# Co-ordinate Geometry II- Co-ordinate Geometry Formulas to Calculate Distance between 2 Points

### Co-ordinate Geometry Problems asked in SSC CGL exam are of various kinds. One such set of co-ordinate geometry problems asked in the SSC CGL Exam needs you to find the distance between two points. Read on for a clear understanding.

Co-ordinate geometry formulas can be extremely helpful in solving co-ordinate geometry problems asked in SSC CGL Exams. These co-ordinate geometry problems are simple and you just need a basic understanding of the Cartesian system to be able to solve them. In the second blog in our series of co-ordinate geometry, we will discuss the various co-ordinate geometry formulas used to solve problems. So we will start with discussing some co-ordinate geometry formulas and then move to questions based on them. But before we move ahead you must quickly revise the basics of co-ordinate geometry.

### Co-ordinate Geometry Formulas: Distance between 2 Points

The distance between two points- A (x1, y1) and B (x2, y2) in the plane xy is
The distance of a point A (x1, y1) form the Origin, O (0, 0), in the plane xy is
The distance between two points- A (x1, y) and B (x2, y) in the plane xy is
The distance between two points- A (x, y1) and B (x, y2) in the plane xy is

### Set 1: Co- ordinate Geometry Problems based on Distance Formula

A lot of problems asked in SSC Exams can be solved by simply using appropriate co-ordinate geometry formulas. These co-ordinate geometry formulas are simple to remember and using smart calculation tricks will get us to the correct answer in just a few seconds.

Problem 1: If the distance between two points is (0, -5) and (x, 0) is 13 units. What is the value of ‘x’?

Solution 1:
We know the formula used to find distance between two points-
Now the above formula can be used to find the value of x, substituting the given values we get-
13 = { (x- 0)2 + (0- -5)2}
13 = (x2 + 25)
By squaring both sides-
132 = x2 + 25
169- 25 = x2
144 = x2
x= +12

Therefore the value of point x is +12

Problem 2: What is the distance between the points (0, 0) and the intersecting point of the graphs x=2 and y=4?

Solution 2:
Just because the question seems to be a little complicated on reading, does not imply that it is actually complicated. Use of co-ordinate geometry formulas will help us get the answer. So the formula to be used in this question from the above co-ordinate geometry formulas is-
We know that one of the points is the Origin. For the other point, let’s start by plotting a graph. Now there are actually two points we are talking about here- (3, 0) and (0, 4). Plot both these points on the xy plane and then extend the lines. Once we do that, they intersect at a common point P, the co-ordinates for which will be (3, 4).
Now since one of the points is the origin, we can use the formula-
Distance= √(32 + 42)
Distance= √(9 +16)
Distance= √25
Distance= 5
So the distance between the origin and point P is 5 units.

This how co-ordinate geometry problems can be solved- by using the appropriate formula, from the list of co-ordinate geometry formulas.

### Co-ordinate Geometry Formulas: Section Formulas

If any point P (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n internally then,
If P is the midpoint, in that case the ratio between m and n will be 1: 1 and the formula will be-
If any point P (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n externally then,

### Set 2: Co- ordinate Geometry Problems based on Section Formula

Co-ordinate geometry problems asked in SSC Exams are also in section formulas. Though they may appear complex and complicated, they can be solved easily by using co-ordinate geometry formulas.

Problem 1: Find the point that divides the line segment joining the points (4, 5) and (-4, 1) in the ratio 1:3 (i) internally (ii) externally.

Solution 1:
So we have to find two sets of points- internal and external. The question is direct and can be easily solved by using these two co-ordinate geometry formulas-

(i) Let us start with the first part of the question, where the division is done internally. Assume the point to be P (x, y)-
PX= [(1x-4) + (3x4)]/ (1+3) = (-4+12)/4 = 8/4 = 2
Py= [(1x1) + (3x5)]/ (1+3) = (1+15)/4 = 16/4 = 4
P= (2, 4)

(ii) Moving on to the second part of the question where the division is done externally. Assume the point to be Q (x, y)-
Qx= [(1x-4) - (3x4)]/ (1-3) = (-4 -12)/ -2 = -16/ -2 = 8
Qy = [(1x1) - (3x5)]/ (1-3) = (1-15)/-2 = -14/ -2 = 7
Q= (8, 7)

So the two set of co-ordinates that we get are: (2,4) divides the line internally and (8,7) divides the line externally.

### Practice Problems based on Co- ordinate Geometry Formulas

Question 1: In the xy-coordinate system, the distance between (2 3,- 2) and (5 3, 3 2) and is approximately-
a) 5.1       b) 7.7       c) 4.3       d) 3.8

Question 2: Consider the three points in the x-y plane: P = (2, 4), Q= (7, 7), and R = (6, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin
a) P, R, Q       b) R, P, Q       c) R, Q, P       d) P,Q,R

Question 3: Find the point that divides the line segment joining the points (4,5) and (-4,1) in the ratio 1:3 (i) internally (ii) externally
a) (1, 2) (4, 3)      b) (2, 3) (5, 8)       c) (2, 4) (8, 7)       d) (3, 2) (6, 7)

Question 4: Find the co-ordinates of the point which divides the join of the points (2, 3) and (5, -3) in the ratio 1:2 (i) internally (ii) externally
a) (3, 1) (-1, 9)       b) (4, 3)(7, 8)        c) (1, 4) (2, 7)       d)(3, 2) (6, 9)

Remember to write the answer to the above questions in the comment section below.

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