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Section Formula and Midpoint

🎓 Class 10 Mathematics CBSE Theory Ch 7 — Coordinate Geometry ⏱ ~35 min 📢 Class 10 ચેનલમાં જોડાઓ
🌐 ભાષા:

આ MCQ મોડ્યુલ આના પર આધારિત છે: Section Formula and Midpoint

આ મૂલ્યાંકન આના પર આધારિત હશે: Section Formula and Midpoint
Class 10 સ્તર, Coordinate Geometry માં, Intermediate કઠિનતા સાથે લક્ષ્ય.

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

7.3 Section Formula

Suppose a telephone company wants to place a relay tower at P between two towns A and B so that it covers both with equal reception. Because A and B use different tariffs, the tower must divide AB in the ratio \(m_1:m_2\). Where exactly is P?

Let A(\(x_1,y_1\)) and B(\(x_2,y_2\)) be given, and let P(x, y) divide AB internally in ratio \(m_1:m_2\). Drop perpendiculars AR, PS, BT to the x-axis. Draw AQ ⊥ PS and PC ⊥ BT (see Fig. 7.9).

XYO A(x₁,y₁) P(x, y) B(x₂,y₂) RST QC m₁m₂
Fig. 7.9 — Deriving the section formula by similar triangles

Derivation

Triangles APQ and PBC are similar (AA). Hence

\[\frac{AP}{PB}=\frac{AQ}{PC}=\frac{PQ}{BC}=\frac{m_1}{m_2}.\]

Also \(AQ=RS=OS-OR=x-x_1\) and \(PC=ST=OT-OS=x_2-x\). Similarly \(PQ=y-y_1\) and \(BC=y_2-y\). Therefore

\[\frac{x-x_1}{x_2-x}=\frac{m_1}{m_2}\Rightarrow m_2(x-x_1)=m_1(x_2-x)\Rightarrow x=\frac{m_1x_2+m_2x_1}{m_1+m_2}.\]

Similarly,

\[y=\frac{m_1y_2+m_2y_1}{m_1+m_2}.\]
Section Formula (Internal Division)
If P divides the segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) internally in the ratio \(m_1:m_2\), then \[\boxed{\;P\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)\;}\]
Special Case — Midpoint
If P is the midpoint of AB, the ratio is 1 : 1. Put \(m_1=m_2=1\): \[M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right).\]
Activity — Experience the Ratio
Materials: graph paper, ruler
Predict: If P divides AB in the ratio 2 : 1 and A(0, 0), B(6, 6), where is P?
  1. Plot A(0, 0) and B(6, 6).
  2. Divide AB visually into three equal parts.
  3. Mark P at two-thirds of the way from A to B.
  4. Apply the section formula and compare.
x = (2·6 + 1·0)/3 = 4, y = (2·6 + 1·0)/3 = 4. So P(4, 4). Visually this is the second gridline from B.

Example 6 — Internal division

Find the coordinates of the point which divides the line segment joining A(4, −3) and B(8, 5) in the ratio 3 : 1 internally.

Solution. Here \(m_1=3,m_2=1\):

\[x=\frac{3(8)+1(4)}{3+1}=\frac{28}{4}=7,\;\;y=\frac{3(5)+1(-3)}{3+1}=\frac{12}{4}=3.\]

Required point: (7, 3).

Example 7 — Find the ratio

In what ratio does the point (−4, 6) divide the line segment joining A(−6, 10) and B(3, −8)?

Solution. Let the ratio be \(m_1:m_2\). Using the x-coordinate:

\[-4=\frac{3m_1-6m_2}{m_1+m_2}\Rightarrow -4m_1-4m_2=3m_1-6m_2\Rightarrow 7m_1=2m_2\Rightarrow m_1:m_2=2:7.\]

(You should verify that the y-coordinate also gives the same ratio.)

Example 8 — On the y-axis

In what ratio does the y-axis divide the line segment joining A(5, −6) and B(−1, −4)? Also find the point of intersection.

Solution. On y-axis, x = 0. Let ratio be k : 1. Then

\[0=\frac{-k+5}{k+1}\Rightarrow k=5.\]

Ratio = 5 : 1. y = \(\frac{5(-4)+1(-6)}{6}=\frac{-26}{6}=-\frac{13}{3}\). Point: (0, −13/3).

Example 9 — Midpoint of a parallelogram diagonal

If A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram ABCD, find p.

Solution. The diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD:

\[\left(\tfrac{6+9}{2},\tfrac{1+4}{2}\right)=\left(\tfrac{8+p}{2},\tfrac{2+3}{2}\right)\Rightarrow\tfrac{15}{2}=\tfrac{8+p}{2}\Rightarrow p=7.\]
Competency-Based Questions
A straight road connects two bus stops A(−2, 4) and B(10, −8). Three shelters are to be placed along the road dividing it into four equal parts.
Q1. Find the coordinates of all three shelters P₁, P₂, P₃.
L3 Apply
P₁ divides AB in 1:3 → (1, 1). P₂ = midpoint → (4, −2). P₃ divides AB in 3:1 → (7, −5).
Q2. Show that the distance from A to P₂ equals the distance from P₂ to B.
L4 Analyse
AP₂ = √(36+36) = 6√2, P₂B = √(36+36) = 6√2. Equal — confirming P₂ is the midpoint.
Q3. Critique: a planner proposes putting the first shelter at (0, 2). Is this P₁? If not, what ratio does (0, 2) divide AB?
L5 Evaluate
For (0, 2): 0 = (10k − 2)/(k+1) ⇒ k = 1/5, so ratio 1:5. Not P₁, which is at 1:3.
Q4. Design a plan with four intermediate shelters (dividing AB in five equal parts). List their coordinates.
L6 Create
Ratios 1:4, 2:3, 3:2, 4:1 ⇒ (0.4, 1.6), (2.8, −0.8), (5.2, −3.2), (7.6, −5.6).
Assertion & Reason
A: The midpoint of the segment joining (1, 2) and (3, 4) is (2, 3).
R: The midpoint formula arises from the section formula with ratio 1 : 1.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. ((1+3)/2, (2+4)/2) = (2, 3). The midpoint is the section formula with m₁ : m₂ = 1 : 1.
A: A point on the x-axis that divides AB internally must satisfy y = 0.
R: Any point on the x-axis has y-coordinate zero.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. The defining property of the x-axis is y = 0; any dividing point on it must have that y-coordinate.
A: The diagonals of a parallelogram always bisect each other.
R: The midpoints of both diagonals of a parallelogram coincide.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. "Diagonals bisect" means they meet at a common midpoint — equivalently the midpoints of both diagonals coincide.

Frequently Asked Questions

What is the midpoint formula?
The midpoint of the segment joining (x1, y1) and (x2, y2) is ((x1 plus x2)/2, (y1 plus y2)/2). It is the section formula with ratio 1:1.
How does the section formula handle external division?
For external division in ratio m : n, replace n with -n in the formula, giving ((m x2 minus n x1)/(m minus n), (m y2 minus n y1)/(m minus n)). Class 10 focuses on internal division.
In what ratio does the y-axis divide a segment?
If a segment joins (x1, y1) and (x2, y2) and the y-axis (x = 0) divides it, the ratio is -x1 : x2. Use the section formula on the x-coordinate set to zero to derive this.
How is the centroid of a triangle found?
The centroid of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is ((x1 plus x2 plus x3)/3, (y1 plus y2 plus y3)/3). It divides each median in the ratio 2:1 from a vertex.
Why is the section formula useful?
It allows dividing a segment in any ratio without geometric construction, a key tool in problems on medians, centroids, trisection points and geometric proofs with coordinates.
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Mathematics Class 10
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નમસ્તે! 👋 હું ગૌરા છું, Section Formula and Midpoint માટે તમારું AI ટ્યુટર. આરામથી પાઠ ભણો — જ્યારે પણ કોઈ શંકા થાય, બસ મને પૂછો! હું મદદ માટે અહીં જ છું.

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