Welcome

JAWAHAR NAVODAYA VIDYALAYA

LEPAKSHI

ANANTAPUR DISTRICT

Coordinate Geometry

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CLASS :X

CBSE

TOPIC: Section Formula

BY G.SUMATHI

TGT MATHEMATICS

JNV ANANTAPUR

A

B

C

(x₁,y₁)

(x₂,y₂)

(x₃ ,y₃)

Prerequisite knowledge

• Area of Trapezium:

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• Area= ½ (Sum of the parallel side)x Distance between them
• Area of ABCD = ½ (AB+CD) x EF
• In figure three
• Area of ABCD = ½ (AB+CD) x AD

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A

A

B

A

B

B

C

C

C

D

D

D

E

F

E

F

fig. 3

fig. 2

fig.1

Area of Triangle

O

(x₃ ,y₃)

(x₂,y₂)

(x₁,y₁)

_C

_Q

_R

_P

_B

_A

area of ΔABC = area of trapezium ABQP + area of trapezium APRC -

area of trapezium BQRC

ar(tpzm ABQP) = ½(BQ + AP) QP

ar(tpzm APRC) = ½( AP+RC) PR

ar(tpzm BQRC) = ½(BQ + RC) QR

= ½(BQ + AP) QP + ½( AP+RC) PR + ½(BQ + RC) QR

= ½(y₂ + y₁) (x₁-x₂) + ½(y₁ + y₃ )(x₃-x₁) + ½(y₂ + y₃ )(x₃-x₂)

= ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₁ – y₂ )]

ar(ΔABC)

Formula for area of Triangle

Area of Triangle

If ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x, y₃) then the Area of ΔABC is given by the formula

ar(ΔABC) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₂ – y₁ )]

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A

B

C

(x₁,y₁)

(x₂,y₂)

(x₃ ,y₃)

Points to remember:

• 1.Points lying on the same line are collinear points .

• 2. If three point are collinear ,then the area of triangle formed by these three points is zero.
• 3. Area of quadrilateral can be found by dividing it into two parts (using any of its diagonals)
• Median is the line segment joining vertex and mid point of its opposite side.

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A

B

C

D

C

D

B

A

Example1:

Q) Find the area of the triangle whose vertices are

( 3,2), (11,8), (8,12)?

Sol: let A(3,2), B(11,8), C(8,12)

ar(ΔABC) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₁ – y₂ )]

ar(ΔABC) = ½[ 3 (8 – 12) + 11(12 – 2) + 8 (2 -8 )]

= ½ [ 3(-4) + 11(10) + 8(-6) ]

= ½ [ -12 +110-48]

= ½ [ 50]

= 25 sq units

(x₁,y₁)

(x₂,y₂)

(x₃ ,y₃)

Example2:

Q) Prove that the points (2,-2), (-3,8), (-1,4) are collinear.

Sol: let A(2,-2), B(-3,8), C(-1,4)

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ar(ΔABC) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₁ - y₃ )]

ar(ΔABC) = ½[ 2 (8 – 4) + (-3)(4 – (-2)) + (-1) (-2 -4 )]

= ½ [ 3(4) + (-3)(6) + (-1)(-6) ]

= ½ [ 12 -18+6 ]

= ½ [0 ]

= 0 sq units

Hence the given points are collinear

(x₁,y₁)

(x₂,y₂)

(x₃ ,y₃)

Example3:

• Q) Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5),(3, – 2) and (2, 3).
• Sol: ar(ΔABD) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₁ – y₂ )]
• ar(ΔABC) = ½[ -4 (-5 – 3) + -3(3 – -2) + 2 (-2 --5)]
• = ½ [ -4(-8) + -3(5) + 2(3) ]
• = ½ Ι 32 -15+6 Ι
• = 23/2 sq units
• ar(ΔBCD) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₁ – y₂ )]

ar(ΔABC) = ½[ -3 (-2 – 3) + 3(3 – -5) + 2 (-5 --2 )]

• = ½ [-3(-5) + 3(8) + 2(-3) ]
• = ½ [ 15 +24 -6 ]
• = 33/2sq units

ar(ABCD) = ar(ΔABD) + ar(ΔBCD) = 23/2 sq units + 33/2 sq units =28 sq units

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A(-4,-2)

B(-3,-5)

C(3,-2)

D(2,3)

Summery

Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x, y₃) then the Area of ΔABC is given by the formula

ar(ΔABC) = ½[ x₁ (y₂ – y₃) + x₂(y₃ – y₁ ) + x₃ (y₂ - y₃ )]

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A

B

C

(x₁,y₁)

(x₂,y₂)

(x₃ ,y₃)

Home Work:

• 1.Prove that the points (-2, -1), (1,0), (4,3) and ( 1,2 ) are the vertices of a parallelogram . Is it a rectangle?
• 2. Find the co-ordinates of the point of trisection of line segment joining (4,-1) and(-2,-3)?
• 3.Find the ratio in which the line segment joining A (1,-5) and B (-4, 5) is divided by the X-axis also find the coordinates of the point of division?
• Exercise 7.2 of NCERT Text Book

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THANK

YOU

ALL…

BY G.SUMATHI

TGT MATHEMATICS

JNV, ANANTAPUR