###
**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 (x

_{1}, y_{1}) and B (x_{2}, y_{2}) in the plane xy is
The distance
of a point A (x

_{1}, y_{1}) form the Origin, O (0, 0), in the plane xy is
The distance
between two points- A (x

_{1}, y) and B (x_{2}, y) in the plane xy is
The distance
between two points- A (x, y

_{1}) and B (x, y_{2}) 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 = √(x

^{2}+ 25)
By squaring
both sides-

13

^{2}= x^{2}+ 25
169- 25 = x

^{2}
144 = x

^{2}
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= √(3

^{2}+ 4^{2})
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 (x

_{1}, y_{1}) and B (x_{2}, y_{2}) 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 (x

_{1}, y_{1}) and B (x_{2}, y_{2}) 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)-

P

_{X}= [(1x-4) + (3x4)]/ (1+3) = (-4+12)/4 = 8/4 = 2
P

_{y}= [(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)-

Q

_{x}= [(1x-4) - (3x4)]/ (1-3) = (-4 -12)/ -2 = -16/ -2 = 8
Q

_{y }= [(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.

## 0 comments:

## Post a Comment