user81435: Please tell me if there is a formula for A-B, (That is, subtraction of two points / minus). I know how two points are added, A + B this addition is described in detail 
(Please tell me how you can change the formula to subtract two points / minus) A-B

user82248: As other's have said, the essential point is the algebraic definition for additive inverses in elliptic arithemetic.
 -(x,y)=(x,-y)

But if it helps, there are also some nice geometric illustrations like this one from Vitalik Buterin's Exploring Elliptic Curve Pairings:

Suppose R = (x,y). Since the elliptic curve is symmetric with respect to the x-axis, we can guarantee that the inverse -R = -(x,y) = (x,-y) will also be a point on the curve. And since R and -R have the same x-coordinate, the line that connects them is vertical representing their sum as the identity O or "the point at infinity". So we have R + (-R) = O as required.

user6091: You can negate a point (x, y) by simply changing it to (x, −y).
The document that defines ECDSA reminds us of this fact:
https://www.secg.org/sec1-v2.pdf
Here's a screenshot:

So once you have negated one of your points, just add it to the other one, and you have achieved subtraction.

81435: <p>Please tell me if there is a formula for <code>A-B</code>, (That is, subtraction of two points / minus). I know how two points are added, <code>A + B</code> this addition is described in detail </p>

<p>(Please tell me how you can change the formula to subtract two points / minus) <code>A-B</code></p>
 82248: <p>As other's have said, the essential point is the algebraic definition for additive inverses in elliptic arithemetic.</p>

<pre><code> -(x,y)=(x,-y)
</code></pre>

<p>But if it helps, there are also some nice geometric illustrations like this one from Vitalik Buterin's <a href="https://medium.com/@VitalikButerin/exploring-elliptic-curve-pairings-c73c1864e627" rel="nofollow noreferrer">Exploring Elliptic Curve Pairings</a>:</p>

<p><a href="https://i.stack.imgur.com/vX5Ed.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vX5Ed.jpg" alt="Elliptic arithmetic"></a></p>

<p>Suppose <code>R = (x,y)</code>. Since the elliptic curve is symmetric with respect to the <code>x</code>-axis, we can guarantee that the inverse <code>-R = -(x,y) = (x,-y)</code> will also be a point on the curve. And since <code>R</code> and <code>-R</code> have the same <code>x</code>-coordinate, the line that connects them is vertical representing their sum as the identity <code>O</code> or "the point at infinity". So we have <code>R + (-R) = O</code> as required.</p>
 6091: <p>You can negate a point (x, y) by simply changing it to (x, −y).</p>

<p>The document that defines ECDSA reminds us of this fact:</p>

<p><a href="https://www.secg.org/sec1-v2.pdf" rel="nofollow noreferrer">https://www.secg.org/sec1-v2.pdf</a></p>

<p>Here's a screenshot:</p>

<p><a href="https://i.stack.imgur.com/N07pS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N07pS.png" alt="Definition of point negation"></a></p>

<p>So once you have negated one of your points, just add it to the other one, and you have achieved subtraction.</p>
