eAge Tutor
Login

Articles

Distance between two points

Print

co-geo_distance1The distance between any two points in the plane is the length of the line segment joining them.

The distance between two points P(x1, y1) and Q(x2, y2) is given by

PQ = √(x2 – x1)2 + (y2 – y1)2
PQ = √ (Difference of abscissae)2 + (Difference of ordinates)2

Let P(x1, y1) and Q(x2, y2) are two given points in the plane.



co-geo_distance2
Draw PM and QL perpendicular from P and Q on x – axis. From P draw PN perpendicular to QL on x – axis. Then,


OM = x1

OL = x2
PM = y1
QL = y2
PN = ML = OL – OM = x2 – x1
QN = QL – NL = QL – PM = y2 – y¬¬1



So, ΔPNQ is a right triangle right angled at N. Therefore, by Pythagoras theorem, we have

PQ2 = PN2 + QN2
PQ2 = (x2 – x1)2 + (y2 – y¬¬1)2
PQ = √(x2 – x1)2 + (y2 – y¬¬1)2

Hence, distance between any two points is given by

√ (Difference of abscissae)2 + (Difference of ordinates)2

Important Note:
•    If O is the origin and P(x, y) is any point, then from the above formula, we have
OP = √(x – 0)2 + (y – 0)2
OP = √x2 + y2
•    In order to prove that a given figure is a:
i) Square, prove that four sides are equal and the diagonals are also equal.
ii) Rhombus, prove that four sides are equal.
iii) Rectangle, prove that opposite sides are equal and the diagonals are also equal.
iv) Parallelogram, prove that opposite sides are equal.
v) Parallelogram but not a rectangle, prove that its opposite sides are equal but the diagonals are not equal.
vi) Rhombus but not a square, prove that its all sides are equal but the diagonals are not equal.
•    For three points to be collinear, prove that the sum of the distances between two pairs of points is equal to the third pair of points.

Example: Find the distance between the points: P (-6, 7) and Q (-1, -5)

Here x1 = -6, y1 = 7 and x2 = -1, y2 = -5
PQ = √(x2 – x1)2 + (y2 – y¬¬1)2

PQ = √(-1 + 6)2 + (-5 - 7)2
PQ = √25 + 144
PQ = √169 = 13 units

Example: Find a point on x – axis which is equidistant from A (2, -5) and

B (-2, 9).

We know that a point on x – axis is of the form (x, 0). SO, let P(x, 0) be the point equidistant from A (2, -5) and B (-2, 9). Then,

PA = PB
√(x – 2)2 + (0 + 5)2 = √(x + 2)2 + (0 – 9)2
(x – 2)2 + 25 = (x + 2)2 + 81
x2 – 4x + 4 + 25 = x2 + 4x + 4 + 81
-8x = 56
x = -7

Hence, the required point is (-7, 0)

Now try it yourself!  Should you still need any help, click here to schedule live online session with e Tutor!

About eAge Tutoring :

eAgeTutor.com is the premium online tutoring provider.  Using materials developed by highly qualified educators and leading content developers, a team of top-notch software experts, and a group of passionate educators, eAgeTutor works to ensure the success and satisfaction of all of its students. 

Contact us today to learn more about our guaranteed results and discuss how we can help make the dreams of the student in your life come true!

Reference Links :


    

Archives

Blog Subscription