**SECTION FORMULAE**

**Formula for Internal Division: **The coordinates of the point which divides the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) internally in the ration m : n are given by \(x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\), \(y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\).

If P is the mid-point of AB, then it divides AB in the ratio 1:1, so its coordinates are \(\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\).

The following diagram will help to remember the section formula.**Formula for External Division:** The coordinates of the points (x_{1}, y_{1}) and (x_{2}, y_{2}) externally in the ration m : n are given by \(x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\), \(y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\).

**Example: 1:** Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the point (1, 3) and (2, 7).

**Solution:** Suppose the line 3x + y – 9 = 0 divides the line segment joining A (1, 3) and B (2, 7) in the ratio k:1 at point C.

Then, the coordinates of C are \(\left( \frac{2k+1}{k+1},\,\frac{7k+3}{k+1} \right)\)

But, C lies on 3x + y – 9 = 0.

Therefore, \(3\left( \frac{2k+1}{k+1} \right)+\frac{7k+3}{k+1}-9=0\)

=> 6k + 3 + 7k + 3 – 9k – 9 = 0

=> k = ¾

So, the required ratio is 3:4 internally.

**Example 2**: Prove that (4, -1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square?

**Solution: **Let the given points be A, B, C and D respectively.

Then, coordinates of the mid-points of AC are \(\left( \frac{4+7}{2},\,\frac{-1+2}{2} \right)=\left( \frac{11}{2},\,\frac{1}{2} \right)\)

Coordinates of the mid-points of BD are \(\left( \frac{6+5}{2},\,\frac{0+1}{2} \right)=\left( \frac{11}{2},\,\frac{1}{2} \right)\)

Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now, \(AB=\sqrt{{{\left( 6-4 \right)}^{2}}+{{\left( 0+1 \right)}^{2}}}=\sqrt{5}\),

AB = BC

So, ABCD is a parallelogram whose adjacent sides are equal.

Hence, ABCD is a rhombus.

We have, \(AC=\sqrt{{{\left( 7-4 \right)}^{2}}+{{\left( 2+1 \right)}^{2}}}=3\sqrt{2}\)

And, \(BD=\sqrt{{{\left( 6-5 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{2}\)

Clearly, AC ≠ BD.

So, ABCD is not a square.

**CO-ORDINATES OF CENTROID, IN-CENTRE, EX-CENTRES, ORTHOCENTRE AND CIRCUMCENTRE:**

**Centroid of Triangle:** The coordinates of the centroid of the triangle whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are

\(\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\).

Also, deduce that the medians of a triangle are concurrent.

**IN-Centre of a Triangle: **The coordinates of the in-center of a triangle whose vertices are

(x_{1}, y_{1}) , (x_{2}, y_{2}) and (x_{3},y_{3}) are \(\left( \frac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\frac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)\).

**EX-Centres of a Triangle:** Let A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) be the vertices of the triangle ABC

And let a, b, c be the lengths of the sides BC, CA, AB respectively.

The coordinates of I_{1} are given by \(\left( \frac{-a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{-a+b+c},\frac{-a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{-a+b+c} \right)\)

The coordinates of I_{2} and I_{3} (centers of described circles opposite to the angles B and C respectively) are given by \({{I}_{2}}\left( \frac{a{{x}_{1}}-b{{x}_{2}}+c{{x}_{3}}}{a-b+c},\frac{a{{y}_{1}}-b{{y}_{2}}+c{{y}_{3}}}{a-b+c} \right)\) and \({{I}_{3}}\left( \frac{a{{x}_{1}}+b{{x}_{2}}-c{{x}_{3}}}{a+b-c},\frac{a{{y}_{1}}+b{{y}_{2}}-c{{y}_{3}}}{a+b-c} \right)\) respectively.

**ORTHOCENTRE:** If A (x_{1}, y_{1}) (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of a triangle ABC. Then the point of intersection of intersection of its altitudes is known as the orthocenter and the coordinates are \(\left( \frac{{{x}_{1}}\tan A+{{x}_{2}}\tan B+{{x}_{3}}\tan C}{\tan A+\tan B+\tan C},\frac{{{y}_{1}}\tan A+{{y}_{2}}\tan B+{{y}_{3}}\tan C}{\tan A+\tan B+\tan C} \right)\)

**CIRCITMCENTRE**: If A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of a triangle ABC. Then the point of intersection of perpendicular bisectors of its sides is the circumcentre and its coordinates are \(\left( \frac{{{x}_{1}}\sin 2A+{{x}_{2}}\sin 2B+{{x}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C},\frac{{{y}_{1}}\sin 2A+{{y}_{2}}\sin 2B+{{y}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C} \right)\)

**REMARK**: The circumcentre O, centroid G and orthocenter O’ of a triangle are collinear and G divides O’ O in the ratio 2:1.

**Example:** If α, β, ɣ are the roots of the equation x^{3} – 3px^{2} + 3qx – 1 = 0, find the centroid of the triangle whose vertices are A (α, 1/α), B (β, 1/β) and C (ɣ, 1/ɣ).

**SOLUTION:** Since α, β, ɣ are the roots of the equation x^{3} – 3px^{2} + 3qx – 1 = 0

Therefore, α + β + ɣ = 3p, αβ + βɣ + ɣα = 3q

And, αβɣ = 1 … (i)

Let G(x, y) be the centroid of ΔABC. Then,

\(x=\frac{\alpha +\beta +\gamma }{3}\) and \(y=\frac{1}{3}\left( \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma } \right)\)

=> \(x=\frac{\alpha +\beta +\gamma }{3}\) and \(y=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3\alpha \beta \gamma }\)

=> \(x=\frac{3p}{3}\) and \(y=\frac{3q}{3}\) [Using (i)]

=> x = p and y = q

Hence, the coordinates of G are (p, q).