
How to convert slope intercept form of equation $y = - \dfrac{1}{3}x + 4$ into standard form ?
Answer
557.4k+ views
Hint: In the given problem, we are required to convert the slope intercept form of a line whose equation into the standard equation of a straight line. We can easily tell the slope of a line written in slope intercept form. For converting the line from slope intercept form, we need to have knowledge of algebraic methods like transposition rule. The standard form of a straight line is $ax + by + c = 0$.
Complete step-by-step solution:
We are required to convert the equation of straight line $y = - \dfrac{1}{3}x + 4$ from slope intercept form to standard form of a straight line.
So, we are given the equation of a straight line in slope intercept form as $y = - \dfrac{1}{3}x + 4$ .
For writing in the standard form of the equation of a straight line $ax + by + c = 0$ , we need to have thorough knowledge of algebraic rules such as transposition rules.
So, $y = - \dfrac{1}{3}x + 4$
Now multiplying both the sides of the equation by $3$, we get,
$ \Rightarrow 3y = 3\left( { - \dfrac{1}{3}x + 4} \right)$
Opening bracket and solving further, we get,
$ \Rightarrow 3y = - x + 12$
Rearranging the terms in order to resemble the standard equation of a straight line $ax + by + c = 0$. So, shifting all the terms to the right side of the equation, we get,
$ \Rightarrow 3y + x - 12 = 0$
So, the slope intercept form of equation $y = - \dfrac{1}{3}x + 4$ can be converted into the standard form of equation of straight line as $3y + x - 12 = 0$.
Note: We can find the slope of a line by expressing it in point and slope form as well as slope and intercept form. We can also apply a direct formula for calculating the slope of a line: Slope of line$ = - \left( {\dfrac{{{\text{Coefficient of x}}}}{{{\text{Coefficient of y}}}}} \right)$ when the equation of straight line is written in standard form of a straight line as $ax + by + c = 0$ .
Complete step-by-step solution:
We are required to convert the equation of straight line $y = - \dfrac{1}{3}x + 4$ from slope intercept form to standard form of a straight line.
So, we are given the equation of a straight line in slope intercept form as $y = - \dfrac{1}{3}x + 4$ .
For writing in the standard form of the equation of a straight line $ax + by + c = 0$ , we need to have thorough knowledge of algebraic rules such as transposition rules.
So, $y = - \dfrac{1}{3}x + 4$
Now multiplying both the sides of the equation by $3$, we get,
$ \Rightarrow 3y = 3\left( { - \dfrac{1}{3}x + 4} \right)$
Opening bracket and solving further, we get,
$ \Rightarrow 3y = - x + 12$
Rearranging the terms in order to resemble the standard equation of a straight line $ax + by + c = 0$. So, shifting all the terms to the right side of the equation, we get,
$ \Rightarrow 3y + x - 12 = 0$
So, the slope intercept form of equation $y = - \dfrac{1}{3}x + 4$ can be converted into the standard form of equation of straight line as $3y + x - 12 = 0$.
Note: We can find the slope of a line by expressing it in point and slope form as well as slope and intercept form. We can also apply a direct formula for calculating the slope of a line: Slope of line$ = - \left( {\dfrac{{{\text{Coefficient of x}}}}{{{\text{Coefficient of y}}}}} \right)$ when the equation of straight line is written in standard form of a straight line as $ax + by + c = 0$ .
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

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

