Question

How do you write $y-6=\dfrac{4}{3}\left( x-3 \right)$ in standard form.

Answer 1

Hint: Now the given equation is a linear equation in two variables. We know the standard form of the equation is $ax+by+c=0$ . Now to convert it in standard form we will first multiply the equation by 3 to eliminate the fraction. Now we will use distributive property and simplify the equation. Now we will rearrange the terms to bring the equation in its standard form.

Complete step by step solution:
Now consider the given equation $y-6=\dfrac{4}{3}\left( x-3 \right)$ .
The equation is a linear equation in two variables. Now we know that the standard form of linear equations in two variables is ax + by + c = 0.
Now we will convert the given equation in this form. To do so let us first eliminate the fraction by multiplying the whole equation by the denominator of fraction which is 3.
Hence we get,
$\begin{align}
  & \Rightarrow 3\left( y-6 \right)=4\left( x-3 \right) \\
\end{align}$
Now we know that according to distributive property $a.\left( b+c \right)=a.b+a.c$ . hence using this we get,
$3y-18=4x-12$ .
Now transposing all the terms on RHS we get,
$\begin{align}
  & \Rightarrow -4x+3y-18+12=0 \\
 & \Rightarrow -4x+3y-6=0 \\
\end{align}$
Now the equation is in the standard form $ax+by+c=0$ where a = - 4, b = 3 and c = - 6.
Hence the equation is brought in its standard form.

Note:
Now note that the given equation is in slope point form. The slope point form of a linear equation is given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point on the line and m is the slope of the line. Similarly we can also write the linear equation in slope intercept form which is $y=mx+c$ where m is the slope of line and c is the y intercept.