Junior-Class Mentor, TFI Fellow | Updated on - Jun 29, 2026
The NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 cover all 10 questions on the section formula, midpoint formula, trisection, axis-division, finding unknown vertices, and the area of a rhombus. Every answer is solved step by step for the 2026-27 CBSE syllabus.
Questions covered: 10 in total (Q1 to Q10), ranging from direct section-formula use to real-life flag problems and parallelogram/rhombus problems.
Core formulas: section formula P = m1x2 + m2x1m1 + m2 and midpoint formula M = x1 + x22.
CBSE board value: Exercise 7.2 questions carry 2 to 4 marks in the Class 10 board paper, with the section formula and midpoint applications tested almost every year.
Solved by Collegedunia: Every Exercise 7.2 question below is solved by Mathematics subject experts, checked against the official 2026-27 NCERT textbook, and written with full working so each section formula step and each midpoint computation earns its marks in the CBSE Class 10 board paper.
What Exercise 7.2 of Coordinate Geometry Covers for Class 10
Exercise 7.2 is the section formula and midpoint exercise of Chapter 7. It moves from measuring distances (Exercise 7.1) to locating dividing points on a segment. The exercise builds on two key ideas: the section formula for internal division in a given ratio, and the midpoint formula as a special case when the ratio is 1:1.
Q1 to Q2: direct use of the section formula to find division points and trisection points.
Q3: a real-life word problem on Sports Day flags, combining the distance formula with the midpoint formula.
Q4 to Q5: reverse problems - find the ratio in which a given point divides a segment, and find where an axis cuts a segment.
Q6 to Q7: use the diagonal-bisection property of a parallelogram and the midpoint property of a circle diameter.
Q8 to Q9: locate points that sit a given fraction along a segment, and divide a segment into four equal parts.
Q10: find the area of a rhombus using the diagonal-length formula with coordinates.
How to Apply the Section Formula in Exercise 7.2 Questions
The section formula locates a point that divides a segment from A(x1, y1) to B(x2, y2) internally in the ratio m1 : m2. The steps are the same every time.
Step
What you do
Example: ratio 2:3, A(−1,7), B(4,−3)
1
Identify x1, y1, x2, y2, m1, m2
x1=−1, y1=7, x2=4, y2=−3, m1=2, m2=3
2
Apply x = m1x2 + m2x1m1 + m2
x = 2(4)+3(−1)5 = 55 = 1
3
Apply y = m1y2 + m2y1m1 + m2
y = 2(−3)+3(7)5 = 155 = 3
4
Write the point
Division point = (1, 3)
Key rule: In the ratio m1:m2, the part m1 always multiplies the far endpoint B and m2 multiplies the near endpoint A. Mixing these two up is the most common mistake in Exercise 7.2.
The midpoint formula is just the section formula with ratio 1:1. Substituting m1=m2=1 gives M = x1+x22, y1+y22, which is the formula used in Q3, Q7, Q9, and Q10.
How to Find the Ratio in which a Point Divides a Segment (Reverse Section Formula)
Questions Q4 and Q5 give you the dividing point and ask for the ratio. The trick is to write the ratio as k:1 and solve one equation for k.
Question
Key idea
Answer
Q1
Section formula, ratio 2:3, A(−1,7), B(4,−3)
(1, 3)
Q2
Trisection = two uses of the section formula (1:2 and 2:1)
(2, −5/3) and (0, −7/3)
Q3
Sports Day flags: distance formula + midpoint for blue flag
√61 m apart; blue flag at (5, 22.5)
Q4
Reverse section formula: set ratio as k:1, solve for k from x-coord
2:7
Q5
x-axis forces y = 0; solve for k, then find x
1:1 at (−3/2, 0)
Q6
Diagonals of a parallelogram bisect each other: equal midpoints
x = 6, y = 3
Q7
Centre = midpoint of diameter: reverse midpoint to get far end
A(3, −10)
Q8
AP = 3/7 AB means ratio 3:4; use section formula
(−2/7, −20/7)
Q9
Four equal parts = three midpoints; find Q first, then P and R
(−1, 7/2), (0, 5), (1, 13/2)
Q10
Rhombus area = (1/2) d1 d2; find both diagonals with the distance formula
24 square units
Watch out: In Q4 and Q5, using the x-coordinate to find the ratio is usually simpler. Always verify by substituting back into the y-coordinate formula to confirm the point lies on the segment.
Exercise 7.2 Previous Year Questions and CBSE Board Weightage
Exercise 7.2 concepts appear in the CBSE Class 10 board paper almost every year. The table below shows the typical patterns.
Year
Question type from Exercise 7.2
Marks
2024
Find the ratio of division using the section formula
2
2023
Find the coordinates of a point dividing a segment in a given ratio
3
2022
Midpoint of diameter / parallelogram vertex
2
2021
Trisection of a line segment
3
2020
Axis division problem (where does the x-axis cut a segment)
3
Students who score full marks on Exercise 7.2 board questions consistently do two things: identify the ratio correctly before substituting into the section formula, and verify with the second coordinate when the ratio is the final answer.
Other Resources for This Chapter: Class 10 Maths Coordinate Geometry
Move between the other resources for Chapter 7 and the other exercise of Coordinate Geometry below.
All NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2 with Step-by-Step Solutions
Exercise 7.2
Q 7.1
Find the coordinates of the point which divides the join of
(-1,7) and (4,-3) in the ratio 2:3.
Concept used. The section formula gives the point
P(x,y) that divides the segment from A(x1,y1) to B(x2,y2)
internally in the ratio m1:m2:
P=(m1x2+m2x1m1+m2, m1y2+m2y1m1+m2).
Here A(-1,7), B(4,-3) and m1:m2=2:3.
Identify the values: x1=-1, y1=7, x2=4, y2=-3,
m1=2, m2=3, so m1+m2=5.
Find the x-coordinate:
x=2(4)+3(-1)5=8-35=55=1.
Find the y-coordinate:
y=2(-3)+3(7)5=-6+215=155=3.
The point is (1,3).
IB
Ishita Banerjee
M.Sc Mathematics, University of Calcutta
Verified Expert
Sanity-check the position. A two-to-three split puts the point a
little nearer the first end, so its coordinates should lean that way.
Read the ratio: the two parts total 5, so the point
sits two-fifths of the way along from the first end toward the second.
Walk the fraction: moving two-fifths along gives
x=-1+25(4-(-1))=1 and y=7+25(-3-7)=3.
Confirm: this fraction-of-the-way method lands on the
same point as the section formula, so the two agree.
Why bother: reading the ratio as a fraction of the
journey is a quick mental check that the point lands on the
correct side of the segment and not beyond either end.
(1,3), lying two-fifths of the way from (-1,7) to (4,-3).
Q 7.2
Find the coordinates of the points of trisection of the line
segment joining (4,-1) and (-2,-3).
Concept used. The points of trisection cut a segment
into three equal parts. If P and Q are these points with AP=PQ=QB,
then P divides AB in the ratio 1:2 and Q divides it in 2:1.
Apply the section formula twice.
Take A(4,-1) and B(-2,-3). For P, use m1:m2=1:2:
P=(1(-2)+2(4)1+2, 1(-3)+2(-1)1+2).
Simplify P:
P=(-2+83, -3-23)=(63, -53)=(2, -53).
For Q, use m1:m2=2:1:
Q=(2(-2)+1(4)2+1, 2(-3)+1(-1)2+1).
Simplify Q:
Q=(-4+43, -6-13)=(03, -73)=(0, -73).
The points of trisection are (2,-53) and (0,-73).
SG
Sanjay Gupta
M.Sc Mathematics, University of Delhi
Verified Expert
Picture the two cut points. The first cut sits one-third along
the segment and the second sits two-thirds along, so both coordinates
should step evenly between the ends.
Step the x: the full change in x is -6, so each
third changes it by -2. Starting at 4 the cuts fall at 2 and
then 0, ending at -2, which matches the computed values.
Step the y: the full change in y is -2, so each
third changes it by -23. Starting at -1 the cuts
fall at -53 and then -73.
Pair them: reading the two coordinate lists together
gives the same two trisection points found by the section formula.
The hallmark: equal steps in both coordinates are the
signature of trisection, and checking them catches any ratio
mix-up at once before it costs marks.
(2,-53) and (0,-73).
Q 7.3
To conduct Sports Day activities, in your rectangular shaped
school ground ABCD, lines have been drawn with chalk powder at a
distance of 1 m each. 100 flower pots have been placed at a distance
of 1 m from each other along AD, as shown in Fig. 7.12. Niharika
runs 14th the distance AD on the 2nd line and posts a
green flag. Preet runs 15th the distance AD on the eighth
line and posts a red flag. What is the distance between both the flags?
If Rashmi has to post a blue flag exactly halfway between the line
segment joining the two flags, where should she post her flag?
Concept used. Read the ground as a coordinate grid: the line
number gives the x-coordinate and the distance run along AD gives
the y-coordinate. Then the distance formula finds the gap
between the flags and the mid-point formula locates the blue
flag.
Niharika's green flag: on the 2nd line, 14 of
AD=14× 100=25 m up. So the green flag is at
G(2,25).
Preet's red flag: on the 8th line, 15 of
AD=15× 100=20 m up. So the red flag is at
R(8,20).
Distance between the flags:
GR=√(8-2)2+(20-25)2=√62+(-5)2=√36+25=√61m.
Blue flag is the mid-point of GR:
(2+82, 25+202)=(102, 452)=(5, 22.5).
So Rashmi posts the blue flag on the 5th line, 22.5 m up
AD.
Distance between the flags =√61 m; the blue flag goes at (5, 22.5), that is, on the 5th line at 22.5 m.
LS
Lakshmi Subramaniam
B.Tech Mechanical Engineering, NIT Surathkal
Verified Expert
Set up the model carefully. The only real work is turning the
story into points; after that the formulas are mechanical.
Green flag: one-quarter of the 100 pots is 25 m,
paired with line 2, which fixes the green flag at (2,25).
Red flag: one-fifth of the same length is 20 m,
paired with line 8, which fixes the red flag at (8,20).
Apply the formulas: the horizontal gap is 6 and the
vertical gap is 5, so the flags are √61 m apart, and the
mid-point places the blue flag squarely between the two.
Read the answer: a halfway flag always lands at the
mid-point, so it sits on line 5 at the listed height, equally far
from both runners. Careful modelling of the words earns the method
marks here, not the arithmetic.
Flags are √61 m apart; the blue flag is at (5, 22.5).
Q 7.4
Find the ratio in which the line segment joining the points
(-3,10) and (6,-8) is divided by (-1,6).
Concept used. Let the dividing point split the segment in the
ratio k:1. By the section formula the x-coordinate of the dividing
point is
x=k x2+x1k+1.
Equating this to the known x-coordinate gives one equation in k.
Let (-1,6) divide the join of A(-3,10) and B(6,-8) in the
ratio k:1. Using the x-coordinate:
-1=k(6)+1(-3)k+1.
Clear the denominator:
-1(k+1)=6k-3 -k-1=6k-3.
Collect the k terms:
-1+3=6k+k 2=7k k=27.
So the ratio k:1=27:1=2:7.
The point (-1,6) divides the segment in the ratio 2:7.
DS
Deepak Sharma
M.Sc Mathematics, Banaras Hindu University
Verified Expert
Confirm with the y-coordinate. Finding the ratio from x
alone scores, but the y-coordinate proves the point lies on the segment.
Set up: with the ratio fixed, the y-coordinate from the
section formula is
k(-8)+10k+1=27(-8)+1027+1.
Simplify: the top reduces to 547 and the
bottom to 97 after clearing the small fractions.
Divide: the two parts cancel to give 6, which matches
the given y-coordinate, so the ratio is correct.
The habit: checking the second coordinate is the standard
guard against a point that satisfies x but lies off the line.
Always run it when the ratio is the final answer.
Ratio 2:7, confirmed by the matching y-coordinate 6.
Q 7.5
Find the ratio in which the line segment joining A(1,-5)
and B(-4,5) is divided by the x-axis. Also find the coordinates of
the point of division.
Concept used. A point on the x-axis has y-coordinate 0.
Writing the dividing point as k:1 and setting its y-coordinate to
0 gives k; then the x-formula gives the point.
Let the x-axis cut AB in the ratio k:1. The
y-coordinate of the division point is
y=k(5)+1(-5)k+1.
Set y=0 because the point lies on the x-axis:
5k-5k+1=0 5k-5=0 k=1.
The ratio is k:1=1:1, so the x-axis cuts AB at its
mid-point.
Find the x-coordinate with k=1:
x=k(-4)+1(1)k+1=-4+12=-32.
The x-axis divides AB in the ratio 1:1 at the point (-32, 0).
AD
Anjali Desai
M.Sc Mathematics, Gujarat University
Verified Expert
Read the sign change as a clue. One end sits below the axis and
the other an equal amount above it, so the crossing must be halfway.
Spot the symmetry: one point is 5 below the x-axis
and the other is 5 above it, so the axis sits midway between
them in the vertical direction.
Predict the ratio: midway means a ratio of one-to-one,
which the equation 5k-5=0 confirms with the value k=1.
Find the point: the mid-point x-coordinate is
1+(-4)2=-32, giving the crossing point on the axis.
The shortcut: whenever the two y-values are equal and
opposite, the axis always bisects the segment, so you can predict
the one-to-one ratio and use the calculation purely as a check.
Ratio 1:1; point of division (-32,0).
Q 7.6
If (1,2), (4,y), (x,6) and (3,5) are the vertices of a
parallelogram taken in order, find x and y.
Concept used. The diagonals of a parallelogram bisect
each other. So the mid-point of one diagonal equals the mid-point of
the other. With vertices A,B,C,D in order, the diagonals are AC and
BD, and their mid-points must match.
Label in order: A(1,2), B(4,y), C(x,6), D(3,5). Diagonal
AC joins A and C; diagonal BD joins B and D.
Mid-point of AC=(1+x2,2+62) and
mid-point of BD=(4+32,y+52).
Set them equal.
Equate the x-coordinates:
1+x2=4+32=72 1+x=7 x=6.
Equate the y-coordinates:
2+62=y+52 8=y+5 y=3.
x=6 and y=3.
SK
Suresh Kumar
M.Sc Mathematics, University of Madras
Verified Expert
Why the diagonal property wins here. The mid-point method needs
only one equation per unknown and avoids the side-vector notation that
this class has not yet covered.
Shared centre: the two diagonals meet at one common
mid-point, found from the known coordinates of three vertices, so
both diagonals must pass through it.
Solve for x: forcing the first diagonal to pass through
that shared centre gives a single equation that fixes the value of x.
Solve for y: forcing the second diagonal through the
same centre gives a second equation that fixes the value of y.
The takeaway: the single fact that diagonals bisect each
other resolves both unknowns cleanly, which is why it is the
preferred approach for parallelogram problems at this level.
x=6, y=3, from the equal mid-points of the diagonals.
Q 7.7
Find the coordinates of a point A, where AB is the
diameter of a circle whose centre is (2,-3) and B is (1,4).
Concept used. The centre of a circle is the mid-point
of any diameter. So if O(2,-3) is the centre and B(1,4) is one end
of the diameter, then O is the mid-point of AB. Using the mid-point
formula in reverse gives A.
Let A(x,y). Since O is the mid-point of AB:
(x+12,y+42)=(2,-3).
Equate the x-coordinates:
x+12=2 x+1=4 x=3.
Equate the y-coordinates:
y+42=-3 y+4=-6 y=-10.
The point is A(3,-10).
PM
Pooja Malhotra
M.Sc Mathematics, Panjab University
Verified Expert
Visualise the centre as the balance point. The centre is equally
far from both ends of the diameter, so one step repeated lands on the
far end.
First step: go from the known end to the centre, a
change of one across and seven down between the two points.
Repeat it: apply the same change again starting from the
centre, which lands exactly on the unknown far end.
Cross-check: that endpoint agrees with the value found by
the mid-point calculation, so both methods confirm each other.
Why it works: seeing the centre as the mid-point, and a
diameter as two equal steps, turns this into a one-move problem
that doubles as a built-in check.
A(3,-10).
Q 7.8
If A and B are (-2,-2) and (2,-4), respectively, find
the coordinates of P such that AP=37AB and P lies on the
line segment AB.
Concept used. If AP=37AB, then P covers
37 of the segment from A, so PB=AB-AP=47AB.
Hence AP:PB=3:4, and P divides AB internally in the ratio 3:4.
Apply the section formula.
Convert the fraction to a ratio: AP:PB=37:47=3:4,
so m1=3, m2=4 with A(-2,-2), B(2,-4).
Find the x-coordinate:
x=3(2)+4(-2)3+4=6-87=-27.
Find the y-coordinate:
y=3(-4)+4(-2)3+4=-12-87=-207.
P=(-27, -207).
RP
Ravi Pillai
M.Sc Mathematics, University of Kerala
Verified Expert
Use the fraction-of-the-way form. Because the point sits a
fixed fraction of the way along, you can add that fraction of the
coordinate change directly to the start.
The change: the move from start to end is
(2-(-2), -4-(-2))=(4,-2), the total shift across the segment.
Scale it: take three-sevenths of that shift to get
(127,-67), the partial move to the point.
Add to the start: adding this to the starting point
lands exactly on the required point.
Why use it: this method skips converting the fraction to
a ratio altogether, yet lands on the same point, which makes it a
reliable double-check on the section-formula answer.
P=(-27,-207).
Q 7.9
Find the coordinates of the points which divide the line
segment joining A(-2,2) and B(2,8) into four equal parts.
Concept used. Three points split a segment into four equal
parts. Call them P, Q, R. Then Q is the mid-point of AB, P
is the mid-point of AQ, and R is the mid-point of QB. We use the
mid-point formula three times.
Mid-point Q of A(-2,2) and B(2,8):
Q=(-2+22,2+82)=(0,5).
P is the mid-point of A(-2,2) and Q(0,5):
P=(-2+02,2+52)=(-1,72).
R is the mid-point of Q(0,5) and B(2,8):
R=(0+22,5+82)=(1,132).
The three points are P(-1,72), Q(0,5) and R(1,132).
NA
Neha Agarwal
M.Sc Mathematics, University of Lucknow
Verified Expert
Equal steps confirm the answer. Four equal parts mean each
coordinate rises by the same amount at every cut from start to end.
Step the x: the total x-change of 4 over four parts
makes each step +1, so the values run -2,-1,0,1,2 and the three
interior values are -1,0,1.
Step the y: the total y-change of 6 over four parts
makes each step +32, so the interior values are
72,5,132.
Pair them: matching the interior x and y values in
order gives the three division points, the same ones the mid-point
method produces.
Why it is fast: reading the points as a steady
progression in both coordinates lists all three at once and
verifies them in a single glance.
P(-1,72), Q(0,5), R(1,132).
Q 7.10
Find the area of a rhombus if its vertices are (3,0), (4,5),
(-1,4) and (-2,-1) taken in order.
[Hint: Area of a rhombus =12(product of its diagonals)]
Concept used. The area of a rhombus equals half the
product of its diagonals: Area=12 d1 d2. With
vertices A,B,C,D in order, the diagonals are AC and BD, whose
lengths come from the distance formula.
Use the diagonals, not the four sides. The hint points straight
to half the product of the diagonals, so find those two lengths and skip
the longer route through equal sides and an angle.
First diagonal: the length 4√2 links the opposite
vertices, jumping across the figure rather than along an edge.
Second diagonal: the length 6√2 links the other
pair of opposite vertices in the same way.
Combine: half the product of the two lengths gives the
area as 24 square units, the final answer.
The one decision: the only judgement call is picking the
diagonals as opposite vertices rather than adjacent ones. Once
chosen, the surds collapse neatly since root two times root two is two.
24 square units.
Student Feedback
Out of 14,800 students surveyed before the 2026 boards, 78% said the section formula questions in Exercise 7.2 were straightforward once they memorised which ratio multiplies which endpoint, but the axis-division and parallelogram vertex problems caught many off guard. 4 out of 5 students who scored full marks said they always verified the result by checking the second coordinate.
Source: Collegedunia Class 10 Maths student survey, 2026 board cohort.
Coordinate Geometry Class 10 Maths Exercise 7.2 NCERT Solutions FAQs
Ques. How many questions are there in Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2?
Ans. Exercise 7.2 has 10 questions. The questions cover the section formula, trisection, midpoint formula, finding ratios in which a point divides a segment, axis-division problems, parallelogram vertex problems, diameter-midpoint problems, and the area of a rhombus using coordinates. All questions are based on the 2026-27 CBSE syllabus.
Ques. What is the section formula used in Exercise 7.2?
Ans. The section formula gives the coordinates of a point P that divides the segment from A(x1, y1) to B(x2, y2) internally in the ratio m1:m2. The formula is x = (m1x2 + m2x1) / (m1 + m2) and y = (m1y2 + m2y1) / (m1 + m2). The midpoint formula is a special case with m1 = m2 = 1.
Ques. How do you find the ratio in which a point divides a segment (reverse section formula) in Exercise 7.2?
Ans. To find the ratio, write it as k:1 and substitute the known x-coordinate of the dividing point into the section formula. This gives one equation in k. Solve for k to get the ratio k:1. Always verify the result by checking the y-coordinate separately. This method is used in Q4 and Q5 of Exercise 7.2.
Ques. What is the difference between AP = 3/7 AB and ratio 3:7 in Exercise 7.2 Question 8?
Ans. If AP = (3/7)AB, then P covers 3/7 of the segment from A. The remaining part PB = AB - AP = (4/7)AB. So the ratio AP:PB = 3:4, not 3:7. Using ratio 3:7 is the most common mistake in Q8. Always convert the given fraction to a ratio between AP and PB before applying the section formula.
Ques. Are these Exercise 7.2 solutions based on the 2026-27 CBSE syllabus?
Ans. Yes. These solutions follow the current 2026-27 syllabus for Class 10 Mathematics. Exercise 7.2 on the section formula and midpoint formula is fully retained in the latest NCERT edition and is directly tested in the CBSE Class 10 board paper, with questions typically carrying 2 to 4 marks.
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