Question

Find $x$ and $y$ intercepts of the given line:

$2x+3y-7=0$

• A. $\frac{x}{\frac{7}{2}}+\frac{y}{\frac{7}{3}}=1,a=\frac{7}{2},b=\frac{7}{3}$
• B. $\frac{x}{\frac{7}{2}}-\frac{y}{\frac{7}{3}}=1,a=\frac{7}{2},b=\frac{7}{3}$
• C. $\frac{x}{\frac{2}{7}}+\frac{y}{\frac{3}{7}}=1,a=\frac{2}{7},b=\frac{3}{7}$
• D. $\frac{x}{\frac{7}{3}}+\frac{y}{\frac{7}{2}}=1,a=\frac{7}{3},b=\frac{7}{2}$

Answer 1

Answer: (A)

$\frac{x}{\frac{7}{2}}+\frac{y}{\frac{7}{3}}=1,a=\frac{7}{2},b=\frac{7}{3}$

Given equation is $2x+3y-7=0$ ........ $\left(i\right)$

Any line can be expressed 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.

Let's represent $\left(i\right)$ in two-intercept form

From $\left(i\right)$

$2x+3y=7$

$⟹\frac{2x}{7}+\frac{3y}{7}=1$

$⟹\frac{x}{\frac{7}{2}}+\frac{y}{\frac{7}{3}}=1$

$a=\frac{7}{2},b=\frac{7}{3}$

Hence, option A is correct.