Distance Formula
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- 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.
Questions and Answers
Click on the questions below to reveal the answers
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
-
Find the distance.
(2, 1) and (6, 4)
2
4
5
6
-
Find the distance.
(2, 1) and (6, 4)
5
6
4
1
-
Find the distance.
(5, 6) and (8, 10)
2
5
4
1
-
Find the distance.
(3, -9) and (7, -6)
8
4
1
5
-
Find the distance.
(2, -7) and (6,- 4)
8
5
10
9
-
Find the distance from origin (3, 4)
5
10
9
12
-
Find the distance from origin (-3, 4)
12
4
18
5
-
Find the distance from origin (1, 2(sqrt{6}))
5
6
12
4
-
Find the distance from origin (2, (sqrt{21}))
5
26
14
-5
-
Find the distance from origin (4,3)
9
8
10
5
-
Find the distance.
(-7, -7) and (-3, -4)
13
4
12
5
-
Find distance: (3, 4) and (5, 7)
(sqrt{20})
(sqrt{13})
13
2(sqrt{22})
-
Find distance: (5, 6 ) and (8, 9)
(sqrt[2]{2})
12
3(sqrt{2})
3(sqrt{3})
-
Find distance: (2, 0) and (0, 2)
(sqrt{2})
12
(sqrt[2]{2})
2(sqrt{2})
-
Find distance: (0, 0) and (-6, -4)
(sqrt{13})
(sqrt[2]{13})
2(sqrt{13})
2(sqrt{2})
-
Find the distance between point (0, 0) and (−6, −4)
4
2(sqrt{13})
2(sqrt{14})
3(sqrt{13})
-
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ASK ANY QUESTION ON Distance Formula
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Sajeena
If the points (-1,b) and (b,a) are at equidistant from the point (m,n), then show that m(a b) = n(b-a)
Jan 24, 2017
0 Replies
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