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
No comments:
Post a Comment