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Wednesday, October 26, 2016

Cartesian co-ordinate system in plane - II

SECTION FORMULAE

Formula for Internal Division:

The coordinates of the point which divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ration m : n are given by
x = (mx₂ + nx₁)/ m+n, y = (my₂ + ny₁)/ m+n.

If P is the mid-point of AB, then it divides AB in the ratio 1:1, so its coordinates are 
[(x₁ + x₂)/2, (y₁ + y₂)/2].

The following diagram will help to remember the section formula.

Formula for External Division:

The coordinates of the points (x1, y1) and (x2, y2) externally in the ration m : n are given by
x = (mx₂ + nx₁)/ m+n, y = (my₂ + ny₁)/ 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 [(2k + 1)/ (k + 1), (7k + 3)/ (k + 1)]

But, C lies on 3x + y - 9 = 0.

Therefore, 3(2k + 1)/ (k + 1) + (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 [(4 + 7)/2, (-1 + 2)/2] = (11/2, ½)

Coordinates of the mid-points of BD are [(6 + 5)/ 2, (0 + 1)/2] = (11/2, ½)

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

Hence, ABCD is a parallelogram.

Now, AB = √ [(6 - 4)² + (0 + 1)²] = √5,

AB = BC

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

Hence, ABCD is a rhombus.

We have, AC = √ [(7-4)² + (2 + 1)² = 3√2

And, BD = √ [(6 - 5)² + (0 - 1)²] = √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 (x1, y1), (x2, y2) and (x3, y3) are
[(x₁ + x₂ + x₃)/ 3, (y₁ + y₂ + y₃)/ 3].

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
(x1, y1) , (x2, y2) and (x3,y3) are [(ax₁ + bx₂ + cx₃)/ a+b+c, (ay₁ + by₂ + cy₃)/ a+b+c].

Ex-Centres of a Triangle:

Let A (x1, y1), B (x2, y2) and C (x3, y3) 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 I1 are given by

[(- ax₁ + bx₂ + cx₃)/ - a + b + c, (- ay₁ + by₂ + cy₃)/ - a + b + c]

The coordinates of I2 and I3 (centers of described circles opposite to the angles B and C respectively) are given by

I₂ [(ax₁ - bx₂ + cx₃)/ a - b + c, (ay₁ - by₂ + cy₃)/ a - b + c]

And, I₃ [(ax₁ + bx₂ - cx₃)/ a + b - c, (ay₁ + by₂ - cy₃)/ a + b - c] respectively.

Orthocentre:

If A (x1, y1) (x2, y2) and C (x3, y3) 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

[(x₁tanA+x₂tanB+x₃tanC)/ tanA+tanB+tanC, (y₁tanA+y₂tanB+y₃tanC)/ tanA+tanB+tanC]

Circumcentre:

If A (x1, y1), B (x2, y2) and C (x3, y3) 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

[(x₁sin2A+x₂sin2B+x₃sin2C)/ sin2A+sin2B+sin2C, (y₁sin2A+y₂sin2B+y₃sin2C)/ sin2A+sin2B+sin2C]

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 x3 - 3px2 + 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

x3 - 3px2 + 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 = α + β + ɣ / 3 and y = ⅓ (1/α + 1/β + 1/ɣ)

=> x = α + β + ɣ / 3 and y = (αβ + βɣ + ɣα)/3 αβɣ

=> x = 3p/3 and y = 3q/3 [Using (i)]

=> x = p and y = q

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

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