(i) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e., $\frac{x}{a}+\frac{y}{b}$ = 1 ... (1) Now given, 3x + 2y = 12 ⇒ $\frac{3x}{12}+\frac{2y}{12}$ = 1 ⇒ $\frac{x}{4}+\frac{y}{6}$ = 1 ... (2) On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6. (ii) Given equation is 4x – 3y = 6 We have to reduce the given equation into intercept form, i.e., $\frac{x}{a}+\frac{y}{b}$ = 1 ... (1) $\frac{4}{6}x - \frac{3}{6}y$ = 1 or $\frac{x}{\frac{3}{2}}+\frac{y}{-2}$ = 1 ... (2) On comparing (1) and (2), we get a = $\frac{3}{2}$ and b = - 2 Hence, the intercepts of the line are $\frac{3}{2}$ and - 2 iii) Given equation is 3y + 2 = 0 We have to reduce the given equation into intercept form, i.e., $\frac{x}{a}+\frac{y}{b}$ = 1 3y = - 2 ⇒ y = $-\frac{2}{3}$ The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis. We observe that y-intercept of the line is $-\frac{2}{3}$ , but there is no intercept on x-axis.