Question

How do you write the equation $y-1=\dfrac{5}{6}\left( x-4 \right)$ in its standard form?

Answer 1

Hint: We start solving the problem by recalling the fact that the slope point form of equation of line is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ which resembles the given equation. We then recall the fact that the standard form of equation of a line is $ax+by+c=0$ to proceed through the problem. We then make the necessary calculations involving multiplication, addition and subtraction operations to get the required answer for the given problem.

Complete step by step answer:
According to the problem, we are asked to write the equation $y-1=\dfrac{5}{6}\left( x-4 \right)$ in its standard form.
We have given the equation as $y-1=\dfrac{5}{6}\left( x-4 \right)$.

We can see that the given equation resembles the form $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ which is slope point form of equation of a line.
We know that the standard form of the equation of a line is $ax+by+c=0$. Let us convert the given equation into this form.
We have $y-1=\dfrac{5}{6}\left( x-4 \right)$.
$\Rightarrow 6\left( y-1 \right)=5\left( x-4 \right)$.
$\Rightarrow 6y-6=5x-20$.
\[\Rightarrow 5x-20-6y+6=0\].
\[\Rightarrow 5x-6y-14=0\].
So, we have found the standard form of the given equation $y-1=\dfrac{5}{6}\left( x-4 \right)$ as $5x-6y-14=0$.

$\therefore $ The standard form of the given equation $y-1=\dfrac{5}{6}\left( x-4 \right)$ is $5x-6y-14=0$.

Note: Whenever we get this type of problem, we first try to find what does the given equation represent to proceed through the solution. We can also convert the given equation of line to the slope form of the line which is $y=mx+c$. We should not make calculation mistakes while solving this type of problem. Similarly, we can expect problems to convert the given equation of line $5x+4y=40$ to the intercept form.