Co-ordinate Geometry - NCERT Questions

Find the distance between the following pairs of points :

(i) (2, 3), (4, 1)

(ii) (-5, 7), (-1, 3)

(iii) (*a*, *b*)(-*a*, -*b*)

**Soln. :** (i) Here *x*_{1} = 2, *y*_{1} = 3 and *x*_{2} = 4 and *y*_{2} = 1

∴ The required distance

(ii) Here *x*_{1} = -5, *y*_{1} = 7 and *x*_{2} = -1, *y*_{2} = 3

∴ The required distance

(iii) Here *x*_{1} = *a*, *y*_{1 }= *b* and *x*_{2} = -*a*, *y*_{2} = -*b*

∴ The required distance

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns *A* and *B* discussed below as following : '*A*town *B* is located 36 km east and 15 km North of the town *A*'.

**Soln. :** Part-I

Let the points be *A*(0, 0) and *B*(36, 15)

Part-II

We have *A*(0, 0) and *B*(36, 15) as the positions of two towns

Here *x*_{1} = 0, *x*_{2} = 36 and *y*_{1} = 0,* y*_{2} = 15

.

Determine if the points (1, 5), (2, 3) and

(-2, -11) are collinear.

**Soln. :** Let the points be *A*(1, 5), *B*(2, 3) and *C*(-2, -11) *A*, *B* and *C* are collinear, if
*AB* + *BC* = *AC*, *AC* + *CB* = *AB*, *BA* + *AC* = *BC*

But *AB* + *BC* ≠ *AC*, *AC* + *CB* ≠ *AB*, *BA* + *AC* ≠ *BC*

∴ *A*, *B* and *C* are not collinear.

Check whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.

SOLUTION:
**Soln. :** Let the points be *A*(5, -2), *B*(6, 4) and *C*(7, -2).

We have *AB* = *BC *≠ *AC*.

∴ ∆*ABC* is an isosceles triangle.

In a classroom, 4 friends are seated at the point *A*, *B*, *C *and *D* as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think *ABCD* is a square?" Chameli disagrees. Using distance formula, find which of them is correct.

**Soln. :** Let the number of horizontal columns represent the *x*-coordinates whereas the vertical rows represent the *y*-coordinates.

∴ The points are : *A*(3, 4), *B*(6, 7), *C*(9, 4) and *D*(6, 1)

Since, *AB* = *BC* = *CD* = *AD* *i*.*e*., All the four sides are equal.

*i*.*e*., *BD* = *AC *⇒ Both the diagonals are also equal.

∴ *ABCD* is a square.

Thus, Champa is correct.

Name the type of quadrilateral formed if any, by the following points, and give reasons for your answer :

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

**Soln. :** (i) Let the points be *A*(-1, -2), *B*(1, 0), *C*(-1, 2) and *D*(-3, 0) of a quadrilateral *ABCD*.

⇒ *AB* = *BC* = *CD* = *AD* *i*.*e*., All the sides are equal. And *AC* = *BD*

Also, *AC* and *BD* (the diagonals) are equal.

∴ *ABCD* is a square.

(ii) Let the points be *A*(-3, 5), *B*(3, 1), *C*(0, 3) and *D*(-1, -4).

We see that
*i*.*e*., *AC* + *BC* = *AB*

⇒ *A*, *B* and *C* are collinear. Thus, *ABCD* is not a quadrilateral.

(iii) Let the points be *A*(4, 5), *B*(7, 6), *C*(4, 3) and *D*(1, 2).

Since, *AB* = *CD*, *BC* = *DA* [opposite sides of the quadrilateral are equal]

And *AC* ≠ *BD* ⇒ Diagonals are unequal.

∴ *ABCD* is a parallelogram.

Find the point on the *x*-axis which is equidistant from (2, -5) and (-2, 9).

**Soln. :** We know that any point on *x*-axis has its ordinate = 0.

Let the required point be *P*(*x*, 0).

Let the given points be *A*(2, -5) and *B*(-2, 9)

Since, *A* and *B* are equidistant from *P*,

∴ *AP* = *BP*

⇒ *x*^{2} - 4*x* + 29 = *x*^{2} + 4*x* + 85

⇒ *x*^{2} - 4*x* - *x*^{2} - 4*x* = 85 - 29

⇒ -8*x* = 56

∴ The required point is (-7, 0)

Find the values of *y* for which the distance between the points *P*(2, -3) and *Q*(10, *y*) is 10 units.

**Soln. :** The given points are *P*(2, -3) and *Q*(10, *y*).

But *PQ* = 10

Squaring both sides, *y*^{2} + 6*y* + 73 = 100

⇒ *y*^{2} + 6*y* - 27 = 0

⇒ *y*^{2} - 3*y* + 9*y* - 27 = 0

⇒ (*y* - 3)(*y* + 9) = 0

⇒ Either *y* - 3 = 0 ⇒ *y* = 3

or *y* + 9 = 0 ⇒* y* = -9

∴ The required value of *y* is 3 or -9.

If *Q*(0, 1) is equidistant from *P*(5, -3) and *R*(*x*, 6), find the values of *x*. Also find the distances *QR* and *PR*.

**Soln. :**

∴ *QP* = *QR*

Squaring both sides, we have *x*^{2} + 25 = 41

⇒ *x*^{2} + 25 - 41 = 0

⇒ *x*^{2} - 16 = 0

Thus, the point *R* is (4, 6) or (-4, 6)

Now,

and

Find a relation between *x* and *y* such that the point (*x*, *y*) is equidistant from the point

(3, 6) and (-3, 4).

**Soln. :** Let the points be *A*(*x*, *y*), *B*(3, 6) and *C*(-3, 4).

Since, the point (*x*, *y*) is equidistant from (3, 6) and (-3, 4).

∴ *AB* = *AC*

Squaring both sides,

(3 - *x*)^{2} + (6 -* y*)^{2} = (-3 - *x*)^{2} + (4 - *y*)^{2}

⇒ (9 + *x*^{2} - 6*x*) + (36 + *y*^{2} - 12*y*)

= (9 + *x*^{2} + 6*x*) + (16 + *y*^{2} - 8*y*)

⇒ 9 + *x*^{2} - 6*x* + 36 + *y*^{2} - 12*y* - 9 - *x*^{2} - 6*x*

- 16 - *y*^{2} + 8*y* = 0

⇒ - 6*x* - 6*x* + 36 - 12*y* - 16 + 8*y* = 0

⇒ - 12*x* - 4*y* + 20 = 0

⇒ - 3*x* - *y* + 5 = 0

⇒ 3*x* +* y* - 5 = 0 which is the required relation between* x* and *y*.

Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

SOLUTION:
**Soln. :** Let the required point be *P*(*x*, *y*).

Here the end points are (-1, 7) and (4, - 3)

∴ Ratio = 2 : 3 = *m*_{1} : *m*_{2}

Thus, the required point is (1, 3).

Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

SOLUTION:
**Soln. :** Let the given points be *A*(4, -1) and *B*(-2, -3).

Let the points *P* and *Q* trisect *AB*.
*i*.*e*., *AP* = *PQ* = *QB*
*i*.*e*.,* P* divides *AB* in the ratio of 1 : 2
*Q* divides *AB* in the ratio of 2 : 1

Let the coordinates of *P* be (*x*, *y*)

∴ The required co-ordinates of *P* are

Let the co-ordinates of *Q* be (*X*, *Y*)

∴ The required coordinates of *Q* are

To conduct Sports Day activities, in your rectangular shaped school ground *ABCD*, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along *AD*, as shown in the figure. Niharika runs 1/4^{th} the distance* AD* on the 2nd line and posts a green flag. Preet runs 1/5^{th} the distance *AD* on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly half way between the line segment joining the two flags, where should she post her flag ?.

**Soln. :** Let us consider '*A*' as origin, then

*AB* is the *x*-axis.
*AD* is the *y*-axis.

Now, the position of green flag-post is

And, the position of red flag-post is

⇒ Distance between both the flags

Let the mid-point of the line segment joining the two flags be *M*(*x*, *y*).

or *x* = 5 and *y* = 22.5

Thus, the blue flag is on the 5th line at a distance 22.5 m above *AB*.

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by

(-1, 6).

**Soln. :** Let the given points are *A*(-3, 10) and *B*(6, -8).

Let the point *P*(-1, 6) divides *AB* in the ratio *m*_{1} : *m*_{2}.

∴ Using the section formula, we have

⇒ -1(*m*_{1} + *m*_{2}) = 6*m*_{1} - 3*m*_{2}

and 6(*m*_{1} + *m*_{2}) = - 8*m*_{1} + 10*m*_{2}

⇒ -*m*_{1} - *m*_{2} - 6*m*_{1} + 3*m*_{2} = 0

and 6*m*_{1} + 6*m*_{2} + 8*m*_{1} - 10*m*_{2} = 0

⇒ -7*m*_{1} + 2*m*_{2} = 0

and 14*m*_{1} - 4*m*_{2} = 0 or 7*m*_{1} - 2*m*_{2} = 0

⇒ 2*m*_{2} = 7*m*_{1} and 7*m*_{1} = 2*m*_{2}

⇒ *m*_{1} : *m*_{2} = 2 : 7 and *m*_{1} : *m*_{2} = 2 : 7

Thus, the required ratio is 2 : 7.

Find the ratio in which the line segment joining *A*(1, -5) and *B*(-4, 5) is divided by the *x*-axis. Also find the coordinates of the point of division.

**Soln. :** The given points are : *A*(1, -5) and *B*(-4, 5).

Let the required ratio = *k* : 1 and the required point be *P*(*x*, *y*)

Part-I : To find the ratio

Since the point *P* lies on *x*-axis,

∴ Its *y*-coordinate is 0.

and 0

⇒ *x*(*k* + 1) = - 4*k* + 1

and 5*k* - 5 = 0 ⇒ *k* = 1

⇒ *x*(*k* + 1) = - 4*k* + 1

⇒ *x*(1 + 1) = - 4 + 1 [∴ *k* = 1]

⇒ 2*x* = -3

∴ The required ratio *k* : 1 = 1 : 1

Coordinates of *P* are (*x*, 0)

If (1, 2), (4, *y*), (*x*, 6) and (3, 5) are the vertices of a parallelogram taken in order, find *x* and *y*.

**Soln. :** The given points are : *A*(1, 2), *B*(4, *y*), *C*(*x*, 6) and *D*(3, 5)

Since, the diagonals of a parallelogram bisect each other.

∴ The coordinates of *P* are :

⇒ *x* + 1 = 7 ⇒ *x* = 6

⇒ 5 + *y* = 8 ⇒ *y* = 3

∴ The required values of *x* and *y* are : *x* = 6, *y* = 3.

Find the coordinates of a point *A*, where *AB* is the diameter of a circle whose centre is (2, -3) and *B* is (1, 4)

**Soln. :** Here, centre of the circle is *O*(2, -3)

Let the end points of the diameter be *A*(*x*, *y*) and *B*(1, 4)

The centre of a circle bisects the diameter.

And

Here the coordinates of *A* are (3, -10)

If *A* and *B* are (-2, -2) and (2, -4) respectively, find the coordinates of *P* such that *AP* = *AB* and *P* lies on the line segment *AB*.

**Soln. :**

Here the given points are *A*(-2, -2) and *B*(2, -4)

Let the coordinates of *P* are (*x*, *y*)

Since, the point *P* divides* AB* such that

⇒ *AB* = *AP* + *BP*

* *⇒ *AP* :* PB* = 3 : 4 *i.e*., *P*(*x*, *y*) divides *AB* in the ratio 3 : 4.

Thus, the coordinates of *P* are

Find the coordinates of the points which divide the line segment joining *A*(-2, 2) and *B*(2, 8) into four equal parts.

**Soln. :** Here the given points are* A*(-2, 2) and *B*(2, 8)

Let *P*_{1}, *P*_{2} and *P*_{3} divide *AB* in four equal parts.

*AP*_{1} = *P*_{1}*P*_{2} = *P*_{2}*P*_{3} = *P*_{3}*B*

Obviously, *P*_{2} is the mid-point of *AB*

∴ Coordinates of *P*_{2} are

Again, *P*_{1} is the mid-point of *AP*_{2}.

∴ Coordinates of *P*_{1} are

Also *P*_{3} is the mid-point of *P*_{2}*B*.

∴ Coordinates of *P*_{3} are

Thus, the coordinates of *P*_{1}, *P*_{2} and *P*_{3} are respectively.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

[Hint : Area of a rhombus = (product of its diagonals)]

**Soln. :** Let the vertices of the given rhombus are : *A*(3, 0), (4, 5), (-1, 4) and (-2, -1)

∴ *AC* and *BD* are the diagonals of rhombus *ABCD*.

∴ Diagonal

Diagonal

∴ For a rhombus,

Area × (Product of diagonals)

= 4 × 6 = 24 square units.

Find the area of the triangle whose vertices are :

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

**Soln. :** (i) Let the vertices of the triangles be *A*(2, 3), *B*(-1, 0) and *C*(2, - 4)

Here *x*_{1} = 2, *y*_{1} = 3,

*x*_{2} = -1, *y*_{2} = 0

*x*_{3} = 2, *y*_{3} = -4

∴ Area of a ∆

∴ Area of a ∆*ABC*

(ii) Let the vertices of the triangles be *A*(-5, -1), *B*(3, -5) and *C*(5, 2)
*i*.*e*., *x*_{1} = -5, *y*_{1} = -1
*x*_{2} = 3, *y*_{2} = -5
*x*_{3} = 5, *y*_{3} = 2

∵ Area of a

∆

∴ Area of a ∆*ABC *

In each of the following find the value of '*k*', for which the points are collinear.

(i) (7, -2), (5, 1), (3, *k*)

(ii) (8, 1), (*k*, -4), (2, -5)

**Soln. :** The given three points will be collinear if the ∆ formed by them has equal to zero area.

(i) Let *A*(7, -2), *B*(5, 1) and *C*(3, *k*) be the vertices of a triangle.

∴ The given points will be collinear, if

ar (∆*ABC*) = 0

or 7(1 - *k*) + 5(*k* + 2) + 3(-2 - 1) = 0

⇒ 7 - 7*k* + 5*k* + 10 + (-6) - 3 = 0

⇒ 17 - 9 + 5*k* - 7*k* = 0

⇒ 8 - 2*k* = 0 ⇒ 2*k* = 8 ⇒

The required value of *k* = 4.

(ii)

⇒ 8 - 6*k* + 10 = 0 ⇒ 6*k* = 18 ⇒ *k* = 3.

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the given triangle.

SOLUTION:
**Soln. :** Let the vertices of the triangle be *A*(0, -1), *B*(2, 1) and *C*(0, 3).

Let *D*, *E* and *F* be the mid-points of the sides *BC*, *CA* and *AB* respectively. Then :

Coordinates of *D* are

Coordinates of *E *are

Coordinates of *F *are

Now, ar(∆*ABC*)

Now, ar(∆*DEF*)

∴ ar(∆*DEF*) : ar(∆*ABC*) = 1 : 4.

Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

SOLUTION:
**Soln. :** Let *A*(- 4, - 2), *B*(- 3, - 5), *C*(3, - 2) and *D*(2, 3) be the vertices of the quadrilateral.

Let us join diagonal *BD*.

Now, ar(∆*ABD*)

Now, ar(∆*CBD*)

Since, ar(quad *ABCD*) = ar(∆*ABD*) + ar∆*CBD*

You have studied in class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆*ABC* whose vertices are *A*(4, -6), *B*(3, -2) and *C*(5, 2).

**Soln. :** Here, the vertices of the triangles are *A*(4, -6), *B*(3, -2) and *C*(5, 2).

Let *D* be the midpoint of *BC*.

∴ The coordinates of the mid point *D* are

Since *AD* divides the triangle *ABC* into two parts *i*.*e*., ∆*ABD* and ∆*ACD*,

Now, ar(∆*ABD*)

= 3 sq.units (numerically) ........ (1)

ar(∆*ADC*)

= 3 sq.units (numerically) ........ (1)

From (1) and (2)

ar(∆*ABD*) = ar(∆*ADC*)
*i*.*e*., A median divides the triangle into two triangles of equal areas.

Determine the ratio in which the line 2*x *+ *y* - 4 = 0 divides the line segment joining the points *A*(2, -2) and *B*(3, 7).

**Soln. :** Let the required ratio be *k* : 1 and the point *C* divide them in the above ratio.

∴ Coordinates of *C* are

Since the point *C* lies on the given line

2*x* + *y* - 4 = 0,

∴ We have

⇒ 2(3*k* + 2) + (7*k* - 2) = 4 × (*k* + 1)

⇒ 6*k* + 4 + 7*k* - 4*k* - 4 - 2 = 0

⇒ (6 + 7 - 4)*k* + (-2) = 0 ⇒ 9*k* - 2 = 0

∴ The required ratio = *k* : 1

Find a relation between *x* and *y* if the points

(*x*,* y*), (1, 2) and (7, 0) are collinear.

**Soln. :** The given points are *A*(*x*,* y*), *B*(1, 2) and *C*(7, 0) are collinear

The points *A*, *B* and *C* will be collinear if area of

∆*ABC* = 0 [*x*(2 - 0) + 1 (0 - *y*) + 7(*y* - 2) = 0

or if 2*x* - *y* + 7*y* - 14 = 0

or if 2*x* + 6*y* - 14 = 0 or if *x* + 3*y* - 7 = 0 which is the required relation between *x* and *y*.

Find the centre of circle passing through the points (6, -6), (3, -7) and (3, 3)

SOLUTION:
**Soln. :** Let *P*(*x*, *y*) be the centre of the circle since the circle is passing through
*A*(6, -6), *B*(3, -7) and *C*(3, 3)

∴ *AP* = *BP* = *CP*

Taking *AP* = *BP*, we have *AP*^{2} = *BP*^{2}

⇒ (*x* - 6)^{2} + (*y* + 6)^{2} = (*x* - 3)^{2} + (*y* + 7)^{2}

⇒ *x*^{2} - 12*x* + 36 + *y*^{2} + 12*y* + 36

= *x*^{2} - 6*x* + 9 + *y*^{2} + 14*y* + 49

⇒ - 12*x* + 6*x* + 12*y* - 14*y* + 72 - 58 = 0

⇒ - 6*x* - 2*y* + 14 = 0

⇒ 3*x* + *y* - 7 = 0 ..... (1) [Dividing by (-2)]

Taking *BP* = *CP*, we have *BP*^{2} = *CP*^{2}

⇒ (*x* - 3)^{2} + (*y* + 7)^{2} = (*x* - 3)^{2} + (*y* - 3)^{2}

⇒ *x*^{2} - 6*x* + 9 + *y*^{2} + 14*y* + 49

= *x*^{2} - 6*x* + 9 + *y*^{2} - 6*y* + 9

⇒ - 6*x* + 6*x* + 14*y* + 6*y* + 58 - 18 = 0

⇒ 20*y* + 40 = 0

........ (2)

From (1) and (2),

3*x* - 2 - 7 = 0

⇒ 3*x* = 9 ⇒ *x* = 3
*i*.*e*., *x* = 3 and *y* = -2

∴ The required centre is (3, -2).

The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

SOLUTION:
**Soln. :** Let us have a square *ABCD* such that *A*(-1, 2) and *C*(3, 2) are the opposite vertices.

Let *B*(*x*, *y*) be an unknown vertex.

Since all sides of a square are equal,

∴ *AB* = *BC* ⇒ *AB*^{2} = *BC*^{2}

⇒ (*x* + 1)^{2} + (*y* - 2)^{2} = (*x* - 3)^{2} + (*y* - 2)^{2}

⇒ 2*x* + 1 = -6*x* + 9

⇒ 8*x* = 8 ⇒ *x* = 1 ........ (1)

Since each angle of a square = 90°,

∴ *ABC* is a right angled triangle.

∴ Using Pythagoras theorem, we have :
*AB*^{2} + *BC*^{2} = *AC*^{2}

⇒ [(*x* + 1)^{2} + (*y* - 2)^{2}] + [(*x* - 3)^{2} + (*y* - 2)^{2}]

= [(3 + 1)^{2} + (2 - 2)^{2}]

⇒ 2*x*^{2} + 2*y*^{2} + 2*x* - 4*y* - 6*x* - 4*y* + 1 + 4 + 9 + 4 = 16

⇒ 2*x*^{2} + 2*y*^{2} - 4*x* - 8*y* + 2 = 0

⇒ *x*^{2} + *y*^{2} - 2*x* - 4*y* + 1 = 0 ........ (2)

Substituting the value of *x* from (1) into (2) we have

1 + *y*^{2} - 2 - 4*y* + 1 = 0

⇒ *y*^{2} - 4*y* + 2 - 2 = 0

⇒ *y*^{2} - 4y = 0 ⇒ *y*(*y* - 4) = 0

⇒ *y* = 0 or *y* = 4

Hence, the two required other vertices are

(1, 0) and (1, 4).

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinats of the vertices of ∆PQR if C is the origin ?

Also calculate the areas of the triangles in these cases. What do you observe ?

**Soln. :** (i) By taking *A *as the origin and *AD* and *AB *as the coordinate axes. We have *P*(4, 6), *Q*(3, 2) and *R*(6, 5) as the vertices of ∆*PQR*.

(ii) By taking *C* as the origin and *CB* and *CD* as the coordinate axes, then the vertices of ∆*PQR* are *P*(- 12, - 2), *Q*(- 13, - 6) and *R*(- 10, - 3)

Now ar(∆*PQR*) [when *P*(4, 6), *Q*(3, 2) and *R*(6, 5) are the vertices]

sq. units

[taking numerical value]

ar(∆*PQR*) [when *P*(-12, -2),* Q*(-13, -6) and *R*(-10, -3) are the vertices]

Thus, in both cases, the area of ∆*PQR* is the same.

The vertices of a ∆*ABC* are *A*(4, 6), *B*(1, 5) and *C*(7, 2). A line is drawn to intersect sides *AB* and *AC* at *D* and *E* respectively, such that Calculate the area of the ∆*ADE* and compare it with the area of ∆*ABC*. [Recall "The converse of basis proportionality theorem", and "theorem of similar triangles taking their areas and corresponding sides"]

**Soln. :** We have

⇒ *AD* : *DB* = 1 : 3

Thus, the point *D* divides *AB* in the ratio

1 : 3

∴ The coordinates of *D* are :

Similarly, *AE* : *EC* = 1 : 3 *i*.*e*., *E* divides *AC* in the ratio 1 : 3

⇒ Coordinates of *E *are

Now, ar(∆*ADE*)

Area of ∆*ABC *

⇒ ar(∆*ADE*) : ar(∆*ABC*) = 1 : 16.

Let *A*(4, 2), *B*(6, 5) and *C*(1, 4) be the vertices of ∆*ABC*.

(i) The median from *A* meets *BC* at *D*. Find the coordinates of the point *D*.

(ii) Find the coordinates of the point *P* on *AD *such that *AP* :* PD *= 2 : 1.

(iii) Find the coordinates of points *Q* and *R* on medians *BE* and *CF* respectively such that

*BQ* : *QE* = 2 : 1 and *CR* : *RF* = 2 : 1

(iv) What do you observe ?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If *A*(*x*_{1}, *y*_{1}), *B*(*x*_{2}, *y*_{2}) and *C*(*x*_{3}, *y*_{3}) are the vertices of ∆*ABC*, find the coordinates of the centroid of the triangle.

**Soln. :** We have the vertices of ∆*ABC* as *A*(4, 2), *B*(6, 5) and *C*(1, 4).

(i) Since *AD* is a median

∴ Coordinates of *D* are

(ii) Since *AP* : *PD* = 2 : 1 *i*.*e*., *P* divides *AD* in the ratio 2 : 1

∴ Coordinates of *P* are

(iii) *BQ* : *QE* = 2 : 1 ⇒ [The point *Q* divides *BE *in the ratio 2 : 1]

[coordinates of *E* are

∴ Coordinates of *Q* are

Also, *CR* : *RF* = 2 : 1

⇒ The point *R* divides *CF* in the ratio 2 : 1

∴ Coordinates of *F* are

So, Coordinates of *R* are

(iv) We observe that* P*,* Q* and *R* represent the same point.

(v) Here, we have *A*(*x*_{1}, *y*_{1}), *B*(*x*_{2}, *y*_{2}) and *C*(*x*_{3}, *y*_{3}) are the vertices of ∆*ABC*. Also *AD*, *BE *and *CE* are its medians.

∴ *D*, *E* and *F* are the mid points of *BC*,* CA* and *AB* respectively.

We know, the centroid is a point on a median, dividing it in the ratio 2 : 1.

Considering the median *AD*, coordinates of *D* are

Let *G* be the centroid.

∴ Coordinates of the centroid are

Similarly, considering the other medians we find that in each the coordinates of *G* are

*i*.*e*., The coordinates of the centroid are

*ABCD* is a rectangle formed by the points *A*(-1, -1), *B*(-1, 4), *C*(5, 4) and *D*(5, -1). *P*, *Q*, *R* and *S* are the mid points of *AB*, *BC*, *CD* and *DA* respectively. Is the quadrilateral *PQRS* is square? a rectangle? or a rhombus? Justify your answer.

**Soln. :** We have a rectangle whose vertices are *A*(-1, -1), *B*(-1, 4), *C*(5, 4) and *D*(5, -1).

∵ *P* is mid-point of *AB*

∴ Coordinates of *P* are

Similarly, Coordinates of *Q* are

Coordinates of *R* are

Coordinates of *S* are

Now,

We see that *PQ* = *QR* = *RS *= *SP* *i*.*e*., all sides of *PQRS* are equal.

∴ It can be a square or a rhombus.

But its diagonals are not equal.
*i*.*e*., *PR* ≠ *QS*

∴ *PQRS* is a rhombus.