Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula Exercise 13B Solutions
Section and Mid-Point Formula
Exercise 13B
(Q1) Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3) and (-1, 7)
Solution:
Co-ordinate of P = (x1 + x2/2, y1 + y2/2)
If we have to find mid-point then we have to use this formula.
(i)
Let A (-6, 7) and B (3, 5) be the co-ordinate at P.
A (-6, 7) = (x1, y1) and B (3, 5) = (x2, y2)
Now, we have to find co-ordinate at P = (x1 + x2/2, y1 + y2/2)
= (-6+3/2, 7+5/2)
= (-3/2, 12/2)
= (-3/2, 6)
(ii)
Let A (5, -3) and B (-1, 7) be the co-ordinate at P
A (5, -3) = (x1, y1) and B (-1, 7) = (x2, y2)
Now, we have to find co-ordinate at P (x1 + x2/2, y1 + y2/2)
P = (5 + (-1)/2, -3+7/2)
= (5-1/2, -3+7/2)
= (4/2, 4/2)
P = (2, 2)
(Q2) Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Solution:
Given that, the co-ordinate of A is (3, 5) and B is (x, y)
∴ P is the mid-point of co-ordinate A and B.
∴ P = (2, 3)
A (3, 5) = (x1, y1) and B (x, y) = (x2, y2)
Now, we have to find the value of (x, y) –
By using mid-point formula,
P = (x1 + x2/2, y1 + y2/2)
(2, 3) = (3+x/2, 5+y/2)
Equating the both sides,
2 = 3+x/2
2×2 = 3+x
4 = 3+x
4-3 = x
1 = x
∴ x = 1
3 = 5+y/2
3×2 = 5+y
6 = 5+y
6-5 = y
1 = y
∴ y = 1
∴ The co-ordinate of B (x, y) = (1, 1)
(Q3) A (5, 3), B (-1, 1) and C (7, -3) are the vertices of triangles ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = 1/2 BC.
Solution:
Given that A, B and C be the vertices of triangle ABC.
A (5, 3), B (-1, 1) and C (7, -3)
L is the mid-point of AB ad M is the mid-point of AC
To show: LM = ½ BC
(i) L is the mid-point of AB.
By mid-point formula,
A (5, 3) = (x1, y1) and B (-1, 1) = (x2, y2)
L = (x1 + x2/2, y1 + y2/2)
= (5+ (-1)/2, 3+1/2)
L = (5-1/2, 4/2)
= (4/2, 4/2)
L = (2, 2)
(ii) M is the mid-point of AC
By mid point formula,
A (5, 3) = (x1, y1)
B = (7, -3) = (x2, y2)
M = (x1 + x2)/2, y1 + y2/2)
= (5+7/2, 3 + (-3)/2)
= (12/2, 3-3/2)
M = (6, 0)
Now, we have to find distance of BC, so we have to use distance formula,-
BC = √(x2 – x1)2 + (y2 – y1)2
B (-1, 1) = (x1, y1) and C (7, -3) = (x2, y2)
BC = √(7 – (-1)2 + (-3- 1)2
= √(7 + 1)2 + (-4)2
= √(8)2 + (-4)2
BC = √64 + 16
BC = √80
BC = √16×5
BC = 4√5 —— (i)
After we have to find distance of LM.
LM = √(x2 – x1)2 + (y2 – y1)2
L = (2, 2) = (x1, y1) and
M (6, 0) = (x2, y2)
LM = √(6 – 2)2 + (0 – 2)2
= √(4)2 + (-2)2
= √16 + 4
= √20
= √4×5
LM = 2√5 —– (ii)
From (i) and (2), we can see that, the length of LM is half of the length of BC –
LM = ½ BC (Hence Proved)
(Q4) Given M is the mid-point of AB. Find the co-ordinate of:
(i) A; if M = (1, 7) and B = (-5, 10)
(ii) B; if A = (3, -1) and M = (-1, 3)
Solution:
(i) Given that, M is the mid-point of AB –
M (1, 7) and B (-5, 10)
Let A = (x, y)
We have to find the co-ordinate of A (x, y) –
By using mid-point formula,
M = (x1 + x2/2, y1 + y2/2)
A (x, y) = (x1, y1) and B (-5, 10) = (x2, y2)
M = (x + (-5)/2, y + 10/2)
(1, 7) = (x – 5/2, y + 10/2)
Equating both sides,
1 = x-5/2
2 × 1 = x – 5
2 = x – 5
2+5 = x
x = 7
7 = y+10/2
7×2 = y+10
14 = y + 10
14 – 10 = y
4 = y
∴ y = 4
∴ The co-ordinate of A = (7, 4)
(ii)
Given that, the M is the mid-point of co-ordinates A and B.
∴ A (3, -1) = (x1, y1) and Let B (x, y) = (x2, y2)
We have to find the co-ordinates of B –
By using mid-point formula,
= (x1 + x2/2, y1 + y2/2)
(-1, 3) = (3 + x/2, (y) + y/2)
(-1, 3) = (3+x/2, -1+y/2)
Equating both sides,
-1 = 3+x/2
2× (-1) = 3+x
-2 = 3+x
-2-3 = x
-5 = x
x = -5
3 = -1+y/2
3×2 = -1+y
6 = -1+y
6+1 = y
7 = y
∴ y = 7
The co-ordinate of B is (-5, 7)
(Q5) P (-3, 2) is the mid-point of line segment AB as shown in the give figure, find the co-ordinate of points A and B
Solution:-
Given that P is the mid-point of line segment AB.
∴ B is on x-axis so, the co-ordinate of B is (x, 0) and A is on y-axis so, the co-ordinate of A is (o, y)
Now, we have to find co-ordinate of A –
B (x, 0) = (x1, y1) and A (0, y) = (x2, y2)
By using mid-point formula,
P = (x1 + x2/2, y1 + y2/2)
(-3, 2) = (x+0/2, 0+y/2)
(-3, 2) = (x/2, y/2)
Equating both sides,
-3 = x/2
-3×2 = x
-6 = x
∴ x = -6
2 = y/2
2×2 = y
y = 4
∴ The co-ordinate of ( )
A (o, y) = (0, 4) and B (x, 0) = (-6, 0)
(Q6) In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.
=
Solution:-
Given that, P is the mid-point of line segment AB.
A (x, 0) = (x1, y1) and B (o, y) = (x2, y2)
We have to find the co-ordinate of A and B –
By using mid-point formula,
P = (x1 + x2/2, y1 + y2/2)
(4, 2) = (x + 0/2, 0 + y/2)
(4, 2) = (x/2, y/2)
Equating both sides,
4 = x/2
4×2 = x
x = 8
And 2 = y
2×2 = y/2
4 = y
y = 4
(Q7) (-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6)
Solution:
Let A (+3, -6), B (-5, 2) and C (7, 4)
Given that AD is the median.
So, D is a mid-point of AC.
To find = The length of median AD
By using the distance formula,
First we have to find the co-ordinate of mid-point D.
So, by using mid-point formula,
D = (x1 + x2/2, y1 + y2/2)
B (-5, 2) = (x1, y1) and C (7, 4) = (x2, y2)
D = (-5+7/, 2+4/2)
D = (2/2, 6/2)
D = (1, 3) = (x2, y2)
A = (3, -6) = (x1, y1)
Now, we have to find the length of median AD –
AD = √(x2 – x1)2 + (y2 – y1)2
= √(1 – 3)2 + (3 – (-6)2
= √(-2)2 + (3 + 6)2
= √4 + (9)2
= √4 + 81
AD = √85
AD = 9.22
(Q8) Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
Solution:
Given that, AB = BC = CD
Let A = (x, y), B = (0, 3), C = (1, 8) and D = (x1, y1)
We have to find, the co-ordinate of A –
B is the mid-point of co-ordinates A and C.
B = (x1 + x2/2, y1 + y2/2)
(0, 3) = (x1 + x2/2, y1 + y2/2)
A (x, y) = (x1, y1 and C (1, 8) = (x2, y2)
(0, 3) = (x+1/2, y+3/2)
Equating both sides,
0 = x+1/2
2×0 = x+1
0 = x+1
x = -1 and
3 = y+3/2
3×2 = y+3
6 = y+3
6-3= y
3 = y
∴ y = 3
∴ The co-ordinates of A is (-1, 3)
Now, we have to find the co-ordinates of D –
C is the mid-point of the co-ordinate B and D.
B (0, 3) = (x1, y1) and D (x1, y1) = (x2, y2)
By mid-point formula,
C = (x1 + x2/2, y1 + y2/2)
(1, 8) = (0 + x1/2, 3 + y1/2)
Equating both sides,
1 = x1/2
1×2 = x1
x1 = 2 and
8 = 3+y1/2
8×2 = 3+y1
16 = 3+y1
16 – 3 = y1
13 = y1
∴ y1 = 13
∴ The co-ordinates of D is (2, 13)
(Q9) One end of the diameter of a circle is (-2, 5) Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1)
Solution:
Let AB is the diameter of a circle.
Let A (-2, 5) and B (x, y)
Let P is the mid-point of diameter of AB.
Now, we have to find the co-ordinate of B (x, y)
By mid-point formula –
A (-2, 5) = (x1, y1) and B (x, y) = (x2, y2)
P = (x1 + x2/2, y1 + y2/2)
(2, -1) = (-2+x/2, 5+y/2)
Equating both sides,
2 = -2+x/2
2×2 = -2+x
4 = – 2 +x
4+2 = x
6 = x
∴ x = 6 and
5+y/2 = -1
5+y = -1×2
5+y = -2
y = -2-5
y = -7
∴ The co-ordinates of the other end is B (x, y) = (6, -7) or B (6, -7)
Here is your solution of Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula
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