#### Meaning

Co-ordinate Geometry is a branch of geometry which is used to identify a point on a plane. It was invented by RENE DESCARTES.

#### Rectangular Co-ordinate Axis

The two mutually perpendicular number lines which are used to find the position of a point on a plane is called rectangular axis. In the graph, XOX' is called x-axis and the point YOY' is called y-axis. The two lines XOX' and YOY' are also called rectangular co-ordinate axis which divide the plane into four equal parts which are called quadrant.

#### Distance Between Two Points

If the elements or co-ordinates of any two points are given, the distance between them can be found with the help of distance formulae.
Suppose that,
P(x1, y1) and Q(x2, y2) are any two points in the co-ordinates plane and 'd' is the distance between them.
Now,
Draw PT and QK perpendicular on x-axis and PM perpendicular to QK.
Then,
OK = x2, KQ = y2, OT = x, PT = y.
PM = TK= OK - OT = (x2- x1)
Also,
OM = OK - MK = QK - PT
= (y2- y1) [∴ MK = PT]
Since, ∠PMQ is a right angle, so $$\triangle$$PMQ is a right angle triangle.
Now,
$$\triangle$$PMQ using Pythagoras Theorem,
PQ2= PM2= QM2
= (x2- x1)2+ (y2- y1)2
or, PQ = (x2− x1)2 + (y2− y1)2 ................... $$\sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}$$
∴ Distance (d) = (x2− x1)2+ (y2− y1)2 ...................$$\sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}$$
Again,
The distance of a point A(x, y) from the origin O(0, 0) is,
or, OA = (x−0)2+ (y−0)2 .................. $$\sqrt{(x−0)^2 + (y−0)^2}$$
∴ OA = x2+ y2 .................... $$\sqrt{x^2 + y^2}$$
Again,
Slope of the line = PQ = Tanθ
$$\frac{OM}{PQ}$$ = $$\frac{y_2 − y_1}{x_2 − x_1}$$

#### Types of triangle using the co-ordinates of the vertices

 a) Scalene No sides are equal i.e. in ΔABC, AB≠BC, BC≠CA and CA≠AB. b) Isosceles Two sides are equal i.e. in ΔABC, AB=BC or BC=CA or AB=AC. c) Equilateral All sides are equal i.e. in ΔABC, AB = BC = CA. d) Right-Angled Triangle Sum of squares of two shorter sides is equal to the square of the longest side. e) Right Angled Isosceles Triangle Two shorter sides are equal and the sum of the squares of two shortest sides is equal to the square of the longest side.

#### Types of quadrilateral using the co-ordinates of the vertices

 a) Parrallelogram Opposite sides are equal. In quadrilateral ABCD, AB=CD and BC=AD b) Rectangle Opposites sides are equal and diagonals are equal i.e. in quadrilateral, ABCD, AB=CD, AD=BC and AC=BD. c) Rhombus All sides are equal but diagonals are not equal i.e. in quadrilateral, ABCD, AB=BC=CD=DA and AC≠BD. d) Square All sides are equal and diagonals are also equal i.e. in quadrilateral ABCD, AB=BC=CD=DA and AC=BD.

• Co-ordinate Geometry is a branch of geometry which is used to identify a point on a plane.
• Co-ordinate Geometry was invented by RENE DESCARTES.
• The two mutually perpendicular number lines which are used to find the position of a point on a plane is called rectangular axis.
• Plotting points is the process of locating a point whose co-ordinates are given.

Solution:

A(−4, 0) = (x1, y1)
B(0,−3) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2−y_1)^2}$$
or, AB = $$\sqrt{(0 + 4)^2 + (− 3 − 0)^2}$$
or, AB = $$\sqrt{4)^2 + (3)^2}$$
or, AB = $$\sqrt{16 + 9}$$
or, AB = $$\sqrt{25}$$
∴ AB = 5 units.

Solution:

Here,
P(−8, 0) = (x1, y1)
B(0, 6) =(x2, y2)
PB = ?
We know that,
or, PB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, PB = $$\sqrt{(0 + 8)^2 + (6 − 0)^2}$$
or, PB = $$\sqrt{(8)^2 + (6)^2}$$
or, PB = $$\sqrt{64 + 36}$$
or, PB = $$\sqrt{100}$$
∴ PB = 10 units.

Solution:

Here,
A(0,−5) = (x1, y1)
B(4 , 8).= (x2, y2)
Slope (m) = ?

We know that,
Slope (m) = $$\frac{y_2− y_1}{x_2− x_1}$$
=$$\frac{8 + 5}{4 − 0}$$
= $$\frac{13}{4}$$

Solution:

Here,
E(4,−7) = (x2, y1)
F(−3, 4) = (x2, y2)
Slope (m) = ?

we know that,
Slope (m) = $$\frac{y_2− y_1}{x_2− x_1}$$
=$$\frac{4 + 7}{−3 − 4}$$
= $$\frac{11}{−7}$$

Solution:

Let,
A (−1, 7) = (x1, y1)
B (3, 10) = (x2, y2)
Distance of AB = ?

We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB = $$\sqrt{(3 + 1)^2 + (10 − 7)^2}$$
or, AB = $$\sqrt{(4)^2 + (3)^2}$$
or, AB = $$\sqrt{16+ 9}$$
or, AB = $$\sqrt{25}$$
∴ AB = 5 units

Solution:

Let, Q(0, 0) and R(−6,−4)
Distance of QR = ?
We know that,
or, QR = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, QR = $$\sqrt{(− 6 − 0)^2 + (− 4 − 0)^2}$$
or, QR = $$\sqrt{(6)^2 + (4)^2}$$
or, QR = $$\sqrt{36 + 16}$$
or, QR = $$\sqrt{52}$$
∴ QR = 2$$\sqrt{3}$$

Solution:

Here,
A(4, 2) = (x1, y1)
B(6, 8) = (x2, y2)
Slope (m) = ?
We know that,
Slope (m) = $$\frac{y_2− y_1}{x_2− x_1}$$
=$$\frac{8 − 2}{6 − 4}$$
=$$\frac{6}{2}$$
= 3

Solution:

For, AB
A(2, 3) = (x, y)
B(2, 0) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB =$$\sqrt{(2− 2)^2 + (0 − 3)^2}$$
or, AB =$$\sqrt{(0)^2 + (3)^2}$$
or, AB =$$\sqrt{3^2}$$
∴ AB = 3 units.

For BC
B(2, 0) = (x1, y1)
C(−1, 0) = (x2, y2)
BC = ?
We know that,
or, BC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC = $$\sqrt{(−1 − 2)^2 + 0 − 0)^2}$$
or, BC = $$\sqrt{(3)^2}$$
∴ BC = 3 units

For CA
C(− 1, 0) = (x1, y1)
A(2, 3) =(x2, y2)
CA = ?
We know that,
or, CA =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, CA =$$\sqrt{(2 + 1)^2 + (3 − 0)^2}$$
or, CA =$$\sqrt{(3)^2 + (3)^2}$$
or, CA =$$\sqrt{9 + 9}$$
or, CA =$$\sqrt{18}$$
∴ 3$$\sqrt{2}$$ units

Solution:

For AB
A(−3,−4) = (x1, y1)
B(−1,−2) =(x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2−x_1)^2 + (y_2− y_1)^2}$$
or, AB = $$\sqrt{(−1 + 3)^2 + (−2 + 4)^2}$$
or, AB = $$\sqrt{(2)^2 + (2)^2}$$
or, AB = $$\sqrt{4 +4}$$
or, AB = $$\sqrt{8}$$
or, AB = 2$$\sqrt{2}$$ units.

For CD
C(8, 6) = (x1, y1)
D(10, 8) =(x2, y2)
CD = ?
We know that,
or, CD = $$\sqrt{(x_2−x_1)^2 + (y_2− y_1)^2}$$
or, CD = $$\sqrt{(10 − 8)^2 + (8 − 6)^2}$$
or, CD = $$\sqrt{(2)^2 + (2)^2}$$
or, CD = $$\sqrt{4+4}$$
or, CD =$$\sqrt{8}$$
or, CD = 2$$\sqrt{2}$$ units.

∴ AB = CD =2$$\sqrt{2}$$ units. proved.

Solution:

For PQ
P(2,−5) = (x1, y1)
Q(4, 1) = (x2, y2)
PQ = ?
We know that,
or, PQ = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, PQ = $$\sqrt{(4 − 2)^2 + (1 +5)^2}$$
or, PQ = $$\sqrt{(2)^2 + (6)^2}$$
or, PQ = $$\sqrt{4 + 36}$$
or, PQ = $$\sqrt{40}$$
∴ PQ = 2$$\sqrt{10}$$ units.

For QR
Q(4, 1) =(x1, y1)
R(10, 3) =(x2, y2)
QR = ?
We know that,
or, QR = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, QR = $$\sqrt{(10− 4)^2 + (3 − 1)^2}$$
or, QR = $$\sqrt{(6)^2 + (2)^2}$$
or, QR = $$\sqrt{36 + 4}$$
or, QR = $$\sqrt{40}$$
∴ QR = 2$$\sqrt{10}$$
∴ PQ = QR proved.

Solution:

Let, A(1, 2), B(4, 5) and C(7, 2) be the three points
For AB
A(1, 2) = (x1, y1)
B(4, 5) = (x2, y2)
Distance of AB = ?
We knoe that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB = $$\sqrt{(4 − 1)^2 + (5 − 2)^2}$$
or, AB = $$\sqrt{(3)^2 + (3)^2}$$
or, AB = $$\sqrt{18}$$
or, AB = 3$$\sqrt{2}$$ units

For BC
B(4, 5) = (x1, y1)
C(7, 2) = (x2, y2)
BC = ?
we know that,
or, BC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC = $$\sqrt{(7− 4)^2 + (2 − 5)^2}$$
or,BC = $$\sqrt{(3)^2 + (− 3)^2}$$
or, BC = $$\sqrt{9 + 9}$$
or, BC = 3$$\sqrt{2}$$ units

For CA
C(7, 2) = (x1, y1)
A(1, 2) = (x2, y2)
CA = ?
We know that,
or, CA = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, CA = $$\sqrt{(1 − 7)^2 + (2− 2)^2}$$
or, CA = $$\sqrt{( − 6)^2 + (0)^2}$$
or, CA = $$\sqrt{36}$$
or, CA = 6 units

Now,
CA has the longest side
or, 62 = (3$$\sqrt{2}$$)2+(3$$\sqrt{2}$$)2
or, 36 = 18 + 18
or, 36 = 36

AB = BC = 3$$\sqrt{2}$$ units and is satisfied the pythogorus theorem (i.e. h2 = p2 + b2). So, the given points are the vertices of isosceles right angled triangle

Solution:

For AB
A(1,−1) = (x1, y1)
B(−1, 1) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB = $$\sqrt{(− 1− 1)^2 + (1 +1)^2}$$
or, AB = $$\sqrt{(−2)^2 + 2^2}$$
or, AB = $$\sqrt{4 + 4}$$
or, AB = $$\sqrt{8}$$
or, AB = 2$$\sqrt{2}$$ units.

For BC
B(−1, 1) = (x1, y1)
C($$\sqrt{3}$$, $$\sqrt{3}$$) = (x2, y2)
BC = ?
We know that,
or, BC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC = $$\sqrt{(\sqrt{3}+1)^2 + (\sqrt{3} − 1)^2}$$
or, BC = $$\sqrt{(\sqrt{3})^2 + 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 − 2 . \sqrt{3} . 1 + 1^2}$$
or, BC = $$\sqrt{3 +2 . \sqrt{3} + 1 + 3− 2 . \sqrt{3} + 1}$$
or, BC = $$\sqrt{8}$$
or, BC = 2$$\sqrt{2}$$

For AC
A(1,−1) = (x1, y1)
C($$\sqrt{3}$$,$$\sqrt{3}$$) = (x2, y2)
AC = ?
We know that,
or, AC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AC = $$\sqrt{(\sqrt{3} − 1)^2 + (\sqrt{3}+1)^2}$$
or, AC = $$\sqrt{(\sqrt{3})^2 − 2 . \sqrt{3} .1 + 1^2 + (\sqrt{3})^2 +2 . \sqrt{3} . 1 + 1^2}$$
or, AC = $$\sqrt{3 − 2 . \sqrt{3} + 1 + 3 +2 . \sqrt{3} + 1}$$
or, AC = $$\sqrt{8}$$
or, AC = 2$$\sqrt{2}$$

Here, AB = BC = AC. So, the given points are the vertices of an equilateral triangle.

Solution:

For AB
A(1, 1) = (x1, y1)
B(4, 4) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or,AB = $$\sqrt{(4 − 1)^2 + (4 − 1)^2}$$
or,AB = $$\sqrt{(3)^2 + (3)^2}$$
or, AB = $$\sqrt{9 + 9}$$
or, AB = $$\sqrt{18}$$
or, AB = 3$$\sqrt{2}$$ units

For BC
B(4, 4) =(x1, y1)
C(4, 8) = (x2, y2)
BC = ?
We know that,
or, BC =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC =$$\sqrt{(4− 4)^2 + (8− 4)^2}$$
or, BC = $$\sqrt{4^2}$$
or, BC = 4 units.

For CD
C(4, 8) =(x1, y1)
D(1, 5) =(x2, y2)
CD = ?
We know that,
or, CD =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, CD =$$\sqrt{(1 − 4)^2 + (5 − 8)^2}$$
or, CD =$$\sqrt{(− 3)^2 + (− 3)^2}$$
or, CD = $$\sqrt{9 + 9}$$
or, CD = 3$$\sqrt{2}$$ units

A(1, 1) =(x1, y1)
D(1, 5) =(x2, y2)
We know that,
or, AD =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AD =$$\sqrt{(1− 1)^2 + (5 − 1)^2}$$
or, AD = $$\sqrt{4^2}$$

Here, AB = CD = 3$$\sqrt{2}$$ units
BC = AB = 4 units
So, the given points are the vertices of a parallelogram.

Solution:

For AB
A(0,−1) = (x1, y1)
B(2, 1) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB = $$\sqrt{2− 0)^2 + (1+1)^2}$$
or, AB = $$\sqrt{2^2 + 2^2}$$
or, AB = $$\sqrt{4 + 4}$$
or, AB = $$\sqrt{8}$$
or, AB = 2$$\sqrt{2}$$ units

For BC
B(2, 1) = (x1, y1)
C(0, 3) = (x2, y2)
BC = ?
We know that,
or, BC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC = $$\sqrt{(0− 2)^2 + (3 − 1)^2}$$
or, BC = $$\sqrt{(− 2)^2 + (2)^2}$$
or, BC = $$\sqrt{4 + 4}$$
or, BC = $$\sqrt{8}$$
or, BC = 2$$\sqrt{2}$$ units

For CD
C(0, 3) = (x1, y1)
D(−2, 1) = (x2, y2)
CD = ?
We know that,
or, CD = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, CD = $$\sqrt{(− 2− 0)^2 + (1 − 3)^2}$$
or, CD = $$\sqrt{(− 2)^2 + 2^2}$$
or, CD = $$\sqrt{4 + 4}$$
or, CD = $$\sqrt{8}$$
or, CD = 2$$\sqrt{2}$$ units

A(0,−1) = (x1, y1)
D(−2, 1) = (x2, y2)
We know that,
or, AD = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AD = $$\sqrt{(− 2− 0)^2 + (1 − 1)^2}$$
or, AD = $$\sqrt{(− 2)^2 + (2)^2}$$
or, AD = $$\sqrt{4 + 4}$$
or, AD = $$\sqrt{8}$$
or, AD = 2$$\sqrt{2}$$ units

For AC
A(0,−1) = (x1, y1)
C(0, 3) = (x2, y2)
or, AC = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AC = $$\sqrt{(0 − 0)^2 + (3+1)^2}$$
or, AC = $$\sqrt{4^2}$$
or, AC = 4 units.

For DB
D(−2, 1) = (x1, y1)
B(2, 1) = (x2, y2)
DB = ?
We know that,
or, DB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, DB = $$\sqrt{(2+2)^2 + (1 − 1)^2}$$
or, DB = $$\sqrt{4^2}$$
or, DB = 4 units

Here, AB = BC = CD = AD = 2$$\sqrt{[2}$$ units
AC = BD = 4 units
Since the given points are the vertices of a square.

Solution:

Let, A(2,−1), B(7, 4) and C(8, 11) be the vertices of an isoscles trinagle.
For AB
A(2,−1) = (x1, y1)
B(7, 4) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AB =$$\sqrt{(7 − 2)^2 + (4 +1)^2}$$
or, AB =$$\sqrt{(5)^2 + (5)^2}$$
or, AB = $$\sqrt{25 + 25}$$
or, AB = $$\sqrt{50}$$
or, AB = 5$$\sqrt{2}$$ units

For BC
B(7, 4) =(x1, y1)
C(8, 11) =(x2, y2)
BC = ?
we know that,
or, BC =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, BC =$$\sqrt{(8 − 7)^2 + (11 − 4)^2}$$
or, BC = $$\sqrt{1 + 7^2 }$$
or, BC = $$\sqrt{50}$$
or, BC = 5$$\sqrt{2}$$ units

For AC
A(2,−1) = (x1, y1)
C(8, 11) =(x2, y2)
AC = ?
We know that,
or, AC =$$\sqrt{(x_2− x_1)^2 + (y_2− y_1)^2}$$
or, AC =$$\sqrt{(8 − 2)^2 + (11+1)^2}$$
or, AC =$$\sqrt{(6)^2 + (12)^2}$$
or, AC = $$\sqrt{36 + 144}$$
or, AC = $$\sqrt{180}$$
or, AC = 6$$\sqrt{5}$$ units

Here, AB = BC = 5$$\sqrt{2}$$ units
So, the given points are the vertices of an isosceles triangle

Solution:

For AB
A((1,?1) = (x1, y1)
B(3, 2) = (x2, y2)
AB = ?
We know that,
or, AB = $$\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}$$
or, AB = $$\sqrt{(3 ? 1)^2 + (2 ? 1)^2}$$
or, AB = $$\sqrt{(2)^2 + (3)^2}$$
or, AB = $$\sqrt{4 + 9}$$
or, AB = $$\sqrt{13}$$ units

For BC
B(3, 2) = (x1, y1)
C(1, 4) = (x2, y2)
BC = ?
We know that,
or, BC = $$\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}$$
or, BC = $$\sqrt{(1 ? 3)^2 + (4 ? 2)^2}$$
or, BC = $$\sqrt{(? 2)^2 + (2)^2}$$
or, BC = $$\sqrt{4 + 4}$$
or, BC = $$\sqrt{8}$$ units

For CD
C(1, 4) = (x1, y1)
D(?2, 2) = (x2, y2)
CD = ?
We know that,
or, CD = $$\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}$$
or, CD = $$\sqrt{(? 2 ? 1)^2 + (2 ? 4)^2}$$
or, CD = $$\sqrt{(? 3)^2 + (? 2)^2}$$
or, CD = $$\sqrt{9 + 4}$$
or, CD = $$\sqrt{13}$$ units

A(1,?1) = (x1, y1)
D(?2, 2) = (x2, y2)
We know that,
or, AD = $$\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}$$
or, AD = $$\sqrt{(1 ? 1)^2 + (4+1)^2}$$
or, AD = $$\sqrt{(0)^2 + (5)^2}$$

For BC
B(3, 2) = (x1, y1)
D(?2, 2) = (x2, y2)
BD = ?
We know thats,
or, BD = $$\sqrt{(x_2? x_1)^2 + (y_2? y_1)^2}$$
or, BD = $$\sqrt{(? 2 ? 3)^2 + (2? 2)^2}$$
or, BD = $$\sqrt{(?5)^2 + (0)^2}$$
or, BD = 5 units

Here,
AC = BD = 5 units
But other points cannot be proved. So it is not rectangle.

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4
5
6
2

6
5
1
4

1
5
4
2

4
1
5
8

10
8
5
9

12
9
10
5

18
12
4
5

12
4
5
6

14
26
5
-5

8
9
10
5

5
4
13
12

13
(sqrt{13})
2(sqrt{22})
(sqrt{20})

12
3(sqrt{3})
(sqrt[2]{2})
3(sqrt{2})

(sqrt{2})
12
(sqrt[2]{2})
2(sqrt{2})
• ### Find distance: (0, 0) and (-6, -4)

2(sqrt{13})
(sqrt[2]{13})
(sqrt{13})
2(sqrt{2})

4
2(sqrt{14})
2(sqrt{13})
3(sqrt{13})