
How to convert slope intercept form of equation $y = - \dfrac{1}{3}x + 4$ into standard form ?
Answer
550.5k+ 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 $ax + by + c = 0$ .
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