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Hint: Infinite number of lines can pass through a single point. Similarly, an infinite number of curves can pass through 2 points, which are not straight lines. Only 1 straight line can pass through 2 points.

Complete step-by-step answer:

For a given point, an infinite number of lines passes through this point, and an infinite number of lines passes through two given points if we are not considering straight lines.

So, the lines passing through 2 points will be infinite.

But only one straight line can pass through two given points.

So, geometrically only 1 straight line can pass through 2 distinct points.

But if we are considering 2 points where an infinite line can pass through them each.

Let us consider the 2 points as A and B.

Now, infinite lines can pass through the point A.

Similarly, infinite lines can pass through point B.

Now, A and B together will have one line passing through it.

Point A\[\to \]Infinite lines\[\to \](1)

Point B\[\to \]Infinite lines\[\to \](2)

A and B\[\to \]1\[\to \](3)

Number of straight lines that can pass through 2 points A and B = 1.

Note:

Let us consider a case of non-collinear points. If we have been given 4 points and need to find how many lines can pass through these points.

The method is to find the number of straight lines that can be formed.

3 Lines 5 Lines 6 Lines

\[\therefore \](4 - 1) (4 - 1) + (4 - 2) (4 - 1) + (4 - 2) + (4 - 3) + (4 - 4)

\[\therefore \]They are of the form:

\[\begin{align}

& h=\sum\limits_{i=1}^{n}{\left( n-i \right)}=\left( n-1 \right)+\left( n-2 \right)+\left( n-3 \right)+.....+\left( n-n \right) \\

& =\sum\limits_{i=1}^{n}{n}-\sum\limits_{i=1}^{n}{i}\Rightarrow L={{n}^{2}}-\dfrac{n\left( n+1 \right)}{2}=\dfrac{2{{n}^{2}}-{{n}^{2}}-n}{2} \\

& L=\dfrac{{{n}^{2}}-n}{2}=\dfrac{n\left( n-1 \right)}{2} \\

\end{align}\]

\[\therefore L=\dfrac{n\left( n-1 \right)}{2}\], where L = number of lines.

So, for 4 points, n=4,

\[L=\dfrac{4\left( 4-1 \right)}{2}=\dfrac{4\times 3}{2}=6\]lines.

Where n=1, \[L=\dfrac{1\left( 1-1 \right)}{2}=\dfrac{0}{2}\]i.e. Infinite number of lines.

Where n=2, \[L=\dfrac{2\left( 2-1 \right)}{2}=1\]etc.

Complete step-by-step answer:

For a given point, an infinite number of lines passes through this point, and an infinite number of lines passes through two given points if we are not considering straight lines.

So, the lines passing through 2 points will be infinite.

But only one straight line can pass through two given points.

So, geometrically only 1 straight line can pass through 2 distinct points.

But if we are considering 2 points where an infinite line can pass through them each.

Let us consider the 2 points as A and B.

Now, infinite lines can pass through the point A.

Similarly, infinite lines can pass through point B.

Now, A and B together will have one line passing through it.

Point A\[\to \]Infinite lines\[\to \](1)

Point B\[\to \]Infinite lines\[\to \](2)

A and B\[\to \]1\[\to \](3)

Number of straight lines that can pass through 2 points A and B = 1.

Note:

Let us consider a case of non-collinear points. If we have been given 4 points and need to find how many lines can pass through these points.

The method is to find the number of straight lines that can be formed.

3 Lines 5 Lines 6 Lines

\[\therefore \](4 - 1) (4 - 1) + (4 - 2) (4 - 1) + (4 - 2) + (4 - 3) + (4 - 4)

\[\therefore \]They are of the form:

\[\begin{align}

& h=\sum\limits_{i=1}^{n}{\left( n-i \right)}=\left( n-1 \right)+\left( n-2 \right)+\left( n-3 \right)+.....+\left( n-n \right) \\

& =\sum\limits_{i=1}^{n}{n}-\sum\limits_{i=1}^{n}{i}\Rightarrow L={{n}^{2}}-\dfrac{n\left( n+1 \right)}{2}=\dfrac{2{{n}^{2}}-{{n}^{2}}-n}{2} \\

& L=\dfrac{{{n}^{2}}-n}{2}=\dfrac{n\left( n-1 \right)}{2} \\

\end{align}\]

\[\therefore L=\dfrac{n\left( n-1 \right)}{2}\], where L = number of lines.

So, for 4 points, n=4,

\[L=\dfrac{4\left( 4-1 \right)}{2}=\dfrac{4\times 3}{2}=6\]lines.

Where n=1, \[L=\dfrac{1\left( 1-1 \right)}{2}=\dfrac{0}{2}\]i.e. Infinite number of lines.

Where n=2, \[L=\dfrac{2\left( 2-1 \right)}{2}=1\]etc.

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