
(i)  \(3x+2y-12=0\)
(ii)  \(4x-3y=6\)
(iii)  \(3y+2=0. \)

Answer 1

(i) The given equation is \(3x + 2y -12 = 0.\)
It can be written as 
 \(3x + 2y = 12\)

\(\frac{3x}{12} + \frac{2y}{12} = 1\)

\(i.e,\frac{x}{4} +\frac{ y}{6} = 1........(1)\)
This equation is of the form\( \frac{x}{a} + \frac{y}{b} = 1\) , where \(a = 4\) and \(b = 6. \)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is \(4x - 3y = 6. \)
It can be written as 

\(\frac{4x}{6} –\frac{ 3y}{6} = \frac{6}{6}\)

\(\frac{2x}{3} –\frac{ y}{2} = 1\)

\(\frac{x}{(\frac{3}{2})} + \frac{y}{(-2)} = 1.....(2)\)
This equation is of the form \(\frac{x}{a} +\frac{ y}{b} = 1\), where \(a = \frac{3}{2}, b = -2\)
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are \(\frac{3}{2}\) and \(-2\) respectively.

(iii) The given equation is \(3y + 2 = 0.\)
It can be written as
\(3y = -2\)

\(i.e,\frac{y}{(\frac{-2}{3})} = 1............(3)\)
This equation is of the form \(\frac{x}{a} + \frac{y}{b} = 1\), where \(a = 0, b = \frac{-2}{3}\)
Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is \(\frac{-2}{3}\) and it has no intercept on the x-axis.