**Line segment** dividing the point:** Let A (x ₁, y ₁) and B(x ₂, y ₂)** be the given two point in a plane. The coordinate of the point dividing the **line segment AB in the ratio K ₁:k ₂** are

**(k ₁ x ₂+k ₂ x ₁÷ k ₁ + k ₂ , k ₁ y ₂+ k ₂ y ₁÷k ₁ +k ₂)**

** **

**PROOF:**

**Let P(x, y)** be the point that divide AB in the ratio **k ₁ +k ₂.**

From** A, B and P draw** perpendicular to** x-axis** as shown in the figure

**Also draw BC perpendicular A Q**

Since LP is parallel to CA, in the triangle ABC we have

after checking ratio

**K ₁ x ₂ – k ₁ x = k ₂ x – k ₂ x ₁**

**or**

**(k ₁ + K ₂) x = k ₁ x ₂+ k ₂ x ₁**

**or**

**x = k ₁ x ₂ – k ₂ x ₁/ k ₁ + K ₂**

**Similarly**, by drawing **perpendicular from A, B and P** to the **y-axis** and proceeding as before **we can show that.**

** y = k ₁ y ₂ – k ₂ y ₁ / k ₁ + K ₂**

**(1) Directed distance internally:**

If the directed distance AP and BP have the same sign , then the ratio is positive and P is said to divide AB internally.

**(2) Directed distance Externally:**

If the directed distance AP and BP have the opposite sign ,that is P is beyond AB, then the ratio is negative and P is said to divide AB Externally.

Proceeding as before, we can show in this case

**x = k ₁ x ₂ – k ₂ x ₁/ k ₁ + K ₂**

**y = k ₁ y ₂ – k ₂ y ₁ / k ₁ + K ₂**

Thus, P is said to divide the line segment** AB in ratio k ₁ : K ₂**

**Externally,** according as P lies between **AB or beyond AB.**

(3) If ** k ₁ : K ₂= 1:1, then P the mid-point of AB and the coordinate of P are **

**such that**

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