Question

How do you write the equation of the line in the form AX+BY = C if two points on the line are (-6,6) and (3,-4).

Answer 1

Hint: In this question, we are given two points on a line and we need to form an equation of line using these points. For this, we will use the two points form of writing the equation of the line when we are given any two points $ \left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right) $ on a line, equation of line is given by the formula $ y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right) $ . After finding the equation we will rearrange it into the form AX+BY = C to find our required equation of line.

Complete step by step answer:
Here we are given the two points on a line which are (-6,6) and (-3,4). We need to form the equation of the line in which these points lie. For this, we will use the two point’s formula.
Let us first understand the formula, for a line passing through one point $ y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) $ we generally have the equation of line as $ \left( {{x}_{1}},{{y}_{1}} \right) $ where m is the slope of line. But when we are not given slope, but another point on line, we find slope using that point. Slope for two points $ \left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right) $ becomes $ \left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right) $ . So our equation becomes $ y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right) $ which is the two point form of the equation.
Here we have $ \left( {{x}_{1}},{{y}_{1}} \right) $ as (-6,6) and $ \left( {{x}_{2}},{{y}_{2}} \right) $ as (-3,4).

Putting in the values in the equation, we get,
\[\begin{align}
  & y-6=\left( \dfrac{4-6}{-3-\left( -6 \right)} \right)\left( x-\left( -6 \right) \right) \\
 & \Rightarrow y-6=\left( \dfrac{-2}{-3+6} \right)\left( x+6 \right) \\
 & \Rightarrow y-6=\dfrac{-2}{3}\left( x+6 \right) \\
\end{align}\]
We want our equation of line in the form AX+BY = C. So let us simplify the found equation of line.
We have an equation of line as $ y-6=\dfrac{-2}{3}\left( x+6 \right) $ .
Cross multiplying we get $ 3\left( y-6 \right)=-2\left( x+6 \right) $ .
Multiplying constant inside the bracket we get $ 3y-\left( 3 \right)\left( 6 \right)=-2x+\left( -2 \right)\left( 6 \right) $ .
Simplifying we get $ 3y-18=-2x-12 $ .
Taking variables x and y to one side and the constant to the other side we get $ 3y+2x=-12+18 $.
Simplifying it we get $ 3y+2x=6 $ .
Rearranging we get $ 2x+3y=6 $ .
This equation is of the form AX+BY = C where A = 2, B = 3 and C = 6. Hence this is our required equation.
Note:
 Students should take care of the signs while solving the equation. Note that the order in the two points form should not be changed. Do not forget to simplify the equation to convert it into a general equation of a line. Students can check their equation of the line by putting in the given points to see if they satisfy the equation.