Question #c3471

Jan 31, 2017

$\textcolor{red}{5} x - \textcolor{b l u e}{3} y = \textcolor{g r e e n}{0}$

Explanation:

First, we can obtain an equation for the line using the point-slope formula. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the values from the problem gives:

$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{5}{3}} \left(x - \textcolor{red}{3}\right)$

The standard form of a linear equation is:

$\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

We can begin solving for this form by multiplying each side of the equation by $\textcolor{p u r p \le}{3}$ to eliminate the fraction:

$\textcolor{p u r p \le}{3} \left(y - \textcolor{red}{5}\right) = \textcolor{p u r p \le}{3} \times \textcolor{b l u e}{\frac{5}{3}} \left(x - \textcolor{red}{3}\right)$

$3 y - 15 = \cancel{\textcolor{p u r p \le}{3}} \times \textcolor{b l u e}{\frac{5}{\cancel{3}}} \left(x - \textcolor{red}{3}\right)$

$3 y - 15 = 5 x - 15$

Next we can add $\textcolor{red}{15}$ and subtract $\textcolor{b l u e}{5 x}$ to each side of the equation to isolate the constant on the right side of the equation while maintaining the balance of the equation.

$- \textcolor{b l u e}{5 x} + 3 y - 15 + \textcolor{red}{15} = - \textcolor{b l u e}{5 x} + 5 x - 15 + \textcolor{red}{15}$

$- 5 x + 3 y - 0 = 0 - 0$

$- 5 x + 3 y = 0$

Now, we can multiple each side of the equation by $\textcolor{red}{- 1}$ to make the coefficient of $x$ positive or non-negative.

$\textcolor{red}{- 1} \left(- 5 x + 3 y\right) = \textcolor{red}{- 1} \times 0$

$5 x - 3 y = 0$