Question

How do you find the equation for a curve between two points?

Answer 1

Hint: Since in the above question, we are given that the curve passes through two points, we need to take the example of a curve which includes two parameters. So we can choose the curve to be a straight line, whose general equation is given by $y=mx+c$. By taking the coordinates of the two points as $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ and substituting them into the equation $y=mx+c$, we will get the values of the parameters $m$ and $c$, and hence the equation for the curve.

Complete step by step answer:
An equation for a curve is determined by the parameters that it includes. The parameters for a curve are basically the independent constants which can uniquely determine a curve passing between two points. Therefore, for finding out the equation for a curve between two points, we need to substitute the coordinates of the two points into the general equation of the curve. For instance, let us take the curve to be a straight line. We know that the general equation for a straight line is given by
$\Rightarrow y=mx+c.......\left( i \right)$
Let the two points between which the assumed curve, the straight line passes have the coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$. On substituting $x={{x}_{1}}$ and $y={{y}_{1}}$ in the above equation (i) we get
$\Rightarrow {{y}_{1}}=m{{x}_{1}}+c.......\left( ii \right)$
Similarly, we substitute \[x={{x}_{2}}\] and \[y={{y}_{2}}\] in the equation (i) to get
$\Rightarrow {{y}_{2}}=m{{x}_{2}}+c.......\left( iii \right)$
Subtracting the equation (ii) from the equation (iii) we get
$\Rightarrow {{y}_{2}}-{{y}_{1}}=m\left( {{x}_{2}}-{{x}_{1}} \right)$
On dividing the above equation by $\left( {{x}_{2}}-{{x}_{1}} \right)$, we get
$\Rightarrow m=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$
Substituting the above equation into the equation (ii) we get
$\Rightarrow {{y}_{1}}=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}}+c$
Subtracting $\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}}$ from both the sides, we get
\[\begin{align}
  & \Rightarrow {{y}_{1}}-\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}}=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}}+c-\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}} \\
 & \Rightarrow c={{y}_{1}}-\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}{{x}_{1}} \\
 & \Rightarrow c=\dfrac{\left( {{x}_{2}}-{{x}_{1}} \right){{y}_{1}}-\left( {{y}_{2}}-{{y}_{1}} \right){{x}_{1}}}{\left( {{x}_{2}}-{{x}_{1}} \right)} \\
 & \Rightarrow c=\dfrac{\left( {{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{1}} \right)-\left( {{y}_{2}}{{x}_{1}}-{{y}_{1}}{{x}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)} \\
 & \Rightarrow c=\dfrac{{{x}_{2}}{{y}_{1}}-{{y}_{2}}{{x}_{1}}}{\left( {{x}_{2}}-{{x}_{1}} \right)} \\
\end{align}\]
Thus, we found out the two parameters for the straight line. Therefore, the equation of the line can be given as
$\Rightarrow y=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}x+\dfrac{{{x}_{2}}{{y}_{1}}-{{y}_{2}}{{x}_{1}}}{\left( {{x}_{2}}-{{x}_{1}} \right)}$
Similarly, we can find out the equation for any other curve by substituting the coordinates.

Note: We must note that we can determine the equation of that curve only which consists of two parameters only. This is because in the above question, we are told that the curve passes between two points only. Therefore, we can get two equations in terms of the two parameters from which the parameters can be obtained. We can also take the example of a circle passing through the origin, whose general equation is given by ${{x}^{2}}+{{y}^{2}}+2gx+2fy=0$ in which the two parameters are $f$ and $g$.