How do you write a quadratic equation when given two points?

Hint: We begin solving the above question by considering a graph $y=a{{x}^{2}}+bx+c$. Let (p,0) and (q,0) be two points passing through the graph. The quadratic expression of the same can be written as $y=\left( x-p \right)\left( x-q \right)$ . Lastly, we need to multiply $y=\left( x-p \right)\left( x-q \right)$ with any real number to represent all the possible outcomes.

Complete step-by-step solution:
The quadratic equation passing through 2 points is given by $y=c\left( x-a \right)\left( x-b \right)$
This equation represents a parabola passing through 2 points.
The points (a,0), (b,0) represent the points on the parabola and ‘c’ represents coefficient and belongs to a set of real numbers.
The value of coefficient (c) is not fixed so we can have infinite parabolas passing through the points (a,0) and (b,0), and here comes the necessity of the third point.
 For Example:
Given that $y=f\left( x \right)=a{{x}^{2}}+bx+c$ passes through points (1,0) and (2,0) find the quadratic equation of the above given that f (1)=0 and f(2)=0.
The polynomial y passes through points (1,0) and (2,0). It can be written as $y=\left( x-1 \right)\left( x-2 \right)$
Solving the above expression,
we get,
$\Rightarrow y={{x}^{2}}-3x+2$
The final expression derived is,
$\Rightarrow y=\left( x-1 \right)\left( x-2 \right)={{x}^{2}}-3x+2$
Any quadratic equation passing through (1,0) and (2,0) should be divisible by (x-1) and (x-2) respectively.
Hence, we need to multiply the above polynomial with any real number to represent all the possible equations.
Let us assume 'a' as the real number.
On Multiplying,
we get,
 $\Rightarrow y=a\left( x-1 \right)\left( x-2 \right)=a\left( {{x}^{2}}-3x+2 \right)$

Note: The quadratic equation written with 2 points is an equation that represents a plane curve parabola which is passing through 2 points (a,0) and (b,0) with real coefficient c. The equation is given by $y=c\left( x-a \right)\left( x-b \right)$