Distance Formula
Meaning
Coordinate Geometry is a branch of geometry which is used to identify a point on a plane. It was invented by RENE DESCARTES.
Rectangular Coordinate 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 xaxis and the point YOY' is called yaxis. The two lines XOX' and YOY' are also called rectangular coordinate axis which divide the plane into four equal parts which are called quadrant.
Plotting points in coordinate plane
Distance Between Two Points
If the elements or coordinates of any two points are given, the distance between them can be found with the help of distance formulae.
Suppose that,
P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) are any two points in the coordinates plane and 'd' is the distance between them.
Now,
Draw PT and QK perpendicular on xaxis and PM perpendicular to QK.
Then,
OK = x_{2}, KQ = y_{2}, OT = x, PT = y.
PM = TK= OK  OT = (x_{2} x_{1})
Also,
OM = OK  MK = QK  PT
= (y_{2} y_{1}) [∴ MK = PT]
Since, ∠PMQ is a right angle, so \(\triangle\)PMQ is a right angle triangle.
Now,
\(\triangle\)PMQ using Pythagoras Theorem,
PQ^{2}= PM^{2}= QM^{2}
= (x_{2} x_{1})^{2}+ (y_{2} y_{1})^{2}
or, PQ = (x_{2}− x_{1})^{2} + (y_{2}− y_{1})^{2} ................... \(\sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2}\)
∴ Distance (d) = (x_{2}− x_{1})^{2}+ (y_{2}− y_{1})^{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 = x^{2}+ y^{2} .................... \(\sqrt{x^2 + y^2}\)
Again,
Slope of the line = PQ = Tanθ
\(\frac{OM}{PQ}\) = \(\frac{y_2 − y_1}{x_2 − x_1}\)
Types of triangle using the coordinates 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)  RightAngled 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 coordinates 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. 
 Coordinate Geometry is a branch of geometry which is used to identify a point on a plane.
 Coordinate 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 coordinates 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) = (x_{1}, y_{1})
B(0, 6) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(4 , 8).= (x_{2}, y_{2})
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) = (x_{2}, y_{1})
F(−3, 4) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
B (3, 10) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(6, 8) = (x_{2}, y_{2})
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) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(−1, 0) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
A(2, 3) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(−1,−2) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
D(10, 8) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
Q(4, 1) = (x_{2}, y_{2})
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) =(x_{1}, y_{1})
R(10, 3) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(4, 5) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(7, 2) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
A(1, 2) = (x_{2}, y_{2})
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, 6^{2} = (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. h^{2} = p^{2} + b^{2}). So, the given points are the vertices of isosceles right angled triangle
Solution:
For AB
A(1,−1) = (x_{1}, y_{1})
B(−1, 1) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(\(\sqrt{3}\), \(\sqrt{3}\)) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(\(\sqrt{3}\),\(\sqrt{3}\)) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(4, 4) = (x_{2}, y_{2})
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) =(x_{1}, y_{1})
C(4, 8) = (x_{2}, y_{2})
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) =(x_{1}, y_{1})
D(1, 5) =(x_{2}, y_{2})
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
For AD
A(1, 1) =(x_{1}, y_{1})
D(1, 5) =(x_{2}, y_{2})
AD = ?
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 = \(\sqrt{4^2}\)
or, AD = 4 units
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) = (x_{1}, y_{1})
B(2, 1) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(0, 3) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
D(−2, 1) = (x_{2}, y_{2})
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
For AD
A(0,−1) = (x_{1}, y_{1})
D(−2, 1) = (x_{2}, y_{2})
AD = ?
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) = (x_{1}, y_{1})
C(0, 3) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(2, 1) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(7, 4) = (x_{2}, y_{2})
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) =(x_{1}, y_{1})
C(8, 11) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(8, 11) =(x_{2}, y_{2})
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) = (x_{1}, y_{1})
B(3, 2) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
C(1, 4) = (x_{2}, y_{2})
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) = (x_{1}, y_{1})
D(?2, 2) = (x_{2}, y_{2})
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
For AD
A(1,?1) = (x_{1}, y_{1})
D(?2, 2) = (x_{2}, y_{2})
AD = ?
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 = \(\sqrt{(0)^2 + (5)^2}\)
or, AD = 5 units
For BC
B(3, 2) = (x_{1}, y_{1})
D(?2, 2) = (x_{2}, y_{2})
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.

Find the distance.
(2, 1) and (6, 4)
4
5
6
2

Find the distance.
(2, 1) and (6, 4)
6
5
1
4

Find the distance.
(5, 6) and (8, 10)
1
5
4
2

Find the distance.
(3, 9) and (7, 6)
4
1
5
8

Find the distance.
(2, 7) and (6, 4)
10
8
5
9

Find the distance from origin (3, 4)
12
9
10
5

Find the distance from origin (3, 4)
18
12
4
5

Find the distance from origin (1, 2(sqrt{6}))
12
4
5
6

Find the distance from origin (2, (sqrt{21}))
14
26
5
5

Find the distance from origin (4,3)
8
9
10
5

Find the distance.
(7, 7) and (3, 4)
5
4
13
12

Find distance: (3, 4) and (5, 7)
13
(sqrt{13})
2(sqrt{22})
(sqrt{20})

Find distance: (5, 6 ) and (8, 9)
12
3(sqrt{3})
(sqrt[2]{2})
3(sqrt{2})

Find distance: (2, 0) and (0, 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})

Find the distance between point (0, 0) and (−6, −4)
4
2(sqrt{14})
2(sqrt{13})
3(sqrt{13})

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SajeenaIf the points (1,b) and (b,a) are at equidistant from the point (m,n), then show that m(a b) = n(ba) 
Jan 24, 2017 
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