## Note on Distance Formula

• Note
• Things to remember
• Exercise
• Quiz

#### 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.

#### Plotting points in co-ordinate plane

Plotting points is the process of locating a point whose co-ordinates are given. The plotting of points can be done on graph paper.

#### 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.
.

### Very Short Questions

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

0%

4
6
2
5

6
4
1
5

5
4
1
2

1
5
4
8

10
5
9
8

10
12
9
5

18
12
5
4

4
12
6
5

5
26
14
-5

10
8
5
9

4
12
13
5

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

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

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

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

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