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Co-ordinate Geometry - NCERT Questions
Q 1.

Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b)(-a, -b)

SOLUTION:

Soln. : (i) Here x1 = 2, y1 = 3 and x2 = 4 and y2 = 1
∴ The required distance
ncert
ncert
(ii) Here x1 = -5, y1 = 7 and x2 = -1, y2 = 3
∴ The required distance
ncert
ncert
(iii) Here x1 = a, y1 = b and x2 = -a, y2 = -b
∴ The required distance
ncert
ncert

Q 2.

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 : 'Atown B is located 36 km east and 15 km North of the town A'.

SOLUTION:

Soln. : Part-I
Let the points be A(0, 0) and B(36, 15)
ncert
ncert
Part-II
We have A(0, 0) and B(36, 15) as the positions of two towns
ncert
Here x1 = 0, x2 = 36 and y1 = 0, y2 = 15
ncert
ncert.

Q 3.

Determine if the points (1, 5), (2, 3) and
(-2, -11) are collinear.

SOLUTION:

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
ncert
ncert
ncert
ncert
ncert
ncert
But AB + BC AC, AC + CBAB, BA + ACBC
A, B and C are not collinear.

Q 4.

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).
ncert
ncert
ncert
ncert
ncert
ncert
We have AB = BC AC.
∴ ∆ABC is an isosceles triangle.

Q 5.

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.
ncert

SOLUTION:

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)
ncert
ncert
ncert
ncert
ncert
ncert
ncert
ncert
Since, AB = BC = CD = AD i.e., All the four sides are equal.
ncert
ncert
ncert
i.e., BD = AC ⇒ Both the diagonals are also equal.
ABCD is a square.
Thus, Champa is correct.

Q 6.

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)

SOLUTION:

Soln. : (i) Let the points be A(-1, -2), B(1, 0),
C(-1, 2) and D(-3, 0) of a quadrilateral ABCD.
ncert
ncert
ncert
ncert
ncert
ncert
ncert
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).
ncert
ncert
ncert
ncert
ncert
ncert
We see that ncert
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).
ncert
ncert
ncert
ncert
ncert
ncert
Since, AB = CD, BC = DA [opposite sides of the quadrilateral are equal]
And ACBD ⇒ Diagonals are unequal.
ABCD is a parallelogram.

Q 7.

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

SOLUTION:

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)
ncert
ncert
ncert
ncert
Since, A and B are equidistant from P,
AP = BP
ncert
x2 - 4x + 29 = x2 + 4x + 85
x2 - 4x - x2 - 4x = 85 - 29
⇒ -8x = 56 ncert
∴ The required point is (-7, 0)

Q 8.

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

SOLUTION:

Soln. : The given points are P(2, -3) and Q(10, y).
ncert
ncert
But PQ = 10
ncert
Squaring both sides, y2 + 6y + 73 = 100
y2 + 6y - 27 = 0
y2 - 3y + 9y - 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.

Q 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.

SOLUTION:

Soln. : ncert
ncert
ncert ncert
QP = QR
ncert
Squaring both sides, we have x2 + 25 = 41
x2 + 25 - 41 = 0
x2 - 16 = 0 ncert
Thus, the point R is (4, 6) or (-4, 6)
Now,
ncert
and ncert
ncert
ncert
ncert

Q 10.

Find a relation between x and y such that the point (x, y) is equidistant from the point
(3, 6) and (-3, 4).

SOLUTION:

Soln. : Let the points be A(x, y), B(3, 6) and C(-3, 4).
ncert
ncert
Since, the point (x, y) is equidistant from (3, 6) and (-3, 4).
AB = AC
ncert
Squaring both sides,
(3 - x)2 + (6 - y)2 = (-3 - x)2 + (4 - y)2
⇒ (9 + x2 - 6x) + (36 + y2 - 12y)
= (9 + x2 + 6x) + (16 + y2 - 8y)
⇒ 9 + x2 - 6x + 36 + y2 - 12y - 9 - x2 - 6x
- 16 - y2 + 8y = 0
⇒ - 6x - 6x + 36 - 12y - 16 + 8y = 0
⇒ - 12x - 4y + 20 = 0
⇒ - 3x - y + 5 = 0
⇒ 3x + y - 5 = 0 which is the required relation between x and y.

Q 11.

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 = m1 : m2
ncert
ncert
ncert
ncert
Thus, the required point is (1, 3).

Q 12.

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).
ncert
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)
ncert
ncert
ncert
ncert
∴ The required co-ordinates of P are ncert
Let the co-ordinates of Q be (X, Y)
ncert
ncert
ncert
ncert
∴ The required coordinates of Q are ncert

Q 13.

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 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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 ?.

SOLUTION:

Soln. : Let us consider 'A' as origin, then
ncert
AB is the x-axis.
AD is the y-axis.
Now, the position of green flag-post is ncert
And, the position of red flag-post is ncert
⇒ Distance between both the flags
ncert
ncert
Let the mid-point of the line segment joining the two flags be M(x, y).
ncert
ncert
or x = 5 and y = 22.5
Thus, the blue flag is on the 5th line at a distance 22.5 m above AB.

Q 14.

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by
(-1, 6).

SOLUTION:

Soln. : Let the given points are A(-3, 10) and
B(6, -8).
Let the point P(-1, 6) divides AB in the ratio m1 : m2.
∴ Using the section formula, we have
ncert
ncert
ncert
ncert
⇒ -1(m1 + m2) = 6m1 - 3m2
and 6(m1 + m2) = - 8m1 + 10m2
⇒ -m1 - m2 - 6m1 + 3m2 = 0
and 6m1 + 6m2 + 8m1 - 10m2 = 0
⇒ -7m1 + 2m2 = 0
and 14m1 - 4m2 = 0 or 7m1 - 2m2 = 0
⇒ 2m2 = 7m1 and 7m1 = 2m2
ncert
m1 : m2 = 2 : 7 and m1 : m2 = 2 : 7
Thus, the required ratio is 2 : 7.

Q 15.

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.

SOLUTION:

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.
ncert and 0 ncert
ncert
ncert
x(k + 1) = - 4k + 1
and 5k - 5 = 0 ⇒ k = 1
x(k + 1) = - 4k + 1
x(1 + 1) = - 4 + 1 [∴ k = 1]
⇒ 2x = -3
ncert
∴ The required ratio k : 1 = 1 : 1
Coordinates of P are (x, 0) ncert

Q 16.

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

SOLUTION:

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 :
ncert ncert
x + 1 = 7 ⇒ x = 6
ncert
⇒ 5 + y = 8 ⇒ y = 3
∴ The required values of x and y are : x = 6, y = 3.

Q 17.

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)

SOLUTION:

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)
ncert
The centre of a circle bisects the diameter.
ncert
And ncert
Here the coordinates of A are (3, -10)

Q 18.

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

SOLUTION:

Soln. : ncert
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
ncert
AB = AP + BP
ncert
ncertncert
AP : PB = 3 : 4 i.e., P(x, y) divides AB in the ratio 3 : 4.
ncert
ncert
Thus, the coordinates of P are ncert

Q 19.

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

SOLUTION:

Soln. : Here the given points are A(-2, 2) and
B(2, 8)
Let P1, P2 and P3 divide AB in four equal parts.
ncert
AP1 = P1P2 = P2P3 = P3B
Obviously, P2 is the mid-point of AB
∴ Coordinates of P2 are
ncert
Again, P1 is the mid-point of AP2.
∴ Coordinates of P1 are
ncert
Also P3 is the mid-point of P2B.
∴ Coordinates of P3 are
ncert
Thus, the coordinates of P1, P2 and P3 are ncertrespectively.

Q 20.

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 = ncert (product of its diagonals)]

SOLUTION:

Soln. : Let the vertices of the given rhombus are : A(3, 0), (4, 5), (-1, 4) and (-2, -1)
ncert
AC and BD are the diagonals of rhombus ABCD.
∴ Diagonal
ncert
ncert
Diagonal ncert
ncert
∴ For a rhombus,
Area ncert × (Product of diagonals)
ncert
ncert
= 4 × 6 = 24 square units.

Q 21.

Find the area of the triangle whose vertices are :
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)

SOLUTION:

Soln. : (i) Let the vertices of the triangles be A(2, 3), B(-1, 0) and C(2, - 4)
Here x1 = 2, y1 = 3,
x2 = -1, y2 = 0
x3 = 2, y3 = -4
∴ Area of a ∆
ncert
∴ Area of a ∆ABC
ncert
ncert
ncert
(ii) Let the vertices of the triangles be
A(-5, -1), B(3, -5) and C(5, 2)
i.e., x1 = -5, y1 = -1
x2 = 3, y2 = -5
x3 = 5, y3 = 2
∵ Area of a
ncert
∴ Area of a ∆ABC
ncert
ncert
ncert
ncert

Q 22.

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)

SOLUTION:

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 - 7k + 5k + 10 + (-6) - 3 = 0
⇒ 17 - 9 + 5k - 7k = 0
⇒ 8 - 2k = 0 ⇒ 2k = 8 ⇒ ncert
The required value of k = 4.
(ii) ncert
⇒ 8 - 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

Q 23.

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
ncert
Coordinates of E are ncert
Coordinates of F are ncert
ncert
Now, ar(∆ABC)
ncert
ncert
ncert
Now, ar(∆DEF) ncert
ncert
ncert
ncert
∴ ar(∆DEF) : ar(∆ABC) = 1 : 4.

Q 24.

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.
ncert
Now, ar(∆ABD)
ncert
ncert
ncert
Now, ar(∆CBD)
ncert
ncert
ncert
ncert
Since, ar(quad ABCD) = ar(∆ABD) + ar∆CBD
ncert
ncert

Q 25.

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).

SOLUTION:

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
ncert
Since AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD,
ncert
Now, ar(∆ABD)
ncert
ncert ncert
= 3 sq.units (numerically) ........ (1)
ar(∆ADC)ncert
ncertncert
= 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.

Q 26.

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

SOLUTION:

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
2x + y - 4 = 0,
∴ We have
⇒ 2(3k + 2) + (7k - 2) = 4 × (k + 1)
⇒ 6k + 4 + 7k - 4k - 4 - 2 = 0
⇒ (6 + 7 - 4)k + (-2) = 0 ⇒ 9k - 2 = 0
ncert
∴ The required ratio = k : 1

Q 27.

Find a relation between x and y if the points
(x, y), (1, 2) and (7, 0) are collinear.

SOLUTION:

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 ncert [x(2 - 0) + 1 (0 - y) + 7(y - 2) = 0
or if 2x - y + 7y - 14 = 0
or if 2x + 6y - 14 = 0 or if x + 3y - 7 = 0 which is the required relation between x and y.

Q 28.

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 AP2 = BP2
⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
x2 - 12x + 36 + y2 + 12y + 36
= x2 - 6x + 9 + y2 + 14y + 49
⇒ - 12x + 6x + 12y - 14y + 72 - 58 = 0
⇒ - 6x - 2y + 14 = 0
⇒ 3x + y - 7 = 0 ..... (1) [Dividing by (-2)]
Taking BP = CP, we have BP2 = CP2
⇒ (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
x2 - 6x + 9 + y2 + 14y + 49
= x2 - 6x + 9 + y2 - 6y + 9
⇒ - 6x + 6x + 14y + 6y + 58 - 18 = 0
⇒ 20y + 40 = 0
ncert ........ (2)
From (1) and (2),
3x - 2 - 7 = 0
⇒ 3x = 9 ⇒ x = 3
i.e., x = 3 and y = -2
∴ The required centre is (3, -2).

Q 29.

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.
ncert
Since all sides of a square are equal,
AB = BCAB2 = BC2
⇒ (x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1 ........ (1)
Since each angle of a square = 90°,
ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have :
AB2 + BC2 = AC2
⇒ [(x + 1)2 + (y - 2)2] + [(x - 3)2 + (y - 2)2]
= [(3 + 1)2 + (2 - 2)2]
⇒ 2x2 + 2y2 + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x2 + 2y2 - 4x - 8y + 2 = 0
x2 + y2 - 2x - 4y + 1 = 0 ........ (2)
Substituting the value of x from (1) into (2) we have
1 + y2 - 2 - 4y + 1 = 0
y2 - 4y + 2 - 2 = 0
y2 - 4y = 0 ⇒ y(y - 4) = 0
y = 0 or y = 4
Hence, the two required other vertices are
(1, 0) and (1, 4).

Q 30.

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 ?
ncert

SOLUTION:

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]
ncertsq. units
ncert
[taking numerical value]
ar(∆PQR) [when P(-12, -2), Q(-13, -6) and R(-10, -3) are the vertices]
ncert
ncert
ncert
Thus, in both cases, the area of ∆PQR is the same.

Q 31.

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 ncert 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"]

SOLUTION:

Soln. : We have
ncert
ncert
ncert
ncert
ncert
AD : DB = 1 : 3
Thus, the point D divides AB in the ratio
1 : 3
∴ The coordinates of D are :
ncert
ncert
Similarly, AE : EC = 1 : 3 i.e., E divides AC in the ratio 1 : 3
⇒ Coordinates of E are
ncert
ncert
Now, ar(∆ADE)
ncert
ncert
ncert
Area of ∆ABC ncert
ncert
ncert
ncert
⇒ ar(∆ADE) : ar(∆ABC) = 1 : 16.

Q 32.

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(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.

SOLUTION:

Soln. : We have the vertices of ∆ABC as A(4, 2),
B(6, 5) and C(1, 4).
ncert
(i) Since AD is a median
∴ Coordinates of D are
ncert
(ii) Since AP : PD = 2 : 1 i.e., P divides AD in the ratio 2 : 1
∴ Coordinates of P are
ncert
(iii) BQ : QE = 2 : 1 ⇒ [The point Q divides BE in the ratio 2 : 1]
[coordinates of E are ncert
∴ Coordinates of Q are
ncert
ncert
Also, CR : RF = 2 : 1
⇒ The point R divides CF in the ratio 2 : 1
∴ Coordinates of F are ncert
So, Coordinates of R are
ncert
ncert
ncert
(iv) We observe that P, Q and R represent the same point.
(v) Here, we have A(x1, y1), B(x2, y2) and
C(x3, y3) 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 ncert
Let G be the centroid.
∴ Coordinates of the centroid are
ncert
ncert
Similarly, considering the other medians we find that in each the coordinates of G are
ncert
i.e., The coordinates of the centroid are
ncert

Q 33.

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.

SOLUTION:

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
ncert
∴ Coordinates of P are
ncert
Similarly, Coordinates of Q are
ncert
Coordinates of R are ncert
Coordinates of S are ncert
Now, ncert
ncert
ncert
ncert
ncert
ncert
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., PRQS
PQRS is a rhombus.