Question

Reduce the line $2x-3y+5=0$ in intercept form and hence find intercepts on the axes.

• A. $\frac{x}{(-5/2)}+\frac{y}{(-5/3)}=1$, Intercept on x-axis $=-\frac{5}{2}$, Intercept on y-axis $=-\frac{5}{3}$
• B. $\frac{x}{(-5/2)}+\frac{y}{\left(5/3\right)}=1$, Intercept on x-axis $=-\frac{5}{2}$, Intercept on y-axis $=\frac{5}{3}$
• C. $\frac{x}{\left(5/2\right)}+\frac{y}{(-5/3)}=1$, Intercept on x-axis $=\frac{5}{2}$, Intercept on y-axis $=-\frac{5}{3}$
• D. $\frac{x}{(-5/2)}+\frac{y}{\left(5/3\right)}=-1$, Intercept on x-axis $=-\frac{5}{2}$, Intercept on y-axis $=\frac{5}{3}$

Answer 1

Answer: (B)

$\frac{x}{(-5/2)}+\frac{y}{\left(5/3\right)}=1$, Intercept on x-axis $=-\frac{5}{2}$, Intercept on y-axis $=\frac{5}{3}$

Let's represent the given equation in two-intercept form i.e. in the form of $\frac{x}{a}+\frac{y}{b}=1$, where $a$ is $x-$intercept and $b$ is $y-$intercept.

$2x-3y+5=0\Rightarrow 2x-3y=-5\Rightarrow \frac{2x}{-5}+\frac{-3y}{-5}=1\Rightarrow \frac{x}{\frac{-5}{2}}+\frac{y}{\frac{5}{3}}=1$

x-intercept =$\frac{-5}{2}$ and y-intercept =$\frac{5}{3}.$