# Find the x and y-intercepts of the line 4x-y-7 = 0 by converting the given equation in double intercept form $\dfrac{x}{a}+\dfrac{y}{b}=1$

[a] $\dfrac{x}{\dfrac{4}{7}}-\dfrac{y}{7}=1,a=\dfrac{4}{7},b=-7$

[b] $\dfrac{x}{\dfrac{4}{7}}+\dfrac{y}{-7}=1,a=\dfrac{4}{7},b=-7$

[c] $\dfrac{x}{\dfrac{7}{4}}-\dfrac{y}{7}=1,a=\dfrac{7}{4},b=7$

[d] $\dfrac{x}{\dfrac{7}{4}}+\dfrac{y}{-7}=1,a=\dfrac{7}{4},b=-7$

Last updated date: 22nd Mar 2023

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Answer

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Hint: First convert the equation in ax+by = c form. Then divide both sides by c. Make numerators independent of a and b by dividing numerators and denominators of individual fractions by a and b respectively.

We have 4x-y-7=0.

Adding 7 on both sides, we get

4x - y - 7 + 7 = 0 + 7

i.e. 4x - y = 7

Dividing both sides by 7, we get

$\begin{align}

& \dfrac{4x}{7}-\dfrac{y}{7}=1 \\

& \Rightarrow \dfrac{4x}{7}+\dfrac{-y}{7}=1 \\

\end{align}$

Dividing numerator and denominator of $\dfrac{4x}{7}$ by 4 and that of $\dfrac{-y}{7}$ by -1, we get

$\begin{align}

& \dfrac{\dfrac{4x}{4}}{\dfrac{7}{4}}+\dfrac{\dfrac{-y}{-1}}{\dfrac{7}{-1}}=1 \\

& \Rightarrow \dfrac{x}{\dfrac{7}{4}}+\dfrac{y}{-7}=1 \\

\end{align}$

Comparing with $\dfrac{x}{a}+\dfrac{y}{b}=1$, we get

$a=\dfrac{7}{4}$ and $b=-7$.

Hence option [d] is correct.

Note: Alternatively we can find x-intercept by putting y = 0 and y-intercept by putting x = 0 in the equation of the line.

If y = 0, we have

4x-(0)-7=0

i.e. 4x = 7

Dividing both sides by 4, we get

$\begin{align}

& \dfrac{4x}{4}=\dfrac{7}{4} \\

& \Rightarrow x=\dfrac{7}{4} \\

\end{align}$

Hence we have $a=\dfrac{7}{4}$

Similarly, if x = 0, we have

4(0)-y-7=0

-y-7=0

Adding 7 both sides, we get

-y-7+7 = 0+7

i.e. -y = 7

Multiplying by -1 on both sides, we get

-1(-y) = -1(7)

y = -7

Hence b = -7

Which is the same as obtained above.

Alternatively: For lines in general form Ax+By+C = 0, we have x-intercept = $\dfrac{-C}{A}$ and y-intercept = $\dfrac{-C}{B}$.

We have A = 4, B = -1 and C =-7. Using the above result, we have

x-intercept $=\dfrac{-C}{A}=\dfrac{-\left( -7 \right)}{4}=\dfrac{7}{4}$ and y-intercept $=\dfrac{-C}{B}=\dfrac{-\left( -7 \right)}{-1}=-7$

__Complete step-by-step solution -__We have 4x-y-7=0.

Adding 7 on both sides, we get

4x - y - 7 + 7 = 0 + 7

i.e. 4x - y = 7

Dividing both sides by 7, we get

$\begin{align}

& \dfrac{4x}{7}-\dfrac{y}{7}=1 \\

& \Rightarrow \dfrac{4x}{7}+\dfrac{-y}{7}=1 \\

\end{align}$

Dividing numerator and denominator of $\dfrac{4x}{7}$ by 4 and that of $\dfrac{-y}{7}$ by -1, we get

$\begin{align}

& \dfrac{\dfrac{4x}{4}}{\dfrac{7}{4}}+\dfrac{\dfrac{-y}{-1}}{\dfrac{7}{-1}}=1 \\

& \Rightarrow \dfrac{x}{\dfrac{7}{4}}+\dfrac{y}{-7}=1 \\

\end{align}$

Comparing with $\dfrac{x}{a}+\dfrac{y}{b}=1$, we get

$a=\dfrac{7}{4}$ and $b=-7$.

Hence option [d] is correct.

Note: Alternatively we can find x-intercept by putting y = 0 and y-intercept by putting x = 0 in the equation of the line.

If y = 0, we have

4x-(0)-7=0

i.e. 4x = 7

Dividing both sides by 4, we get

$\begin{align}

& \dfrac{4x}{4}=\dfrac{7}{4} \\

& \Rightarrow x=\dfrac{7}{4} \\

\end{align}$

Hence we have $a=\dfrac{7}{4}$

Similarly, if x = 0, we have

4(0)-y-7=0

-y-7=0

Adding 7 both sides, we get

-y-7+7 = 0+7

i.e. -y = 7

Multiplying by -1 on both sides, we get

-1(-y) = -1(7)

y = -7

Hence b = -7

Which is the same as obtained above.

Alternatively: For lines in general form Ax+By+C = 0, we have x-intercept = $\dfrac{-C}{A}$ and y-intercept = $\dfrac{-C}{B}$.

We have A = 4, B = -1 and C =-7. Using the above result, we have

x-intercept $=\dfrac{-C}{A}=\dfrac{-\left( -7 \right)}{4}=\dfrac{7}{4}$ and y-intercept $=\dfrac{-C}{B}=\dfrac{-\left( -7 \right)}{-1}=-7$

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