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Question-1

Find the centroid of the triangle whose vertices are : (3, -5), (-7, 4), (10, -2).

Solution:
The centroid of the triangle whose vertices are (3, -5), (-7, 4), (10, -2) is
∵ The centroid of the triangle ABC =  
 ∴ The centroid of the triangle whose vertices A(3, -5), B(-7, 4), C(10, -2) = 
                                                                                                      = (2, -1).

Question-2

Find the centroid of the triangle whose vertices are : (2, 1), (5, 2), (3, 4).

Solution:
∵ The centroid of the triangle ABC =  
 ∴ The centroid of the triangle whose vertices (2, 1), (5, 2), (3, 4) = = (10/3, 7/3).

Question-3

Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.

Solution:
Given, the two vertices of the triangle are (-3, 1), (0, -2).
Let the third vertex be (x, y)
Also given the centroid of the triangle = (0, 0) 
∵ The centroid of the triangle =   
 ∴ 
⇒ -3 + x = 0, -1 + y = 0
∴ x = 3, y = 1

Question-4

Prove that the diagonals of a rectangle bisect each other and are equal. (Hint: With O as origin, let the vertices of the rectangle be (0, 0), (a, 0), (a, b) and (0, b)).

Solution:


AC and OB are diagonals

In the figure let the intersecting point of OB and AC be P
To show that diagonals bisect each other we have to prove that OP = PB and PA = PC
The co-ordinates of P is obtained by


P is the point

OP = PB
Similarly we can prove that PC = PA
Thus diagonals bisect each other in a rectangle.


∴ The diagonals of a rectangle bisects each other and equal.
 

Question-5

Show that the points A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a parallelogram. (Hint: Diagonals of a parallelogram bisect each other).

Solution:
The midpoint of diagonal AC is =
The midpoint of diagonal BD is  
The diagonals AC and BD bisect each other ABCD is a parallelogram.

Question-6

In what ratio does the y -axis divide the line segment joining the points P(-4, 5) and Q(3, 7)?

Solution:
Let the required ratio be k : 1. Then, the coordinates of the point of division are,
But, it is a point on y-axis on which x-coordinate of every point is zero.
Therefore = 0
3k - 4 = 0
k = 4/3
Thus, the required ratio is k = 4/3 or 4 : 3.

Question-7

Find the cirumcentre of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).

Solution:
Let the centre of the circle be O(x, y). The points are A(-2, -3), B(-1, 0) and C(7, -6).

OA =

OB =
OC =
OA = OB = OC = radius
=

(2 + x)2 + (3 + y)2 = (1 + x)2 + y2
4 + 4x + x2 + 9 + 6y + y2 = 1 + 2x + x2 + y2
4 + 4x + 9 + 6y = 1 + 2x

2x + 6y = -12
x + 3y = -6 ………………………..(i)

=
(1 + x)2 + y2 = (7 - x)2 + (6 + y)2
1 + 2x + x2 + y2 = 49 - 14x + x2 + 36 + 12y + y2
1 + 2x = 49 - 14x + 36 + 12y

16x – 12y = 84
4x – 3y = 21 ………………………..(ii)

Solving (i) and (ii)
x +  3y = -6
4x – 3y = 21
5x        = 15

x = 3

Substituting x = 3 in (i)
4(3) – 3y = 21

12 – 3y = 21
-3y = 9
y = -3
Therefore the centre of the circle is (3, -3).

Question-8

The three vertices of a parallelogram are (1, 1), (4, 4) and (4, 8). Find the fourth vertex.

Solution:
Let A(1, 1), B(4, 4), C(4, 8) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.
Therefore coordinates of the mid-point of AC = Coordinates of the mid-point of BD
=
=
4 + x = 5, 8 + y = 5
x = 1, y = -3
Therefore (1, -3) is the fourth vertex.

Question-9

Show that the points (2, 1), (5, 2), (6, 4) and (3, 3) are the angular points of a parallelogram. Is the figure a rectangle?

Solution:
Let A(2, 1), B(5, 2), C(6, 4) and D(3, 3) be the vertices of a parallelogram ABCD. Since, the diagonals of a parallelogram bisect each other.
AC2 = (6 - 2)2 + (4 - 1)2 = (4)2 + (3)2 = 16 + 9 = 25
BC2 = (6 - 5)2 + (4 - 2)2 = (1)2 + (2)2 = 1 + 4 = 5
AB2 = (5 - 2)2 + (2 - 1)2 = (3)2 + (1)2 = 9 + 1 = 10
DC2 = (6 - 3)2 + (4 - 3)2 = (3)2 + (1)2 = 9 + 1 = 10
AD2 = (3 - 2)2 + (3 - 1)2 = (1)2 + (2)2 = 1 + 4 = 5
Since BC = AD and DC = AB, ABCD is a parallelogram.
AB2 + BC2 = 10 + 5 = 15
AB2 + BC2
AC2
∴ ΔABC is not right angled. Therefore parallelogram ABCD is not a rectangle.

Question-10

Find the third vertex of a triangle, if two of its vertices are (-3, 1) and (0, -2) and the centroid is at the origin.

Solution:
Let the third vertex of the triangle be C(x, y) and the other vertices A(-3, 1) and B(0, -2).
Coordiantes of the centroid of the triangle = (0, 0)

= (0, 0)
-3 + x = 0
x = 3
And –1 + y = 0
y = 1

The third vertex of the triangle is (3, 1).

Question-11

If the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3, 4), find its vertices.

Solution:
Let the vertices of the triangle be A(x1, y1), B(x2, y2) and C(x3, y3). Let the midpoints of the sides of a triangle are
D(1, 1), E(2, -3) and F(3, 4).

= (1, 1)
 

x1 + x2 = 2 ………………………..(i)
y1 + y2 = 2 ………………………..(ii)

= (2, -3)
x2 + x3 = 4 ………………………..(iii)
y2 + y3 = -6
 ………………………..(iv)
= (3, 4)
x3 + x1 = 6 ………………………..(v)
y3 + y1 = 8 ………………………..(vi)

Add the equations (i), (iii), (v) we get,
2(x1 + x2+ x3) = 12
⇒ x1 + x2+ x3 = 6...............(vii)
Substitute eqn.(i) in (vii) then x3 = 4.
Substitute eqn.(iii) in (vii) then x1 = 2. 
Substitute eqn.(v) in (vii) then x2 = 0. 
 
Add the equations (ii), (iv), (vi) we get,
2(y1 + y2+ y3) = 4
y1 + y2+ y3  = 2...............(viii)
Substitute eqn.(ii) in (viii) then y3 = 0.
Substitute eqn.(iv) in (viii) then y1 = 8. 
Substitute eqn.(vi) in (viii) then y2 = -6. 
The vertices A(x1, y1), B(x2, y2) and C(x3, y3) = A(2, 8), B(0, -6) and C(4, 0)
 

Question-12

Find the ratio in which the line points (6, 4) and (1, - 7) is divided internally by the axis of x.

Solution:
Let the required ratio be k : 1. Then, the coordinates of the point of division are,
But, it is a point on y-axis on which x-coordinate of every point is zero.
Therefore = 0
7k - 4 = 0
k = 4/7
Thus, the required ratio is k = 4/7 or 4 : 7.

Question-13

If the points (-2, -1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the values of x and y.

Solution:
Let the vertices of the parallelogram be A(-2, -1), B(1, 0), C(x, 3) and D(1, y). Since the diagonals of a parallelogram bisect each other the coordinates of the mid-point of AC = coordinates of the mid-point of BD.
 
 =
 
 
 

 x = 4, y = 2.

Question-14

If the mid-points of the sides of a triangle PQR are A(-1, -3), B(2, 1) and C(4, 5), find the coordinates of P, Q and R.

Solution:
Coordinate of mid point P is =
Coordinate of mid point Q is =(3, 3)
Coordinate of mid point R is
=
The coordinates of the P, Q(3, 3) and R  .

Question-15

Three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3). Find the fourth vertex. 

Solution:
Let A(-2, -1), B(1, 0), C(4, 3) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.

Therefore coordinates of the mid-point of AC = Coordinates of the mid-point of BD

=
(1, 1)=
x + 1 = 2, y = 2
Therefore (1, 2) is the fourth vertex.

Question-16

Determine the ratio in which 2x + 3y - 30 = 0 divides the line segment joining A (3, 4) and B (7, 8) and the point at which it divides.

Solution:
Let the P(a, b) be the point which divides the line segment joining A (3, 4) and B (7, 8) in the ratio k : 1.

Then coordinates of the point P is .
This point lies on the line 2x + 3y - 30 = 0.


2 + 3 - 30 = 0.
2(7k + 3) + 3(8k + 4) - 30(k + 1) = 0.
14k + 6 + 24k + 12 - 30k - 30 = 0.
8k - 12 = 0.
k = 3/2

The required ratio is 3 : 2.
The coordinates of the point P is =.

Question-17

Prove that the points (2a, 4a), (2a, 6a), (2a +a, 5a) are the vertices of an equilateral triangle.

Solution:
Let A(2a, 4a), B(2a, 6a) and C(2a +a, 5a) be the vertices of an equilateral triangle.
AB = == 2a
BC = == 2a
CA = == = 2a

AB = BC = CA.
 the vertices are of an equilateral triangle.

Question-18

The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a Δ ABC right –angled at A. Find the value of a.

Solution:
Given, the vertices of a Δ ABC are right –angled at A.
∴ AB2 + AC2 = BC2
AB2 = (- 2 - 0)2 + (a - 3)2 = 4 + (a – 3)2
BC2 = (- 1 + 2)2 + (4 - a)2 = 1 + (4 – a)2
AC2 = (- 1 - 0)2 + (4 - 3)2 = 1 + 1 = 2

Since, AB2 + AC2 = BC2
4 + (a – 3)2 + 2 = 1 + (4 – a)2
4 + a2 + 9 – 6a + 2 = 1 + 16 + a2 – 8a
2a = 2

a = 1.

Question-19

The points A(2, 0), B(9, 1), C(11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Solution:
AB = ==

BC = ===

CD = ==

DA = ==

No it is not a rhombus as all sides are not equal.


Question-20

The vertices of a triangle are A(3, 4) , B(7, 2)and C(-2, -5). Find the length of the median through the vertex A.

Solution:
Let D the mid point of BC. Then the coordinate of D is =.
Length of the median is AD =

                                        =
                                        =
                                        =.


Question-21

Two vertices of a triangle are (3, -5) and (-7, 4). If its centroid is (2, -1), find the third vertex.

Solution:
Let P(3, -5) and Q(-7, 4) be the two vertices of a triangle. Centroid of a triangle is G(2, -1).

The coordinates of the centroid of
Δ ABC is = (2, -1)
= (2, -1)

x1 = 6 + 4, y1 = - 3 + 1

x1 = 10, y1 = - 2
 The third vertex is (10, -2).

Question-22

Are the points (-2, 2), (8, -2) and (-4, 3) are the vertices of a right angled triangle.

Solution:
Let the points A(-2, 2), B(8, -2) and C(-4, 3) be the vertices of a triangle ABC.
AB2 = (8 + 2)2 + (-2 - 2)2 = 102 + 42 = 100 + 16 = 116
BC2 = (-4 - 8)2 + (3 + 2)2 = (-12)2 + 52 = 144 + 25 = 169
AC2 = (-4 + 2)2 + (3 - 2)2 = (-2)2 + 1 = 4 + 1 = 5
AB2 + AC2
BC2
The above vertices are not points of a right angled triangle.

Question-23

If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.

Solution:
Let P(-2, 3), Q(4, -3), R(4, 5) be the mid-points of sides AB, BC and CA respectively of a triangle ABC. Let A(x1 , y1), B(x2 , y2) and C(x3, y3) be the vertices of triangle ABC. Then,
P is the midpoint of AB.

= -2, = 3
x1 + x2= - 4 and y1 + y2 = 6 …………………….(i)

Q is the midpoint of BC.

= 4, = -3
x2 + x3 = 8 and y2 + y3 = -6 …………………….(ii)

R is the midpoint of CA.

= 4, = 5
x1 + x3 = 8 and y1 + y3 = 10 …………………….(iii)

From (i), (ii) and (iii)
2(x1 + x2 + x3 ) = - 4 + 8 + 8 = 12
x1 + x2 + x3 = 6

2(y1 + y2 + y3 ) = 6 - 6 + 10 = 10
y1 + y2 + y3 = 5
Therefore the coordinates of the centroid of Δ ABC are or, (6/3, 5/3) = (2, 5/3).

Question-24

If the points (-1, 3), (1, -1) and (5, 1) are vertices of a triangle, find the length of the median through third vertex.

Solution:
Let the points A(-1, 3), B(1, -1) and C(5, 1) be the vertices of a triangle ABC. Let D be the mid point of AB. Then the coordinate of D is =(0, 1).
Length of the median is CD =
                                     =
                                     = 5.

Question-25

Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, -1) and C(4, -6).

Solution:
AB = ===
BC = = ==
AC = ===

Therefore the lengths of the sides of the triangle are , and .

Question-26

A line is of length 10 and one end is at the point (-3, 2). If the ordinate of the other end be 10, prove that the abscissa will be 3 or –9.

Solution:
The two points are (-3, 2) and (x, 10). Their length is 10.
= 10

         = 102
 
x2 + 6x + 9 + 64 = 100
         x2 + 6x – 27 = 0

  x2 + 9x – 3x – 27 = 0
x(x + 9) – 3(x + 9) = 0
        (x – 3)(x + 9) = 0
                          x = 3 or –9.
Therefore the required abscissa will be 3 or –9.


Question-27

Show that the points (-2, 6), (5, 3), (-1, -11) and (-8, -8) are the vertices of a rectangle.

Solution:
Let A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8) are the vertices of a rectangle.

(i) AB = ===

CD = ===
BC = ===

DA = ===

AB = CD and BC = DA.

(ii) AC = ===
     BD = ===
     AC = DB

Since the opposite sides and the diagonals are equal

∴ ABCD is a rectangle.

Question-28

If a point (x, y) is equidistant from (6, -1) and (2, 3), find the relation between x and y.

Solution:
Let the points be P(x, y), A(6, -1) and B(2, 3).
AP2 = (x - 6)2 + (y + 1)2
BP2 = (x - 2)2 + (y - 3)2
Given, 
(x, y) is equidistant from (6, -1) and (2, 3) 
(x - 6)2 + (y + 1)2 = (x - 2)2 + (y - 3)2
x2 – 12x + 36 + y2 + 2y + 1 = x2 – 4x + 4 + y2 – 6y + 9
– 12x + 36 + 2y + 1 = – 4x + 4 – 6y + 9

– 8x + 8y = -24
    – x + y = -3
       x – y = 3.

Question-29

Coordinates of A and B are (-3, a) and (1, a + 4). The mid-point of AB is (-1, 1). Find the value of a.

Solution:
Mid point of AB =

= (-1, 1)

      = (-1, 1)

               a + 2 = 1
                     a = -1.


Question-30

Find a point on the line through A(5, -4) and B(-3, 2), that is, twice as far from A as from B.

Solution:
Let the required point be P(x, y).
Then, AP = 2PB
AP/PB = 2/1
or AP : PB = 2 : 1
Therefore x = and y =

So, the required point is (-1/3, 0).

Question-31

Determine the ratio in which y – x + 2 = 0 divides the line joining (3, -1) and (8, 9).

Solution:
Let the required ratio be k : 1.
Then, the point of division is .

This point must lie on y – x + 2 = 0.
Therefore or k = .
So, the required ratio is :1

i.e. 2 : 3.

Question-32

If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.

Solution:
Let the points A(2, 1) and B(1, -2) be at equidistant from the point P(X, Y).

AP =

AB =

=

(x - 2)2 + (y - 1)2 = (x - 1)2 + (y + 2)2

x2 – 4x + 4 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 4y + 4
 
– 4x + 4 – 2y + 1 = - 2x + 1 + 4y + 4

– 2x – 6y = 0

x + 3y = 0.


Question-33

Find the ratio in which the point (2, y) divides the join of (-4, 3) and (6, 3) and hence find the value of y.

Solution:
Let the required ratio be k : 1.

Then, 2 =
k = .
Therefore the required ratio is : 1 i.e. 3 : 2.
Also, y = = 3.

Question-34

Prove that the diagonals of a rectangle bisect each other and are equal. (Hint: With O as origin, let the vertices of the rectangle be (0, 0), (a, 0), (a, b) and (0, b).

Solution:

ABCO is a rectangle with vertices A(a, 0), B(a, b), C(0, b) and O(0, 0).
The midpoint of AC is = (a/2, b/2)
The midpoint of OB is = (a/2, b/2)
Hence the diagonals bisect each other.
The length of the diagonal AC = =units
The length of the diagonal OB = =units
Hence the diagonals are equal.

Question-35

Show that the points A(1, 0), B(5, 3), C(2, 7) and D(-2, 4) are the vertices of a parallelogram. (Hint: Diagonals of a parallelogram bisect each other).

Solution:
The midpoint of diagonal AC is = (3/2,7/2)
The midpoint of diagonal BD is = (3/2,7/2)
The diagonals AC and BD bisect each other
ABCD is a parallelogram.

Question-36

If the distances of A(x, y) from P(a + b, b - a) and Q(a - b, a + b) are equal, prove that bx = ay.

Solution:
AP2 = (a + b - x)2 + (b - a - y)2
AQ2 = (a - b - x)2 + (a + b - y)2
AP = AQ (Given)

AP2 = AQ2
(a + b - x)2 + (b - a - y)2 = (a - b - x)2 + (a + b - y)2
a2 + b2 + x2 + 2ab – 2ax – 2bx + b2 + a2 + y2 – 2ba – 2by + 2ay = a2 + b2 + x2 – 2ab + 2bx – 2ax + a2 + b2 + y2 + 2ab – 2ay – 2by – 2bx + 2ay = 2bx – 2ay
4ay = 4bx
ay = bx
Hence proved.

Question-37

Prove that the points A (0, 1), B(1, 4), C(4, 3) and D(3, 0) are the vertices of a square.

Solution:
Let A (0, 1), B(1, 4), C(4, 3) and D(3, 0) be the four points.
AB = = =units
BC = ==units
CD = = =units
DA = ==units
Therefore the points A (0, 1), B(1, 4), C(4, 3) and D(3, 0) are the vertices of a square.

Question-38

Show that the points (1, 1), (-2, 7) and (3, -3) are collinear.

Solution:
Let A(1, 1), B(-2, 7) and C(3, -3) be the three points.
AB = = == 3units
BC = = = = 5units
AC = = == 2units
AB + AC = BC. Hence A, B, C are collinear.

Question-39

Find the ratio in which the points (2, 5) divides the line-segment joining the points (-1, 2) and (4, 7).

Solution:
Let A(-1, 2) and B(4, 7) be points. Let the point at which it is divided be C(2, 5).
AC2 = (2 + 1)2 + (5 - 2)2 = (3)2 + (3)2 = 9 + 9 = 18
BC2 = (2 - 4)2 + (5 - 7)2 = (-2)2 + (-2)2 = 4 + 4 = 8
AC/ BC = = 18/8 = 9/4
The ratio in which the line segment is divided is 9 : 4.

Question-40

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

Solution:
Let the centre of the circle be O(x, y). The points are A(5, -8), B(2, -9) and C(2, 1).
OA =
OB =
OC =
OA = OB = OC = radius
=
(5 - x)2 + (8 + y)2 = (2 - x)2 + (9 + y)2
25 – 10x + x2 + 64 + 16y + y2 = 4 - 4x + x2 + 81 + 18y + y2
89 – 10x + 16y = 85 - 4x + 18y
– 6x - 2y = - 4
3x + y = 2 ………………………..(i)
=
(2 - x)2 + (9 + y)2 = (2 - x)2 + (1 - y)2
4 – 4x + x2 + 81 + 18y + y2 = 4 - 4x + x2 + 1 - 2y + y2
81 + 18y = 1 - 2y
20y = -80
y = -4 ………………………..(ii)
Substituting y = -4 in (i)
3x - 4 = 2
3x = 6
x = 2
Therefore the centre of the circle is (2, -4).

Question-41

Find the distance between the origin and (6, 8).

Solution:
Distance between the points (x1, y1) and (x2, y2) =

           Distance between points (x, y) and (6, 8) =

                                                                        =
                                                                        =
                                                                        =
                                                                        = 10 units


Question-42

The centre of a circle is (-6, 4). If the diameter of the circle has its one end at the origin. Find its other end.

Solution:
Mid-point of the diameter is the centre of the circle. If one end of the diameter is the origin let the other end be (x, y)
centre of the circle = = = (-6, 4)
          Thus = (-6, 4)
Comparing we get, x = -12 and y = 8.
 The other end of the diameter is (-12, 8).

Question-43

Find the ratio in which the x -axis divides the line segment joining the points (3, -2) and (-7, -1).

Solution:
Any point on the x –axis will be of the form A(x, 0). Let this point divides the line segment joining (3, -2) and (-7, -1) in the ratio
m : n internally.
Thus the coordinate of A is where (3, -2) and (-7, -1) are (x1, y1) and (x2, y2) respectively.
Thus the coordinate of A = = (x, 0)
Here the y coordinates of A is zero.
Thus = 0. Hence –m – 2n = 0.
=> -m = 2n
=>.
=> m : n = 2 : -1 internally
Thus m : n = 2 : 1 externally.

Question-44

 If the points (1, 4), (r, -2) and (-3, 16) are collinear, find r.

Solution:
The condition for collinearity of three given points (x1, y1), (x2, y2) and (x3, y3) is
x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0.
Thus 1(-2 - 16) + r(16 - 4) + (-3)(4 - (-2)) = 0.
1(-18) + r(12) -3(6) = 0.
-18 + 12r – 18 = 0.
=>12r = 36
    r = 3.

Question-45

Find the fourth vertex of the parallelogram whose vertices are given by (1, 1),
(2, 3) and (2, -2) taken in order.
 

Solution:
If A(1, 1), B(2, 3) and C(2, -2) are the three points of the parallelogram. Let the fourth vertex be D(x,y).
Since ABCD is a parallelogram.
The mid point of the diagonal AC = The mid point of the diagonal BD.
The mid point of (x1, y1) and (x2, y2) is
The mid point of AC = ---------------(1)
The mid point of BD = -------------(2)
Since (1) = (2), =
                               3 = 2 + x and -1 = 3 + y
x = 1 and y = -4
Thus the fourth vertex is (1, -4).

Question-46

If the points (x, y) is collinear with the points (a, 0) and (0, b) prove that
= 1.

Solution:
Let the points be A(x, y), B(a, 0) and C(0, b). As the points are collinear.
Area of triangle ABC is zero.
The condition for collinearity of three given points (x1, y1), (x2, y2) and (x3, y3) is
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0.
   Thus, x(0 – b) + a(b – y) + 0(y – 0) = 0.
 
               => -bx + ab – ay = 0
          => bx + ay = ab
            => = 1.
                => = 1.

Question-47

ABC is a triangle whose vertices are A(2, -1), B(-4, 2) and C(2, 5). Find the length of the median AD.

Solution:
Let D be the mid point of BC.

The mid point of the line segment joining (x1, y1) and (x2, y2) is
Thus the coordinates of D = D = .

As A = (2, -1) and D =
AD =

     =
     =
     = units.

Question-48

The mid point of the line segment PQ is (5, 1). If P is (8, 4) find the coordinate of Q.

Solution:
Let the coordinate of Q be(x, y).

The mid point of the line segment joining (x1, y1) and (x2, y2) is .

The mid point of the line segment PQ is (5, 1) and the coordinate of P is (8,4).
Thus = (5, 1)
 
              = (5, 1)
Hence, = 5 and = 1
Thus 8 + x = 10 and 4 + y = 2. Thus x = 2, y = -2.
Hence the point Q is (2, -2).

Question-49

Calculate the mid point of A(2, 7) and B(10, 1).

Solution:
The mid point of the line segment joining (x1, y1) and (x2, y2) is .
The mid point of AB =
                            =  
                            = (6, 4)

Question-50

What is the distance of the point P(5, 12) from the origin?

Solution:
The distance between the points =
The distance between the points P(5, 12) and the origin is
                 =
                 = = 13 units.




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