Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula Exercise 13C Solutions
Section and Mid-Point Formula
Exercise 13C
(Q1) Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10) A point P lies on BC such that BP:PC = 3:2. Find the length of line segment AP.
Solution:
Point ‘P’ is divide the line segment BC in the ratio 3:2 = m:n
P (x, y) = (mx2 + nx1/m + n, my2 + ny1/m + n)
P (x, y) = (3× (x2) + 2x1/3+2, 3y2 + 2y1/3+2)
B = (0, 5) = (x1, y1) and
C (5, 10) = (x2, y2)
P (x, y) = (3× (5) + 2 (0)/3+2, 3× (10) + 2 × (5)/3+2)
= (15+0/5, 30+10/5)
= (15/3, 40/5)
P (x, y) = (5, 8)
Now, we have to find the length of AP –
By using distance formula,
AP = √(x2 – x1)2 + (y2 – y1)2
A (4, -4) = (x1, y1) and P (5, 8) = (x2, y2)
AP = √(5 – 4)2 + (8 – (-4))2
= √(1)2 + (8 + 4)2
= √1 + (12)2
= √1 + 144
AP = √145
AP = 12.04
(Q2) A (20, 0) and B (10, 120) are two fixed points. Find the co-ordinates of a point P in AB such that: 3PB = AB Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Solution:
P is the mid-point of AB.
Given that A (20, 0) and B (10, -20)
Also, 3PB = AB
PB/AB = 1/3
AB = AP + PB (∵ P is mid-point)
3 = AP + 1
3 – 1 = AP
2 = AP
∴ AP = 2
∴ AP:PB = 2:1
∴ m:n= 2:1
∴ P point P is divided the line segment AB in ratio 2:1
m = 2 and n = 1
P (mx2 + nx1/m+n, my2 + ny1/m+n)
A (20, 0) = (x1, y1) and B (10, -20) = (x2, y2)
= (2× (10) + 1×(20)/2+1, 2×(-20) + 1×(0)/2+1)
= (20+20/3, -40+0/3)
P (x, y) = (40/3, -40/3)
Also,
Q is the mid-point of AB
Now, we have to find the co-ordinates of Q.
Q is divided the line segment AB.
Q (x, y) = (mx2 + nx1/m+n, my2 + ny1/m+n)
Given: AB = 6AQ
AB/AQ = 6/1
∴ AB = AQ + QB (∵ Q is mid-point)
6 = 1 + QB
6 – 1 = QB
5 = QB
∴ QB = 5
AQ:QB = 1:5 = m:n
A (20, 0) = (x1, y1) and B (10, -20) = (x2, y2)
Q (x, y) = (1(10) + 5 (20)/1+5, 1(-20) + 5(0)/1+5)
= (10+100/6, -20+0/6)
= (110/6, -20/6)
Q (x, y) = (55/3, -10/3)
(Q3) A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3:5 and AQ : QC = 3:5
Show that : PQ = 3/8 BC
Solution:
Given that A (-8, 0) B (0, 16) and C (0, 0) are the vertices of a triangle ABC.
Point P lies on AB mid-point Q lies on AC.
So, point ‘P’ is the mid-point of co-ordinate A and B.
Now, we have to find the co-ordinate of P.
P Given that,
AP/PB = 3/5
m/n = 3/5
m:n = 3:5
A (-8, 0) = (x1, y1) and
B (0, 16) = (x2, y2)
P = (mx2 + nx1/m+n, my2 + ny1/m+n)
= (3× (0) + 5× (-8)/3+5, 3×(16) + 5×(0)/3+5)
= (0 – 40/8, 48 + 0/8)
= (-40/8, 48/8)
P = (-5, 6)
Now, we have to find the co-ordinate of Q.
Q is the mid-point of co-ordinate A and C.
Also given that,
AQ/QC = 3/5
m/n = 3/5
m:n = 3:5
A (-8, 0) = (x1, y1) and C (0, 0) = (x2, y2)
Q = (mx2 + nx1/m+n, my2+ ny1/m+n)
= (3× (0) + 5× (-8)/3+5, 3×(0)+ 5×(0)/3+5)
= (0 – 40/8, 0+0/8)
Q = (-5, 0)
The co-ordinate of P(-5, 6)x1 , y1 and Q (-5, 0)
x2, y2
Now, we have to find the length or distance of PQ –
By using distance formula –
PQ = √(x2 – x1)2 + (y2 – y1)2
= √(-5 – (-5))2 + (0 – 6)2
= √(-5 + 5)2 + (-6)2
= √0 + (36)
= √36
PQ = 6 —— (A)
Now, we have to find the length of BC,
By distance formula –
B (0, 16) = (x1, y1) and C (0, 0) = (x2, y2)
BC = √(x2 – x1)2 + (y2 – y1)2
= √(0 – 0)2 + (0 – 16)2
C = √0 + (-16)2
= √(16)2
BC = 16
Now, to show: PQ = 3/8 BC
L.H.S = 3/8 B
= 3/8 × 16
= 6
∴ PQ = 6 (∵ From A)
∴ L.H.S = R.H.S
∴ PQ = 3/8 BC (Hence Proved)
(Q4) Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.
Solution:
Let, A (6, -9), P (x, y), Q(x1, y1) and B (0, 0) be the co-ordinates.
P divides in the ratio 1:2
m:n = 1:2
A (6, -9) = (x1, y1) and B (0, 0) = (x2, y2)
We have to find the co-ordinate of P
P = (mx2 + nx1/m+n, my2 + ny1/m+n)
P (x, y) = (1× (0) + 2× (6)/1+2, 1×(0) + 2×(-9)/1+2)
= (0+12/3, 0-18/3)
P (x, y) = (4, -6)
Now,
Q divide in the ratio 2:1
m:n = 2:1
A (6, -9) = (x1, y1) and B (0, 0) = (x2, y2)
We have to find the co-ordinate of Q –
Q (mx2 + nx1/m+n, my2 + ny1/m+n)
Q (x, y) = (2× (0) + 1× (6)/2+1, 2×(0) + 1 ×(-9)/2+1)
= (0 + 6/3, 0 – 9/3)
Q (x, y) = (2, -3)
∴ The required co-ordinates of trisection are (4, -6) and (2, -3)
(Q5) A line segment joining A (-1, 5/3) and B (a, 5) is divided in the ratio 1:3 at P, point where the line segment AB intersects the y – axis.
(i) Calculate the value of ‘a’
(ii) Calculate the co-ordinates of ‘p’
Solution:
Given that: A (-1, 5/3) and B (a, 5) is divided in the ratio 1:3
(i) Now, we have to find the value of a –
P (o, y) = (mx2 + nx1/m+n, my2 + ny1/m+n)
m:n = 1:3
A (-1, 5/3) = (x1, y2) and B (a, 5) = (x2, y2)
P (0, y) = (1 (a) + 3 (-1)/1+3, 1 (5) + 3 (5/3)/1+3
= (a-3/4, 5+5/4)
= (a – ¾, 10/4)
P (0,y) = (a-3/4, 5/2)
Equating both sides,
0 = a – ¾
0 × 4 = a – 3
0 = a – 3
a = 3
∴ The value of a = 3.
(iii) Now, we have to find the co-ordinate at P –
P (0, y) = (a – ¾, 5/4)
Put a = 3
P (0, y) = (3 – ¾, 5/2)
= (0/4, 5/2)
P (0, y) = (0, 5/2)
∴ The co-ordinates of point P are (0, 5/2)
(Q6) In what ratio is the line joining A (0, 3) and (4, -1) divided by the x-axis? Write the co-ordinates of the point. Where AB intersects the x-axis,
Solution:
Given that: A (0, 3) and B (4, -1) divided by the x –axis.
P (x, 0)
Let, the co-ordinates A and B in the ratio z:1
P (x, 0) = (mx2 + nx1/m+n, my2 + ny1/m+n)
A (0, 3) = (x1, y1) and B (4, -1) = (x2, y2)
m:n = z:1
P (x, 0) = (z (4) + 1 (0)/z+1, z (-1) + 1 (3)/z + 1)
P (x, 0) = (4z/z+1, -z+3/z + 1)
Comparing both sides,
0 = -z +3/z+1
0× (z + 1) = – z + 3
0 = -z + 3
z = +3
∴ x = 4z/z+1
x = 4×(3)/3+1
= 12/4
x = 3
∴ The co-ordinates of the point where AB intersects the x-axis is P (x, 0) = (3, 0)
(Q7) The mid-point of the segment AB as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B.
Solution:
Given that
C is divided into the segment AB.
C is the mid-point of segment AB.
The point A is on x-axis
∴ A (x, 0) and the point B is on y-axis.
B (0, y)
Now, we have to find the co-ordinates of A and B –
A (x, 0) = (x1, y1) and B (0, y) = (x2, y2)
C (4, -3) = (x1 + x2/2, y1 + y2/2)
(4, -3) = (x/2, y/2)
Comparing both sides,
4 = x/2 or -3 = y
4×2 = x, -3×2 = y
x = 8, -6 = y
y = -6
∴ The required co-ordinates of A (x, 0) = (8, 0) and B (0, y) = (0, -6)
(Q8) AB is a diameter of a circle with centre C (-2, 5). If A = (3, -7), Find
(i) The length of radius AC
(ii) The co-ordinates of B
Solution:
Radius: AC =?
The distance between radius AC =?
A (3, -7) = (x1, y1) and C (-2, 5) = (x2, y2)
AC = √(x2 – x1)2 + (y2 – y1)2
= √(-2 – 3)2 + (5 – (-7))2
= √(-5)2 + (5 + 7)2
= √(25) + (12)2
= √25 + 144
= √169
AC = 13 units
Now, we have to find the co-ordinates of B (x, y)
C is the mid-point formula –
C (x1 + x2/2, y1 + y2/2)
A (3, -7) = (x1, y1) and B (x, y) = (x2, y2)
(-2, 5) = (3+x/2, -7+y/2)
Comparing both sides,
-2 = 3+x/2
(-2) ×2 = 3+x
-4 = 3+x
-4 –3 = x
x = -7
Now, 5 = -7+y/2
5×2 = -7+y/2
10 = -7+y
10+7= y
17 = y
y = 17
∴ The required co-ordinates are B (-7, 17)
(Q9) Find the co-ordinates of the centroid of a triangle ABC whose vertices are:
A (-1, 3), B (1, -1) and C (B, 1)
Solution:
Now, we have to find the co-ordinates of the centroid of a triangle ABC –
= G (x, y) = (x1 + x2 + x3/3, y1 + y2 + y3/3)
A (-1, 3), = (x1, y1), B (1, -1) = (x2, y2) and C (5, 1) = (x3, y3)
G (x, y) = (-1+1+5/3, 3+(-1)+1/3)
G (x, y) = (0+5/3, 3-1+1/3)
= (5/3, 2+1/3)
= (5/3, 3/3)
G (x, y) = (5/3, 1)
∴ The co-ordinates of centroid are G (x, y) = (5/3, 1)
(Q10) The mid-point of the line segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a) Find the values of A and B,
Solution:
Point p is divided to the line segment AB.
∴ P is the mid-point of the co-ordinate A and B.
A (4a, 2b-3 = (x1, y1) and
B = (-4, 3b) = (x2, y2)
Now, we have to find the value of a and b –
F = (x1 + x2/2, y1 + y2/2)
(2, -2a) = (4a + (-4)/2, 2b-3+3b/2)
(2, -2a) = (4a-4/2, 5b-3/2)
Comparing both sides,
2 = 4a-4/2
2×2 = 4a-4
4 = 4a-4
4+4 = 4a
8 = 4a
8/4 = a
2 = a
∴ a = 2
Now, -2a = 5b-3/2
(-2a) × 2 = 5b-3
-4a = 5b – 3
Put a = 2
-4×2 = 5b – 3
-8 = 5b – 3
-8+3 = 5b
-5 = 5b
-5/5 = b
b = -1
(Q11) The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a+1). Find the values of a and b.
Solution:
∴ P is the mid-point of the co-ordinate A and B.
A (2a, 4) = (x1, y1) and
B (-2, 2b) = (x2, y2)
P = (x1 + x2/2, y1 + y2/2)
(1, 2a + 1) = (2a + (-2)/2, 4 + 2b/2)
(1, 2a + 1) = (2a-2/2, 4+2b/2)
1 = 2a – 2/2
1×2 = 2a – 2
2 = 2a – 2
2+2 = 2a
4 = 2a
4/2 = a
∴ a = 2
Now, 2a + 1 = 4+2b/2
(2a + 1) × 2 = 4+2b
4a + 2 = 4 + 2b
Put a = 2
4×2+2 = 4 + 2b
8+2 = 4 + 2b
10 – 4 = 2b
6 = 2b
6/2 = b
3 = b
∴ b = 3
Here is your solution of Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula
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