Section and Mid-Point Formula Exe-13B Class-10 Concise ICSE Maths Selina Solution Ch-13. In this article you would learn how to find co-ordinate of mid point of a line segment. 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-13B Class-10 Concise ICSE 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-13AB | how to use Mid Point Formula |
| Edition | 2025-2026 |
Co-ordinate of Mid Point of a Line Segment
Section and Mid-Point Formula Exe-13B Class-10 Concise ICSE Maths Selina Solution Ch-13.
Que-1: Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3), (-1, 7)
Ans: (i) A(-6, 7) and B(3, 5)
Mid-point of AB = (-6+3/2,7+5/2)
= (-3/2 , 6)
(ii) A(5, -3) and B(-1, 7)
Mid-point of AB = (5-1/2,-3+7/2)
= (2, 2)
Que-2: 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.
Ans: Mid-point of AB = (2, 3)
∴ (3+x/2,5+y/2)=(2,3)
3+x/2=2 and 5+y/2=3
3 + x = 4 and 5 + y = 6
x = 1 and y =1
Que-3: A (5, 3), B (-1, 1) and C (7, -3) are the vertices of ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = BC.
Ans: Given, L is the mid-point of AB and M is the mid-point of AC.
Co-ordinates of L are
(5-12,3+12)=(2,2)
Co-ordinates of M are
(5+72,3-32)=(6,0)
Using distance formula, we have:
BC=√(7+1)²+(3-1)²
BC=√64+16
C=√80
C=4√5
LM=√(6-2)²+(0-2)²
LM=√16+4
LM=√20
LM=2√5
Hence, LM = 1/2 BC
Que-4: Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10),
(ii) B; if A = (3, -1) and M (-1, 3).
Ans: (i) M is the mid-point of AB.
Let A = (x, y), M = (1, 7) and B = (–5, 10)
∴ 1=x-5/2
=> x – 5 = 2
∴ x=2+5=7 and 7=y+10/2
=> y + 10 = 14
∴ y = 14 – 10 = 4
∴ Co-ordinates of A are (7, 4)
(ii) Let B = (x, y), M = (–1, 3), A = (3, –1)
∴ -1=x+3/2
=> x + 3 = –2
∴ x = –2 – 3 = –5
3=y-1/2
=> y – 1 = 6
∴ y = 6 + 1 = 7
∴ Co-ordinates of B are (–5, 7)
Que-5: P (-3, 2) is the mid-point of line segment AB as shown in the figure. Find the co-ordinates of points A and B.
Ans: Point A lies on y-axis, so let its co-ordinates be (0, y).
Point B lies on x-axis, so let its co-ordinates be (x, 0).
P(–3, 2) is the mid-point of line segment AB.
∴ (-3,2)=(0+x/2,y+0/2)
(-3,2)=(x/2,y/2)
-3=x/2 and 2=y/2
-6 = x and 4 = y
Thus, the co-ordinates of points A and B are (0, 4) and (−6, 0) respectively.
Que-6: In the given figure, P (4, 2) is the mid point of line segment AB. Find the co-ordinates of A and B.
Ans: Point A lies on x-axis, so let its co-ordinates be (x, 0).
Point B lies on y-axis, so let its co-ordinates be (0, y).
P(4, 2) is mid-point of line segment AB.
∴ (4,2)=(x+0/2,0+y/2)
4=x/2 and 2=y/2
8 = x and 4 = y
Hence, the co-ordinates of points A and B are (8, 0) and (0, 4) respectively.
Que-7: (-5, 2), (3, -6) and (7, 4) arc the vertices of a triangle. Find the length of its median through the vertex (3, -6) and (7, 4).
Ans: Let A(–5, 2), B(3, −6) and C(7, 4) be the vertices of the given triangle.
Let AD be the median through A, BE be the median through B and CF be the median through C.
We know that median of a triangle bisects the opposite side.
Co-ordinates of point F are
(-5+3/2,2-6/2)=(-2/2,-4/2)=(-1,-2)
Co-ordinates of point D are
(3+7/2,-6+4/2)=(10/2,-2/2)=(5,-1)
Co-ordinates of point E are
(-5+7/2,2+4/2)=(2/2,6/2)=(1,3)
The median of the triangle through the vertex B(3, −6) is BE
Using distance formula,
BE=√(1-3)²+(3+6)²
BE=√4+81
BE=√85
BE = 9.22
Que-8: Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
Ans: Given, AB = BC = CD
So, B is the mid-point of AC.
Let the co-ordinates of point A be (x, y).
∴ (0,3)=(x+1/2,y+8/2)
0=x+1/2 and 3=y+8/2
0 = x + 1 and 6 = y + 8
-1 = x and –2 = y
Thus, the co-ordinates of point A are (–1, –2).
Also, C is the mid-point of BD.
Let the co-ordinates of point D be (p, q).
∴ (1,8)=(0+p/2,3+q/2)
1=0+p/2 and 8=3+q/2
2 = 0 + p and 16 = 3 + q
2 = p and 13 = q
Thus, the co-ordinates of point D are (2, 13)
Que-9: 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).
Ans: We know that the centre is the mid-point of diameter.
Let the required co-ordinates of the other end of mid-point be (x, y).
∴ (2,-1)=(-2+x/2,5+y/2)
2=-2+x/2 and -1=5+y/2
4 = –2 + x and –2 = 5 + y
6 = x and –7 = y
Thus, the required co-ordinates are (6, −7)
Que-10: A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Ans: Co-ordinates of A = (2, 5), B = (1, 0), C = (-4, 3) and D = ( 3, 8)
Let the mid-point of AC is P (x1, y1)
∴ Co-ordinates of mid-point of AC will be
(x1,y1)=(2-4/2,5+3/2)
=(-2/2,8/2)
= (−1, 4)
Let the mid-point of BD is Q(x2, y2)
Co-ordinates of mid-point of BD will be
∴ (x2,y2)=(1+(-3)/2,0+8/2)
= (1-3/2,8/2)
= (-2/2,8/2)
= (−1, 4)
∵ Co-ordinates of mid-points AC and BD are the same.
∴ The quadrilateral is a parallelogram.
Que-11: P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of its diagonals. Find the co-ordinates of R and S.
Ans: Let the coordinates of R and S be (x, y) and (a, b) respectively.
Mid-point of PR is O.
∴ O(-3,2)=O(4+x/2,2+y/2)
-3=4+x/2,2=2+y/2
−6 = 4 + x, 4 = 2 + y
x = −10, y = 2
Hence, R = (−10, 2)
Similarly, the mid-point of SQ is O.
∴ O(-3,2)=O(a-1/2,b+5/2)
-3=a-1/2,2=b+5/2
−6 = a – 1, 4 = b + 5
a = −5, b = –1
Hence, S = (−5, −1)
Thus, the coordinates of the point R and S are (−10, 2) and (−5, −1).
Que-12: A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates of vertex C.
Ans: Let the co-ordinates of vertex C be (x, y).
ABCD is a parallelogram.
∴ Mid-point of AC = Mid-point of BD
(-1+x/2,0+y/2)=(1+3/2,3+5/2)
(-1+x/2,y/2)=(2,4)
and-1+x/2=2 and y/2=4
x=5 and y/2=4
x = 5 and y = 8
Thus, the co-ordinates of vertex C is (5, 8).
Que-13: The points (2, -1), (-1, 4) and (-2, 2) are the mid-points of the sides of a triangle. Find its vertices.
Ans: Let A(x1, y1), B(x2, y2) and C(x3, y3) be the coordinates of the vertices of ΔABC.
Midpoint of AB, i.e. D
D(2,-1)=D(x1+x2/2,y1+y2/2)
2=x1+x2/2 , y1+y2/2=-1
x1 + x2 = 4 …(1)
y1 + y2 = –2 …(2)
Similarly
x1 + x2 = –2 …(3)
y1 + y3 = 8 …(4)
x1 + x3 = −4 …(5)
y2 + y3 = 4 …(6)
Adding (1), (3) and (5), we get,
2(x1 + x2 + x3) = –2
x1 + x2 + x3 = –1
4 + x3 = –1 …[From (1)]
x3 = –5
From (3)
x1 – 5 = –2
x1 = 3
From (5)
x2 – 5 = –4
x2 = 1
Adding (2), (4) and (6), we get,
2(y1 + y2 + y3) = 10
y1 + y2 + y3 = 5
–2 + y3 = 5 …[From (2)]
y3 = 7
From (4)
y1 + 7 = 8
y1 = 1
From (6)
y2 + 7 = 4
y2 = –3
Thus, the co-ordinates of the vertices of ΔABC are (3, 1), (1, –3) and (–5, 7).
Que-14: Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Ans: Given, AB = BC, i.e., B is the mid-point of AC.
∴ (y,7)=(-5+1/2,x-3/2)
(y,7)=(-2,x-3/2)
y=-2 and 7=x-32
y = –2 and x = 17
Que-15: Points P (a, -4), Q (-2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of ‘a’ and ‘b’:
Ans: Given, PR = 2QR
Now, Q lies between P and R, so, PR = PQ + QR
∴ PQ + QR = 2QR
PQ = QR
Q is the mid-point of PR.
∴ (-2,b)=(a+0/2,-4+2/2)
(-2,b)=(a/2,-1)
⇒ a = −4 and b = -1.
Que-16: Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C = (-1, 4).
Ans: Co-ordinates of the centroid of triangle ABC are
(7+0-1/3,-2+1+4/3)
= (6/3,3/3)
= (2, 1).
Que-17: The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Ans: Let co-ordinates of P be (x, y)
And of centroid G are (2, –5)
∴ 2 =x1+x2+x3/3
= x-6+11/3
= x+5/3
x + 5 = 6
∴ x = 6 – 5 = 1
And -5=y1+y2+y3/3
= y+5+8/3
= y+13/3
=y + 13 = –15
∴ y = –15 – 13
= –28
Hence, co-ordinates of vertex P are (1, -28).
Que-18: A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Ans: Co-ordinates of centroid of ΔABC are (0, 0)
∴ 0=x1+x2+x3/3
= 5-4+y/3
0=y+1/3
y + 1 = 0
∴ y = –1
Again 0=y1+y2+y3/3
= x+3-2/3
0=x+1/3
x + 1 = 0
∴ x = –1
∴ x = –1, y = –1
–: Section and Mid-Point Formula Exe-13B Class-10 Concise ICSE Maths Solution Ch-13 :–
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