
What is the equation of the line, $y=\dfrac{5}{7}x-12$ in standard form?
Answer
528.6k+ views
Hint: In order to rewrite the given line expression in standard form, we need to transpose the term containing ‘x’ and the constant from R.H.S. to L.H.S., thus making the Right-Hand Side of the equation equal to zero. Now, we will rearrange the terms and compare it with the standard equation of a line to get a standardized line equation.
Complete step-by-step answer:
We have been given our line equation as a linear equation in two variables: $y=\dfrac{5}{7}x-12$ . This is a line equation in two dimensions.
Now, the standard form of writing a line equation in two dimensions is, $ax+by+c=0$, where, a, b and c are constants such that a and b cannot be simultaneously equal to zero.
Thus, to rewrite our equation in standard form, we need to transpose the term containing ‘x’ and the constant from R.H.S. to L.H.S. . We can do that in the following manner:
$\begin{align}
& \Rightarrow y=\dfrac{5}{7}x-12 \\
& \Rightarrow y-\dfrac{5}{7}x+12=0 \\
\end{align}$
Now, we can rationalize this equation by multiplying it with 7 on both sides of the equation. This is done as follows:
$\begin{align}
& \Rightarrow \left( y-\dfrac{5}{7}x+12 \right)\times 7=0\times 7 \\
& \Rightarrow 7y-5x+84=0 \\
& or,5x-7y-84=0 \\
\end{align}$
On comparing our above equation with the standard line equation, we get:
$\begin{align}
& \Rightarrow a=5 \\
& \Rightarrow b=-7 \\
& \Rightarrow c=-84 \\
\end{align}$
Here, we can see that the constant terms a and b are not equal to zero simultaneously. Thus, the equation obtained is the correct equation.
Hence, the equation of the line, $y=\dfrac{5}{7}x-12$ in standard form can be written as $5x-7y-84=0$.
Note: While transposing a term from L.H.S. to R.H.S. or vice versa, we should always make sure to change its sign or else or solution will be wrong. Also, the main aim of transposing terms from one side of the equation to another is to always bring like terms together and separately isolate the variable.
Complete step-by-step answer:
We have been given our line equation as a linear equation in two variables: $y=\dfrac{5}{7}x-12$ . This is a line equation in two dimensions.
Now, the standard form of writing a line equation in two dimensions is, $ax+by+c=0$, where, a, b and c are constants such that a and b cannot be simultaneously equal to zero.
Thus, to rewrite our equation in standard form, we need to transpose the term containing ‘x’ and the constant from R.H.S. to L.H.S. . We can do that in the following manner:
$\begin{align}
& \Rightarrow y=\dfrac{5}{7}x-12 \\
& \Rightarrow y-\dfrac{5}{7}x+12=0 \\
\end{align}$
Now, we can rationalize this equation by multiplying it with 7 on both sides of the equation. This is done as follows:
$\begin{align}
& \Rightarrow \left( y-\dfrac{5}{7}x+12 \right)\times 7=0\times 7 \\
& \Rightarrow 7y-5x+84=0 \\
& or,5x-7y-84=0 \\
\end{align}$
On comparing our above equation with the standard line equation, we get:
$\begin{align}
& \Rightarrow a=5 \\
& \Rightarrow b=-7 \\
& \Rightarrow c=-84 \\
\end{align}$
Here, we can see that the constant terms a and b are not equal to zero simultaneously. Thus, the equation obtained is the correct equation.
Hence, the equation of the line, $y=\dfrac{5}{7}x-12$ in standard form can be written as $5x-7y-84=0$.
Note: While transposing a term from L.H.S. to R.H.S. or vice versa, we should always make sure to change its sign or else or solution will be wrong. Also, the main aim of transposing terms from one side of the equation to another is to always bring like terms together and separately isolate the variable.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

