Question

How do you find a standard form equation from the line with points $P=\left( -5,-5 \right)$ and $Q=\left( -3,-2 \right)$?

Answer 1

Now we are given with two points on the line. We know that the equation of line in two point form is given by $\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are the points on the line. Hence we will get the equation of the required line. Now we will simplify the obtained equation to write it in general form of the line which is $ax+by+c=0$ . hence we have the equation of line in general form.

Complete step by step answer:
Now we know that the equation of line in general form is a linear equation in two variables of the form $ax+by+c=0$.
Now we are given two points on the line.
We know that if $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are the points on the line then the equation of the line in two point form is given by $\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
Now we know that $P=\left( -5,-5 \right)$ and $Q=\left( -3,-2 \right)$are the points on the line
Now let $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -5,-5 \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)=\left( -3,-2 \right)$ .
Hence the equation of the line is given by $\dfrac{y-\left( -5 \right)}{-2-\left( -5 \right)}=\dfrac{x-\left( -5 \right)}{-3-\left( -5 \right)}$
Now on simplifying the equation we get,
$\begin{aligned} & \Rightarrow \dfrac{y+5}{5-2}=\dfrac{x+5}{5-3} \\\ & \Rightarrow \dfrac{y+5}{3}=\dfrac{x+5}{2} \\\ \end{aligned}$
Now to eliminate the fractions in the equation we will multiply the equation by LCM of denominators Hence we get,
$\begin{aligned} & \Rightarrow 6\times \dfrac{y+5}{3}=6\times \dfrac{x+5}{2} \\\ & \Rightarrow 2y+10=3x+15 \\\ \end{aligned}$
Now on rearranging the terms we get,
$\Rightarrow 3x-2y+5=0$ .
Now the equation is in standard form of linear equation in two variables which is $ax+by+c=0$ . Hence we have the equation of the line in general form.

Note: Now note that we can also avoid using the equation of the line in two point form. We know that the slope of line is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Now the equation of the line in Slope point form is $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ where m is the slope of the line.