Section and Mid-Point Formula Exe-13A Class-10 Concise ICSE Maths Solution Ch-13. In this article you would learn how to use section 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-13A Class-10 Concise ICSE Maths Solution Ch-13
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-13 | Section and Mid-Point Formula |
| Writer | R.K. Bansal |
| Exe-13A | how to use section formula |
| Edition | 2025-2026 |
How to Find Coordinate of Point Divide Line Segment?
Section and Mid-Point Formula Exe-13A Class-10 Concise ICSE Maths Solution Ch-13
Que-1: Calculate the co-ordinates-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
Ans: (i) Let the co-ordinates of the point P be (x, y)
x=m1x2+m2x1/m1+m2
= 1×5+2×1/1+2
= 7/3
y=m1y2+m2y1/m1+m2
= 1×9+2×3/1+2
= 15/3
= 5
Thus, the co-ordinates of point P are (7/3,5)
(ii) Let the co-ordinates of the point P be (x, y).
x=m1x2+m2x1/m1+m2
= 3×3+2×(-4)/3+2
= 1/5
y=m1y2+m2y1/m1+m2
= 3×(-5)+2×6/3+2
= -3/5
Thus, the co-ordinates of point P are (1/5 , -3/5)
Que-2: In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis ?
Ans: Let the line joining points A (2, −3) and B (5, 6) be divided by point P (x, 0) in the ratio k : 1.
y=ky2+y1/k+1
0=k×6+1×(-3)/k+1
0=6k-3
k=12
Thus, the required ratio is 1 : 2
Que-3: In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis ?
Ans: Let the line joining points A (2, –4) and B (–3, 6) be divided by point P (0, y) in the ratio k : 1.
x=kx2+x1/k+1
0=k×(-3)+1×2/k+1
0=-3k+2
k=2/3
Thus, the required ratio is 2 : 3
Que-4: In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1)? Also, find the value of ‘a’.
Ans: Let the point P (1, a) divides the line segment AB in the ratio k : 1.
Using section formula, we have:
1=4k-1/k+1
k+1=4k-1
2=3k
k=2/3 …(1)
a=-k+4/k+1
a=-(2/3)+4/(2/3)+1 …(From 1)
a=10/5=2
Hence, the required is 2 : 3 and the value of a is 2
Que-5: In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of ‘a’.
Ans: Let the point P (a, 6) divides the line segment joining A (–4, 3) and B (2, 8) in the ratio k : 1.
Using section formula, we have:
6=8k+3k+1
6k+6=8k+3
3=2k
k=3/2 …(1)
a=2k-4/k+1
a=2×(3/2)-4/(3/2)+1 …(From 1)
a=-2/5
Hence, the required ratio is 3 : 2 and the value of a is -2/5
Que-6: In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Ans: Let P(x, 0) be the point of intersection which divides the line joining the points A(4, 3), B(2, –6) in the ratio of m1 : m2
∴ x=m1×2+m2×4/m1+m2
= 2m1+4m2/m1+m2 ….(i)
And 0=m1×(-6)+m2(3)/m1+m2
– 6m1 + 3m2 = 0
3m2 = 6m1
m1/m2=3/6=1/2
∴ Required ratio be m1 : m2 = 1 : 2
Now, substituting the value of m1 and m2 in (i); we have
x=1×2+2×4/1+2
= 2+8/3
= 10/3
∴ Required point of intersection is (10/3 , 0)
Que-7: 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.
Ans: 0=3k-4/k+1
3k = 4
k=4/3 …(1)
y=0+7/k+1
y=7/(4/3)+1 …(From 1)
y = 3
Hence, the required is 4 : 3 and the required point is S(0, 3).
Que-8: Points A, B, C and D 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.
Ans: Point A divides PO in the ratio 1 : 4.
Co-ordinates of point A are:
(1×0+4×5/1+4,1×0+4×(-10)/1+4)
= (20/5,-40/5)
= (4, –8)
Point B divides PO in the ratio 2 : 3.
Co-ordinates of point B are:
(2×0+3×5/2+3,2×0+3×(-10)/2+3)
= (15/5,-30/5)
= (3, –6)
Point C divides PO in the ratio 3 : 2.
Co-ordinates of point C are:
(3×0+2×5/3+2,3×0+2×(-10)/3+2)
= (10/5,-20/5)
= (2, –4)
Point D divides PO in the ratio 4 : 1.
Co-ordinates of point D are:
(4×0+1×5/4+1,4×0+1×(-10)/4+1)
= (5/5,-10/5)
= (1, –2).
Que-9: The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that PB/AB=1/5, find the co-ordinates of P.
Ans: Let the co-ordinates of P be (x, y) which divides the line joining the points A (–3, –10) and B (–2, 6) in the ratio of AP : PB i.e. (5 – 1) : 1 or 4 : 1
Since 5 PB = PA + PB
4 PB = PA
PA/PB=4/1
∴ x=4×(-2)+1×(-3)/4+1
= -8-3/5
= -11/5
y=4×6+1×(-10)/4+1
= 24-10/5
= 14/5
∴ Co-ordinates of P are (-11/5 , 14/5)
Que-10: 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.
Ans: ∵ 5AP = 2BP
AP/BP=2/5
AP : PB = 2 : 5
Let the co-ordinates of P be (x, y) which divides the line joining the points A(4, 3) and B(–2, 6) in the ratio of 2 : 5
∴ x=2×(-2)+5×4/2+5
= -4+20/7
= 16/7
y=2×6+5×3/2+5
= 12+15/7
= 27/7
∴ Co-ordinates of P are (16/7 , 27/7)
Que-11: 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 point of intersection.
Ans: The co-ordinates of every point on the line x = 2 will be of the type (2, y).
Using section formula, we have:
x=m1×5+m2×(-3)/m1+m2
2=5m1-3m2/m1+m2
2m1+2m2=5m1-3m2
5m2=3m1
m1/m2=5/3
Thus, the required ratio is 5 : 3.
y=m1×7+m2×(-1)/m1+m2
y=5×7+3×(-1)/5+3
y=35-3/8
y=32/8
y=4
Thus, the required co-ordinates of the point of intersection are (2,4) .
Que-12: Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Ans: The co-ordinates of every point on the line y = 2 will be of the type (x, 2).
Using section formula, we have:
y=m1×(-3)+m2×5/m1+m2
2=-3m1+5m2/m1+m2
2m1+2m2=-3m1+5m2
2m1+3m1=5m2-2m2
5m1=3m2
m1/m2=3/5
Thus, the required ratio is 3 : 5.
Que-13: The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.
Ans: From the figure, the line AB intersects x-axis at A and y-axis at B.
Let the co-ordinates of A (x, 0) and B (0, y) and P (5, –4) divides it in the ratio of 2 : 5
∴ 5=2×0+5×x/2+5
5=0+5x/7
∴ 5x = 35
x=35/5
x = 7
Again, -4=2×y+5×0/2+5
-4=2y+0/7
∴ 2y = –4 × 7 = –28
y=-28/2
y = -14
∴ Co-ordinates of A are (7, 0) and of B are (0, -14)
Que-14: Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Ans: Let P and Q be the point of trisection of the line segment joining the points A (–3, 0) and B (6, 6).
So, AP = PQ = QB
We have AP : PB = 1 : 2
Co-ordinates of the point P are
(1×6+2×(-3)/1+2,1×6+2×0/1+2)
= (6-6/3,6/3)
= (0, 2)
We have AQ : QB = 2 : 1
Co-ordinates of the point Q are
(2×6+1×(-3)/2+1,2×6+1×0/2+1)
= (9/3,12/3)
= (3, 4)
Que-15: Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Ans: Let the points A (-5, 8) and B (10, -4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
AP = PQ = QB
∴ AP : PB = 1 : 2 and AQ : QB = 2 : 1
Now, co-ordinates of P will be,
x=1×10+2×(-5)/1+2
= 10-10/3
= 0
y = 1×(-4)+2×8/1+2
= -4+16/3
= 12/3 = 4
∴ Co-ordinates of P are (0, 4)
Co-ordinates of Q will be,
x=2×10+1×(-5)/2+1
= 20-5/3
= 15/3 = 5
y=2×(-4)+1×8/2+1
= -8+8/3 = 0/3 = 0
∴ Co-ordinates of Q are (5, 0)
Que-17: Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Ans: Let A and B are the points of trisection of the line segment joining the points P (2, 1) and
Q (5, -8), then
PA = AB = BQ.
PA : AQ = 1 : 2 and PB : BQ = 2 : 1
Co-ordinates of the point A are
(1×5+2×2/1+2,1×(-8)+2×1/1+2)
= (9/3,-6/3)
= (3, −2)
Hence, A (3, −2) is a point of trisection of PQ.
We have PB : BQ = 2 : 1
Co-ordinates of the point B are
(2×5+1×2/2+1,2×(-8)+1×1/2+1)
(10+2/3,-16+1/3)
= (4, −5)
Que-17: If A = (-4, 3) and B = (8, -6)
(i) find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis ?
Ans:
(i)
A (−4, 3) and B (8, −6)
AB=√(x2-x1)²+(y2-y1)²
= √144+81 = 15 units
(ii)
Let P be the point, which divides AB on the x-axis in the ratio k : 1.
Therefore, y-co-ordinate of P = 0.
-6k+3/k+1=0
-6k+3=0
k=1/2
∴ Required ratio is 1 : 2
Que-18: The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Ans: Since, point L lies on y-axis, its abscissa is 0.
Let the co-ordinates of point L be (0, y).
Let L divides MN in the ratio k : 1.
Using section formula, we have:
x=k×(-3)+1×5/k+1
0=-3k+5/k+1
-3k+5=0
k=5/3
Thus, the required ratio is 5 : 3.
Now, y=k×2+1×7/k+1
= 5/3×2+7/(5/3)+1
= 10+21/5+3
= 31/8
Que-19: A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that:
AP : PB = AQ : QC = 1 : 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = BC.
Ans:
(i)
Co-ordinates of P are
(1×(-1)+2×2/1+2,1×2+2×5/1×2)
= (3/3,12/3)
= (1, 4)
Co-ordinates of Q are
(1×5+2×2/1+2,1×8+2×5/1+2)
= (9/3,18/3)
= (3, 6)
(ii)
Using distance formula, we have:
BC=√(5+1)²+(8-2)²
BC=√36+36
BC=6√2
PQ=√(3-1)²+(6-4)²
PQ=√4+4
PQ=2√2
Hence, PQ = 1/3 BC
Que-20: A (-3, 4), B ( 3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2 : 3.
Ans: BP : PC = 2 : 3
Co-ordinates of P are
(2×(-2)+3×3/2+3,2×4+3×(-1)/2+3)
= (-4+9/5,8-3/5)
= (1, 1) …(i)
Using distance formula, we have:
AP=√(x2-x1)²+(y2-y1)²
Let A (–3 , 4) x1 = –3, y1 = 4,
(1, 1) x2 = 1, y2 = 1 …[From (i) we get]
AP=√(1+3)²+(1-4)²
= √16+9
= √25
= 5 units.
Que-21: The line segment joining A (2, 3)and B(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K. [1990, 2006]
Ans: Since, point K lies on x-axis, its ordinate is 0.
Let the point K (x, 0) divides AB in the ratio k : 1.
We have,
y=k×(-5)+1×3/k+1
0=-5k+3/k+1
k=3/5
Thus, K divides AB in the ratio 3 : 5.
Also, we have:
x=k×6+1×2/k+1
x=3/5×6+2/(3/5)+1
x=18+10/3+5
x=28/8=7/2
Thus, the co-ordinates of the point K are (7/2 , 0)
Que-22: The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Ans: Points A (4, -7), B (-6, -2) are joined which intersects y-axis at K. abscissa of K will be 0
Let the point K (0, y) divides AB in the ratio k : 1
We have,
x=k×(-6)+1×4/k+1
0=-6k+4/k+1
k=4/6=2/3
Thus, K divides AB in the ratio 2 : 3.
Also, we have:
y=k×(-2)+1×7/k+1
y=-2k+7/k+1
y=-2×(2/3)+7/(2/3)+1
y=-4+21/2+3
y=17/5
Thus, the co-ordinates of the point K are (0 , 17/5)
Que-23: The line joining P (-4, 5) and Q (3, 2), intersects they axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) The ratio PR: RQ.
(ii) The co-ordinates of R.
(iii) The areas of the quadrilateral PMNQ. [2004]
Ans:
(i).
Let point R (0, y) divides PQ in the ratio k : 1.
We have:
x=k×3+1×(-4)/k+1
0=3k-4/k+1
0=3k-4
k=4/3
Thus, PR : RQ = 4 : 3
(ii).
Also, we have:
y=k×2+1×5/k+1
y=2k+5/k+1
y=2×(4/3)+5/(4/3)+1
y=8+15/4+3
y=23/7
Thus, the co-ordinates of point R are (0,23/7)
(iii).
Area of quadrilateral PMNQ
= 1/2×(PM+QN)×MN
= 1/2×(5+2)×7
= 1/2×7×7
= 24.5 sq. units
Que-24: In the given figure, line APB meets the x- axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.
Ans: Given, A lies on x-axis and B lies on y-axis.
Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.
Given, P is the point (−4, 2) and AP : PB = 1 : 2.
Using section formula, we have:
-4=1×0+2×x/1+2
-4=2x/3
x=-4×3/2=-6
Also,
2=1×y+2×0/1+2
2=y/3
y=6
Thus, the co-ordinates of points A and B are (−6, 0) and (0, 6) respectively.
Que-25: Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Ans:
(i).
Let the required ratio be m1 : m2
Consider A(−4, 6) = (x1, y1); B(8, −3) = (x2 , y2) and let
P(x, y) be the point of intersection of the line segment and the y-axis
By section formula, we have,
x=m1x2+m2x1/m1+m2,y=m1y2+m2y1/m1+m2
x=8m1-4m2/m1+m2,y=-3m1+6m2/m1+m2
The equation of the y-axis is x = 0
x=8m1-4m2/m1+m2=0
8m1-4m2=0
8m1=4m2
m1/m2=4/8
m1/m2=1/2
(ii).
From the previous subpart, we have,
m1/m2=1/2
m1 = k and m2 = 2k, where k
Is any constant.
Also, we have,
x=8m1-4m2/m1+m2,y=-3m1+6m2/m1+m2
x=8×k-4×2k/k+2k,y=-3×k+6×2k/k+2k
x=8k-8k/3k,y=-3k+12k/3k
x=03/k,y=9k/3k
x = 0, y = 3
Thus, the point of intersection is p (0, 3)
(iii).
The length of AB = distance between two points A and B.
The distance between two given points
A(x1, y1) and B(x2, y2) is given by,
Distance AB = √(x2-x1)²+(y2-y1)²
= √(8+4)²+(-3-6)²
= √144+81
= √225
= 15 units
Que-26: If P(-b, 9a – 2) divides the line segment joining the points A(-3, 3a + 1) and B(5, 8a) in the ratio 3: 1, find the values of a and b.
Ans: Take (x1 , y1) = (-3, 3a + 1) ; (x2 , y2) = B(5, 8a) and
And (x, y) = (–b, 9a – 2)
Here m1 = 3 and m2 =1
Coordinate of P(x, y) = (m1x2+m2x1/m1-m2,m1y2+m2y1/m1+m2)
x=m1x2+m2x1/m1-m2 and y=m1y2+m2y1/m1+m2
-b=3×5+1×(-3)/3+1 and 9a-2=3×8a+1(3a+1)/3+1
-b=15-3/4 and 9a-2=24a+3a+1/4
–4b = 12 and 36a – 8 = 27a + 1
b = –3 and 9a = 9
a = 1 and b = –3) .
— : Section and Mid-Point Formula Exe-13A Class-10 Concise ICSE Maths Solution Ch-13 :–
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