Imagine a person riding his two-wheeler on a straight road towards East from his college to village A and then to village B.

**The** **Mid-point
of a Line Segment**

Imagine a
person riding his two-wheeler on a straight road towards East from his college to
village *A* and then to village *B*. At some point in between *A*
and *B*, he suddenly realises that there is not enough petrol for the journey.
On the way there is no petrol bunk in between these two places. Should he travel
back to *A* or just try his luck moving towards *B*? Which would be the
shorter distance? There is a dilemma. He has to know whether he crossed the half
way mid-point or not.

The above
Fig. 5.21 illustrates the situation. Imagine college as origin *O* from which
the distances of village *A* and village *B* are respectively *x*_{1}
and *x*_{2} (*x*_{1} < *x*_{2}
) . Let *M* be the mid-point of *AB* then *x* can be obtained as
follows.

AM = MB and
so,* x *− *x*_{1} =* x*_{2}
− *x*

and this
is simplified to* x *= [ *x*_{1} + *x*_{2} ] / 2

Now it is
easy to discuss the general case. If *A* ( *x*_{1,}
*y*_{1} ) , *B* (
*x*_{2 ,} *y*_{2} ) are any two points and *M *(*
x*,* y*)*
*is the mid-point of the line segment
AB, then* M*′* *is the mid-point of* AC *(in the* *Fig. 5.22). In a
right triangle the perpendicular bisectors of the sides intersect at the mid-point
of the hypotenuse. (Also, this property is due to similarity among the two coloured
triangles shown; In such triangles, the corresponding sides will be proportional).

**Another way of solving (Using similarity
property)**

Let us take
the point M as M(x,y)

Now, ΔAMM′
and ΔMBD are similar. Therefore,

AM′ / MD = MM′ / BD = AM / MB

* x*-coordinate
of M = the average of x-coordinates of A and C = [ x_{1} + x_{2}]
/ 2 and similarly,

* y*-coordinate
of M = the average of y-coordinates of B and C = [ y_{1} + y_{2}
] / 2

The mid-point M of the line segment joining the points *A* (*x*_{1}
, *y*_{1} ) and *B* (*x* _{2} , *y*_{2}
) is

**Thinking
Corner**

If D is the mid-point of AC and C is the mid-point of AB, then find
the length of AB if AD = 4*cm*.

*For example***, **The mid-point of the line segment joining the points ( −8, −10) and
( 4, −2) is given by ( [

where *x*_{1} = −8 , *x*_{2} = 4, *y*_{1} = −10 and *y*_{2} = −2 .

The required mid-point is ( [−8+4]/2 , [−10−2]/2 ) , or ( −2, −6)
.

Let us now
see the application of mid-point formula in our real life situation, consider the
longitude and latitude of the following cities.

Let us take the longitude and latitude Bay of Chennai (80.27° E, 13.00° N) and Mangaluru (74.85° E , 13.00° N) as pairs. Since Bengaluru is located in the middle of Chennai and Mangaluru, we have to find the average of the coordinates, that is ( [80.27 + 74.85] /2 , [13.00 + 13.00]/2 ). This gives (77.56°E, 13.00°N) which is the longitude and latitude of Bengaluru. In all the above examples, the point exactly in the middle is the mid-point and that point divides the other two points in the same ratio.

**Example 5.12**

The point
(
3, −4)
is the centre of a circle. If *AB* is a diameter of the circle and *B*
is (
5, −6)
, find the coordinates of *A*.

*Solution *

Let the coordinates of* **A*** **be

Therefore,
the coordinates of *A* is (1, −2)

**Progress Check**

(i) Let *X* be the mid-point of the line segment joining *A*(3,
0) and *B*(−5, 4) and *Y* be the
mid-point of the line segment joining *P*(−11, −8) and *Q*(8,* *−2)* *. Find the mid-point of the line segment*
XY*.

(ii) If (3, *x*) is the mid-point of the line segment joining
the points *A*(8, −5) and *B*(−2,11) , then find the value of ‘*x* ’.

**Example 5.13**

If (*x*,3),
(6,*y*), (8,2) and (9,4) are the vertices of a parallelogram taken in order,
then find the value of *x* and *y*.

*Solution** *

Let *A*(*x*,3),
*B*(6,*y*), *C*(8,2) and *D*(9,4) be the vertices of the parallelogram
*ABCD*. By definition, diagonals* AC *and* BD *bisect each other.

Mid-point
of *AC* = Mid-point of *BD*

equating
the coordinates on both sides, we get

Hence, *x*
= 7 and *y* = 1.

*A*(6,1),* B*(8, 2)* *and* C*(9, 4)* *are three
vertices of a parallelogram* ABCD *taken in order. Find the fourth vertex *D*.
If (*x*_{1} , *y*_{1}), (*x* _{2} , *y*_{2}
), (*x* _{3} , *y*_{3} ) and (*x* _{4} ,
*y* _{4} ) are the four vertices of the parallelogram, then using the
given points, find the value of (*x*_{1} + *x*_{3} − *x* _{2} , *y*_{1} + *y* _{3} − *y*_{2} ) and state the reason for
your result.

Find the
points which divide the line segment joining *A*(−11,4) and *B*(9,8) into
four equal parts.

Let *P*,
*Q*, *R* be the points on the line segment joining *A*(−11,4) and
*B*(9,8) such that *AP*=*PQ*=*QR*=*RB*.

Here *Q*
is the mid-point of *AB*, *P* is the mid-point of *AQ* and *R*
is the mid-point of *QB*.

Hence the
points which divides *AB* into four equal parts are *P*(–6, 5), *Q*(–1,
6) and *R*(4, 7).

**Example 5.15**

The mid-points
of the sides of a triangle are (5,1), (3, −5) and (−5,
−1).
Find the coordinates of the vertices of the triangle.

*Solution *

Let the vertices of the** **Δ

Therefore
the vertices of the triangles are *A* (−3, 5), *B* (13, −3)
and *C* (−7 , −7).

If (*a*_{1} , *b*_{1}), (*a*_{2}
, *b*_{2} ) and (*a*_{3} , *b*_{3} ) are
the mid-points of the sides of a triangle, using the mid-points given in example
5.15 find the value of (*a*_{1} + *a*_{3} − *a*_{2} , *b*_{1} + *b*_{3} − *b*_{2} ), (*a*_{1}
+ *a*_{2} − *a*_{3} , *b*_{1} + *b*_{2} − *b*_{3} ) and (*a*_{2}
+ *a*_{3} − *a*_{1} , *b*_{2} + *b*_{3} − *b*_{1}) . Compare the results. What
do you observe? Give reason for your result?

Tags : Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths , 9th Maths : UNIT 5 : Coordinate Geometry

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9th Maths : UNIT 5 : Coordinate Geometry : The Mid-point of a Line Segment | Formula, Steps, Example Solved Problems | Coordinate Geometry | Maths

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