Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula Exercise 13A Solutions
Section and Mid-Point Formula
Exercise 13A
(Q1) Calculate the coordinates of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1:2
(ii) A (-4, 6) and B (3, -5) in the ratio 3:2
Solution:
Let the co-ordinates of point P be (x, y).
We know that, the section formula P (x, y) = x = m1x2 + m2x1/x1 + x2
y = m1y2 + m2y1/m1 + m2
(i) A (1, 3) and B (5, 9) in the ratio 1:3:2
We know find co-ordinates,
A = (1, 3) = (x1, y1)
B = (5, 9) = (x2, y2)
The ratio = 1:2 = m1 : m2
Now, by section formula, we have to find the coordinate (x, y)
x = m1x2 + m2x1/m1 + m2
= 1 (5) 2 (1)/1 + 2
= 5+2/3
x = 7/3
y = m1y2 + m2y1/m1 + m2
= 1(9) + 2(3)/1+2
= 9+6/3
y = 15/3 = 5
∴ The co-ordinates at point P
(ii) A (-4, 6) and B (3, -5) in the ratio 3 we have to find co-ordinates
A (-4, 6) = (x1, y1)
B = (3, -5) = (x2, y2)
Now, we have to find co-ordinates (x, y), by section formula,
x = m1x2 + m2x1/m1 + m2
= 3(3) + 2(-4)/3+2
= 9-8/5
x = 1/5
y = m1y2 + m2y1/m1 + m2
= 3 (-5) + 2(6)/3+2
y = -15+12/5
y = -3/5
∴ The coordinates at point P are (1/5, -3/5)
(Q2) In what ratio is the line joining (2, -3) and (5, 6) divided by the x – axis.
Solution:
A (2, -3) and B (5, 6) are co-ordinates at point P (x, o)
Now, A (2, -3) = (x1, y1)
B = (5, 6) = (x2, y2)
We know that, by section formula,
P = m1x2 + m2x1/m1 + m2 , m1y2 + m2y1/m1 + m2
(x, o) = (m1 (5) + m2 (2)/m1 + m2, m1 (6) + m2 (-3)/m1 + m2)
Comparing both sides,
6m1 + (-3m2)/m1 + m2 = 0
6m1 – 3m2 = 0 × (m1 + m2)
6m1 – 3m2 = 0
6m1 = 3m2
m1 = 3/6 m2
m1/m2 = 3/6
m1/m2 = ½
∴ The ratio is m1 : m2 = 1: 2
(Q3) In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis.
Solution:
Let A (2, -4) and B (-3, 6) are the co-ordinates at point P (o, y)
Now, A (2, -4) = (x1, y1)
B = (-3, 6) = (x2, y2)
We know that, by section formula,
P = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
= (m1 (-3) + m2 (2)/m1 + m2, m1 (6) + m2 (-4)/m1 + m2)
P = (-3m1 + 2m2/m1 + m2, 6m1 – 4m2/m1 + m2)
(o, y) = (-3m1 + 2m2/m1 + m2, 6m1 – 4m2/m1 + m2)
Comparing both sides,
o = -3m1 + 2m2/m1 + m2
o × m1 + m2 = -3m1 + 2m2
o = -3m1 + 2m2
3m1 = 2m2
m1/m2 = 2/3
m1/m2 = 2/3
∴ The required ratio is m1 : m2 = 2:3
(Q4) In what ratio does the point (1, a) divided the join of (-1, 4) and (4, -1)? Also find the value of a ?
Solution:
Let A (-1, 4) and B (4, -1) are the co-ordinate at point P (1, 9)
Now, A (-1, 4) = (x1, y1)
B (4, -1) = (x2, y2)
We know that by section formula,
P = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
(1, a) = (m1 + (4) + m2 (-1)/m1 + m2, m1 (-1) + m2 (4)/m1 + m2)
On comparing both sides,
1 = 4m1 – m2/m1 + m2
1 x (m1 + m2) = 4m1 – m2
m1 + m2 = 4m1 – m2
m2 + m2 = 4m1 – m
2m2 = 3m1
2/3 = m1/m2
∴ m1 : m2 = 2:3
a = -m1 + 4m2/m1 + m2
= -2 + 4(3)/2+3 (∵ m1 = 2, m2 = 3)
= -2+12/5
a = 10/5
a = 2
(Q5) In what does the point (a, 6) divide the join of (-4, 3) and (2, 8) also, find the value of a.
Solution:
Let A (-4, 3) = (x1, y1)
B (2, 8) = (x2, y2)
We know that, by section formula,
P = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
(a, 6) = (m1 (2) + m2 (-4)/m1 + m2, m1 (8) + m2 (3)/m1 + m2)
(a, 6) = (2m1 – 4m2/m1 + m2, 8m1 + 3m2/m1 + m2
On comparing both sides,
6 = 8m1 + 3m2/m1 + m2
6 (m1 + m2) = 8m1 + 3m2
6m1 + 6m2 = 8m1 + 3m2
6m2 – 3m2 = 8m1 – 6m1
+3m2 = 2m1
+3/2 = m1/m2
∴ m1 ! m2 = +3 : 2
a = 2m1 – 4m2/m1 + m2
a = 2 (3) – 4(2)/+3 + 2
= 6-8/5
a = -2/5
(Q6) In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the points of intersection.
Solution:
Let A (4, 3) and B (2, -6) are co-ordinates at point P (x, o)
Now, A (4, 3) = (x1, y1)
B (2, -6) = (x2, y2)
We know that,
By section on formula,
P = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
(x, o) = (m1 (2) + m2 (4)/m1 + m2, m1 (-6) + m2 (3)/m1 + m2)
(x, o) = (2m1 + 4m2/m1 + m2, – 6m1 + 3m2/m1 + m2)
Comparing on both sides,
O = -6m1 + 3m2/m1 + m2
O × (m1 + m2) = – 6m1 + 3m2
O = – 6m1 + 3m2
6m1 = 3m2
m1/m2 = 3/6
m1/m2= ½
∴ The ratio is m1 : m2 = 1:2
∴ x = 2m1 + 4m2/m1 + m2
= 2 (1) + 4 (2)/1 + 2
= 2+8/3
x = 10/3
∴ The required co-ordinates of the point of intersection are (10/3, 0)
(Q7) Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Let A (-4, 7) and B (3, 0) be the co-ordinates of point P (O, y)
Now, A (-4, 7) = (x1, y1)
B = (3, 0) = (x2, y2)
We know that,
By section formula,
P (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
(O, y) = (m1 (3) + m2 (-4)/m1 + m2, m1 (o) + m2 (7)/m1 + m2)
(O, y) = (3m1 – 4m2/m1 + m2, O + 7m2/m1 + m2)
Comparing both sides,
O = 3m1 – 4m2/m1 + m2
O × (m1 + m2) = 3m1 – 4m2
O = 3m1 – 4m2
4m2 = 3m1
4/3 = m1/m2
The ratio is m1 : m2 = 4:3
∴ Y = 7m2/m1 + m2
Y = 7(3)/4+3
Y = 21/7
y = 3
∴ The required co-ordinates of point of intersection are (o, 3)
(Q8) Points A, B, C and C divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Solution:
Let, O (o, o) = (x1, y1)
P (5, -10) = (x2, y2)
First we have to find the co-ordinates of point B.
So, Point B divides op in the ratio 2:3.
We know that,
By section formula
B = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
= (m1 (5) + m2 (0)/m1 + m2, m1 (-10) + m2 (0)/m1 + m1 + m2)
B = (2 (5) + 3 (0)/2+3, 2 (-10) + 3(0)/2+3)
= (10/5, -20+0/5
B = (2, -4)
We have to find the co-ordinates of point A.
So, point A divides op in the ratio 1:4
We know that
By section formula,
A = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
A = (1 (5) + 4(0)/1+4, 1(-10) + 4(0)/1+4)
= (5+0/5, -10+0/5)
A = (1, -2)
Now, we have to find the co-ordinate of point C,
So, point C divides op in the ratio 3:2
We know that, by section formula,
C = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
C = (3 (5) + 2 (0)/3+2, 3(-10) + 2(0)/3+2
= (15+0/5, -30+0/5)
C = (3, -6)
Now, we have to find co-ordinates of point D.
So, point D divides op in the ratio 4:1.
We know that, by section formula,
D = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
= (4 (5) + 1(0)/4+1, 4(-10) + 1 (0)/4+1
= (20/5, -40/5)
D = (4, -8)
(Q9) The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that PB/AB = 1/5 and the co-ordinate of P.
Solution:
∴ AP = AB – PB
= 5 – 1
AP = 4
∴ PB:PA = 1:4
m1:m2 = 1:4
Now, we have to find co-ordinates at point P (x, y) by section formula,
P (x, y) = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
So, A (-3, -10) = (x1, y1)
B (-2, 6) = (x2, y2)
P (x, y) = (1 (-2) + 4(-3)/1+4, 1(6) + 4 (-10)/1+4)
= (-2-12/5, 6-40/5)
P (x, y) = (-14/5, -34/5)
Or
PA:PB = 4:1 = m1:m2
P (x, y) = (4 (-2) + 1 (-3)/5, 4(6) + 1 (-10)/5)
= (-8-3/5, 24-10/5)
P (x, y) = (-11/5, 14/5)
(Q10) P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Solution:
Given that, 5AP = 2BP
AP/BP = 2/5
We know that, by section formula,
We have to find the co-ordinates at point P (x, y) –
P (x, y) = (m1 x2 + m2 x1/m1 + m2, m1y2 + m2y1/m1 + m2)
∴ A (4, 3) = (x1, y1) and B (-2, 6) = (x2, y2)
P (x, y) = (2 (-2) + 5(4)/2+5, 2 (6) + 5 (3)/2+5
P (x, y) = (-4+20/7, 12+15/7)
P (x, y) = (16/7, 27/7)
(Q11) Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also find the co-ordinates of the points of intersection.
Solution:
Let A (-3, -1) and B (5, 7) be the co-ordinates at point P (2, y)
We have to find the co-ordinates of point of intersection
P (2, y) = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
A (-3, -1) = (x1, y1) and B (5, 7) = (x2, y2)
= (m1 (5) + m2 (-3)/m1 + m2, m1 (7) + m2 (-1)/m1 + m2)
Comparing both sides,
2 = 5m1 – 3m2/m1 + m2
2 (m1 + m2) = 5m1 – 3m2
2m1 + 2m2 = 5m1 – 3m2
2m2 + 3m2 = 5m1 – 2m1
5m2 = 3m1
5/3 = m1/m2
∴ The required ratio is m1:m2 = 5:3
∴ y = 7m1 – m2/m1 + m2
= 7 (5) – 3/5+3
= 35+3/8
= 32/3
y = 4
∴ The required co-ordinates of the point of intersection are (2, 4)
(Q12) Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2
Solution:
We have to find the ratio,
Now, A (6, 5) = (x1, y1) and B (4, -3) = (x2, y2)
We know that, by section formula,
P (x, 2) = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
= (m1 (4) + m2 (6)/m1 + m2, m1 (-3) + m2 (5)/m1 + m2)
P (x, 2) = (4m1 + 6m2/m1 + m2, -3m1 + 5m2/m1 + m2)
On comparing both sides,
2 = -3m1 + 5m2/m1 + m2
2 × (m1 + m2) = – 3m1 + 5m2
2m1 + 2m2 = -3m1 – 5m2
2m1 + 3m1 = 5m2 – 2m2
5m1 = 3m2
m1/m2 = 3/5
∴ The required ratio is 3:5.
Here is your solution of Selina Concise Class 10 Math Chapter 13 Section and Mid-Point Formula
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