Section and Mid-Point Formula Exe-13C Class-10 Concise Maths Selina Solution Ch-13. In this article you would learn to solve miscellaneous problems on section and mid point formula. We Provide Step by Step Solutions / Answer of questions for Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Section and Mid-Point Formula Exe-13C Class-10 Concise Maths Selina Solution Ch-13
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-13 | Section and Mid-Point Formula |
| Writer | R.K. Bansal |
| Exe-13C | miscellaneous problems on section and mid point formula |
| Edition | 2025-2026 |
Miscellaneous Problems
Section and Mid-Point Formula Exe-13C Class-10 Concise Maths Selina Solution Ch-13
Que-1: 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.
Ans: Given, BP : PC = 3 : 2
Using section formula, the co-ordinates of point P are
(3×5+2×0/3+2,3×10+2×5/3+2)
= (15/5,40/5)
= (3, 8)
Using distance formula, we have:
AP=√(3-4)²+(8+4)²
= √1+144
= √145
= 12.04
Que-2: A (20, 0) and B (10, – 20) are two fixed points, find the co-ordinates of the point P in AB such that 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Ans: Given, 3PB =AB
AB/PB=3/1
AB-PB/PB=3-1/1
AP/PB=2/1
Using section formula,
Coordinates of P are
P(x,y)=P(2×10+1×20/2+1,2×(-20)+1×0/2+1)
= P(40/3,-40/3)
Given, AB = 6AQ
AQ/AB=1/6
AQ/AB-AQ=1/6-1
AQ/QB=1/5
Using section formula,
Coordinates of Q are
Q(x,y)=Q(1×10+5×20/1+5,1×(-20)+5×0/1+5)
= Q(110/6,-20/6)
= Q(55/3,-10/3)
Que-3: 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 = BC.
Ans: Given that, point P lies on AB such that AP : PB = 3 : 5.
The co-ordinates of point P are
(3×0+5×(-8)/3+5,3×16+5×0/3+5)
= (-40/8,48/8)
= (–5, 6)
Also, given that, point Q lies on AC such that AQ : QC = 3 : 5.
The co-ordinates of point Q are
(3×0+5×(-8)/3+5,3×0+5×0/3+5)
= (-40/8,0/8)
= (–5, 0)
Using distance formula,
PQ=√(-5+5)²+(0-6)²
= √0+36
= 6
BC=√(0-0)²+(0-16)²
= √0+16²
= 16
Now, PQ = 3/8 BC
= 3/8×16 = 6
Hence, proved
Que-4: Find the co-ordinates of points of trisection of the line segment joining the points (6, -9) and the origin.
Ans: Let P and Q be the points of trisection of the line segment joining A(6, –9) and B(0, 0).
P divides AB in the ratio 1 : 2.
Therefore, the co-ordinates of point P are
(1×0+2×6/1+2,1×0+2×(-9)/2+1)
=(12/3,-18/3)
= (4, −6)
Q divides AB in the ratio 2 : 1.
Therefore, the co-ordinates of point Q are
(2×0+1×6/2+1,2×0+1×(-9)/2+1)
= (6/3,-9/3)
= (2, –3)
Thus, the required points are (4, −6) and (2, −3).
Que-5: A line segment joining A (-1, ) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.
(i) Calculate the value of ‘a’.
(ii) Calculate the co-ordinates of ‘P’. (1994)
Ans: Since, the line segment AB intersects the y-axis at point P, let the co-ordinates of point P be (0, y).
P divides AB in the ratio 1 : 3.
∴ (0,y)=(1×a+3×(-1)/1+3,1×5+3×(5/3)/1+3)
(0,y)=(a-3/4,10/4)
0=a-3/4 and y=10/4
a=3 and y=5/2
Thus, the value of a is 3 and the co-ordinates of point P are (0 , 5/2)
Que-6: In what ratio is the line joining A (0, 3) and B (4, -1), divided by the x-axis ? Write the co-ordinates of the point where AB intersects the x-axis. [1993]
Ans:
Let the ratio be m1 : m2 when the x-axis intersects the line AB at P.
∴ Let co-ordinate of P(x, 0)
x=m1x2+m2x1/m1+m2,y=m1y2+m2y1/m1+m2
x=m1×4+m2×0/m1+m2,y=m1(-1)+m2×3/m1+m2
x=4m1/m1+m2,y=-m1+3m2/m1+m2
∵ P lies on x-axis,
∴ y = 0
∴ -m1+3m2/m1+m2=0
– m1 + 3m2 = 0
m1 = 3m2
m1/m2=3/1
m1 : m2 = 3 : 1
Now, x=4×3/3+1
= 12/4 = 3
∴ Required co-ordinates of P will be (3, 0) .
Que-7: The mid point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B. (1996)
Ans: Since, point A lies on x-axis, let the co-ordinates of point A be (x, 0).
Since, point B lies on y-axis, let the co-ordinates of point B be (0, y).
Given, mid-point of AB is C(4, −3).
∴ (4,-3)=(x+0/2,0+y/2)
(4-3)=(x/2,y/2)
4=x/2 and -3=y/2
x = 8 and y = −6
Thus, the co-ordinates of point A are (8, 0) and the co-ordinates of point B are (0, −6).
Que-8: 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 coordinates of B.
Ans:
(i).
Radius AC = √(3+2)²+(-7-5)²
= √5²+(-12)²
= √25+144
= √169
= 13 units
(ii).
Let the co-ordinates of B be (x, y)
Using mid-point formula, we have
-2=3+x/2 and 5=-7+y/2
−4 = 3 + x and 10 = –7 + y
x = –7 and y = 17
Thus, the coordinates of B are (–7, 17)
Que-9: Find the co-ordinates of the centroid of a triangle ABC whose vertices are A (- 1, 3), B (1, – 1) and C (5, 1)
Ans: Co-ordinates of the centroid of triangle ABC are
= (-1+1+5/3,3-1+1/3)
= (5/3 , 1)
Que-10: The mid-point of the line segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Ans: It is given that the mid-point of the line segment joining (4a, 2b – 3) and (−4, 3b) is (2, –2a).
∴ (2,-2a)=(4a-4/2,2b-3+3b/2)
2=(4a-4/2)
4a − 4 = 4
4a = 8
a = 2
Also,
-2a=2b-3+3b/2
-2×2=5b-3/2
5b − 3 = −8
5b = −5
b = -1
Que-11: The mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
Ans: Mid-point of (2a, 4) and (−2, 2b) is (1, 2a + 1),
Therefore using mid-point formula, we have:
x=x1+x2/2
1=2a-2/2
1 = a − 1
a = 2
y=y1+y2/2
2a+1=4+2b/2
2a + 1 = 2 + b
2 × 2 + 1 – 2 = b
b = 5 – 2 = 3
Therefore, a = 2, b = 3.
Que-12:(i) Write down the co-ordinates of the point P that divides the line joining A (-4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ? [ICSE 1995]
Ans:
(i).
Co-ordinates of point P are
(1×17+2×(-4)/1+2,1×10+2×1/1+2)
= (17-8/3,10+2/3)
= (9/3,12/3)
= (3, 4)
(ii).
OP=√(0-3)²+(0-4)²
OP=√9+16
OP=√25
OP = 5 units
(iii).
Let AB be divided by the point P(0, y) lying on y-axis in the ratio k : 1
∴ (0,y)=(k×17+1×(-4)/k+1,k×10+1×1/k+1)
(0,y)=(17k-4/k+1,10k+1/k+1)
0=17k-4/k+1
17k-4=0
k=4/17
Thus, the ratio in which the y-axis divide the line AB is 4 : 17//\.
Que-13: Prove that the points A(-5, 4); B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D. So that ABCD is a square. [1992]
Ans: We have:
AB=√(-1+5)²+(-2-4)²
= √16+36
= √52
BC=√(-1+5)²+(-2-2)²
= √36+16
= √52
AC=√(5+5)²+(2-4)²
= √100+4
= √104
AB2 + BC2 = 52 + 52 = 104
AC2 = 104
∵ AB = BC and AB2 + BC2 = AC2
∵ ABC is an isosceles right-angled triangle.
Let the coordinates of D be (x, y).
If ABCD is a square, then,
Mid-point of AC = Mid-point of BD
(-5+5/2,4+2/2)=(x-1/2,y-2/2)
0=x-1/2,3=y-2/2
x = 1, y = 8
Thus, the co-ordinates of point D are (1, 8).
Que-14: M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1). Find the co-ordinates of point M. Further, if R (2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
Ans: Given, M is the mid-point of the line segment joining the points A(−3, 7) and B(9, −1).
The co-ordinates of point M are
(-3+9/2,7-1/2)
= (6/2,6/2)
= (3, 3)
Also, given that, R(2, 2) divides the line segment joining M and the origin in the ratio p : q.
∴ (2,2)=(p×0+q×3/p+q,p×0+q×3/p+q)
p×0+q×3/p+q=2
3q/p+q=2
3q = 2p + 2q
3q – 2q = 2p
q = 2p
p/q=1/2
Thus the ratio p : q is 1 : 2.
Que-15: Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. (2014)
Ans: Let P (x,3) divides the line segment joining the points
A(-4,2) and B(3,6) in the ratio K:1.
Thus , we have
3k-4/k+1=x; 6k+2/k+1=3
for 6k+2=3(k+1)
6k+2=3k+3
3k=3-2
3k=1⇒k=1/3
∴ Required ratio1:3
a now consider the quation3k-4/k+1=x
Substituting the value of k in the above equation, We have
3×(1/3)-4/(1/3)+1=x => -3/(4/3) = x => -9/4 = x
∴x=-9/4
b.AP=√(-9/4+4)²+(3-2)²=√49/16+1=√49+16/16
AP =√65/16 unit
Que-16: Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3, 7).
Ans: Let the line 2x + y = 4 divides the line segment joining the points P(2, –2) and Q(3, 7) in the ratio k : 1.
Then, we have
(x,y)=(3k+2/k+1,7k-2/k+1)
Since (3k+2/k+1,7k-2/k+1) lines on line 2x + y = 4, we have
2(3k+2/k+1)+7k-2/k+1=4
6k + 4 + 7k – 2 = 4k + 4
13k + 2 = 4k + 4
9k = 2
k=2/9
Hence, required ratio is 2 : 9.
Que-17: If the abscissa of a point P is 2. Find the ratio in which this point divides the line segment joining the points (-4, 3) and (6, 3). Also, find the co-ordinate of point P.
Ans: Abscissa of a point P is 2
Let co-ordinates of point P be (2, y)
Let point P(2, y) divides the line segment joining the points (−4, 3) and (6, 3) in the ratio
k:1{x=kx2+1⋅x1/k+1}
∴ 2=6k-4/k+1
2k + 2 = 6k – 4
6k – 2k = 2 + 4
4k = 6
k=6/4=3/2
∴ Required ratio be 3 : 2
Now, y=ky2+1⋅y1/k+1
= 3k+3/k+1
= 3×3/2+3 / 3/2+1
= 9/2+3 / 3/2+1
= 9+6/2 / 3+2/2
= 15/2 / 5/2
= 15/2×2/5 = 3
∴ Co-ordinates of point P are (2, 3).
Que-18: The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k, Also, find the co-ordinates of point Q.
Ans: Let A(2, 1) and B(5, –8) be the given point trisected by the points P and Q
AP = PQ = QB
For P:
m1 : m2 = AP : PB = 1 : 2
(x1, y1) = (2, 1) and (x2, y2) = (5, 8)
∴ x=1×5+2×2/1+2
= 5+4/3
= 9/3 = 3
y=1×(-8)+2×1/1+2
= -8+2/3
= -6/3 = –2
∴ Coordinates of P are (3, 2)
Since point P line on the line 2x – y + k = 0
2(3) – (–2) + k = 0
6 + 2 + k = 0
For Q:
m1 : m2 = AQ : QB = 2 : 1
(x1, y1) = (2, 1) and (x2, y2) = (5, –8)
∴ x=2×5+1×2/2+1
= 10+2/3
= 12/3 = 4
y=2×(-8)+1×1/1+2
= -16+1/3
= -15/3
= –5
∴ Coordinates of Q are (4, –5).
Que-19: M is the mid-point of the line segment joining the points A (0, 4) and B (6, 0). M also divides the line segment OP in the ratio 1 : 3. Find:
(i) co-ordinates of M
(ii) co-ordinates of P
(iii) length of BP
Ans: M is mid point of the line segment joining the points A(0, 4) and B(6, 0) M divides the line segment OP in the ratio 1 : 3
(i).
Now co-ordinates of M = (0+6/2,4+0/2)=(3,2)
(ii).
Let co-ordinates of P be (x, y), O is (0, 0)
3=m1x2+m2x1/m1+m2
3=1×x+3×0/1+3
= x+0/4
= x/4
x = 12 and 2=1⋅y+3×0/1+3=y+0/4=y/4
∴ y = 2 × 4 = 8
∴ Co-ordinates of P are (12, 8)
(iii).
Length of BP = √(x2-x1)²+(y2-y1)²
= √(2-6)²+(8-0)²
= √6²+8²
= √36+64
= √100
= 10 units.
Que-20: Find the image of the point A (5, -3) under reflection in the point P (-1, 3).
Ans: Image of the point A(5, –3) under reflection in the point P(–1, 3)
Let B(x, y) be the point of reflection of A(5, –3) under P(–1, 3)
∴ Point P is mid-point of AB
∴ -1=5+x/2
–2 = 5 + x
x = –2 – 5 = –7 and 3=-3+y/2
6 = –3 + y
y = 6 + 3 = 9
∴ Co-ordinates of B are (–7, 9) which is the image of point A.
Que-21: A (-4, 2), B (0, 2) and C (-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of PQR is the same as the centroid of ABC.
Ans: Given:
P is mid-point of BC.
∴ P=(0+(-2)/2,2+(-4)/2)
= (-2/2,-2/2)
= (–1, –1)
Q is mid-point of CA.
∴ Q=(-2+(-4)/2,-4+2/2)
= (-6/2,-2/2)
= (–3, –1)
R is mid-point of AB.
∴ R=(-4+0/2,2+2/2)
= (-4/2,4/2)
= (–2, 2)
Centroid of the triangle is given by (G)
= (x2+x2+x3/3,y1+y2+y3/3)
Let G1 and G2 be the centroid of △ABC and △PQR.
Substituting values we get,
G1=(-4+0+(-2)/3,2+2+(-4)/3)
= (-6/3,0/3)
= (–2, 0)
G2=((-1)+(-3)+(-2)/3,(-1)+(-1)+2/3)
= (-6/3,0/3)
= (–2, 0)
Since, G1 = G2.
Hence, proved that the centroid of △PQR is the same as the centroid of △ABC.
Que-22: A(3, 1), B(y, 4) and C(1, x) are vertices of a triangle ABC. P, Q and R are mid – points of sides BC, CA and AB respectively. Show that the centroid of ΔPQR is the same as the centroid ΔABC.
Ans: Centroid of ΔABC = (3+y+1/3,1+4+x/3)=(4+y/3,5+x/3)
P, Q and R are the mid points of the sides BC, CA and AB.
By mid – point formula, we get
and => P=(y+1/2,4+x/2),Q=(4/2,1+x/2) and R=(3+y/2,5/2)
Centroid of a ΔPQR = (y+1/2+4/2+3+y/2 / 3 , 4+x/2 – 1+x/2 + 5/2 / 3)
=(y+1+4+3+y/2 / 3,4+x+1+x+5 / 2 /3)
=(8+2y/6,10+2x/6)
=(4+y/3,5+x/3) ……..(ii)
From (i) and (ii), we get
Centroid of a ∆ABC = Centroid of a ∆PQR
–: Section and Mid-Point Formula Exe-13C Class-10 Concise Maths Selina Solution Ch-13 :–
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