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Section Formula and Midpoint

🎓 Class 10 Mathematics CBSE Theory Ch 7 — Coordinate Geometry ⏱ ~35 min
🌐 Language: [gtranslate]

This MCQ module is based on: Section Formula and Midpoint

This mathematics assessment will be based on: Section Formula and Midpoint
Targeting Class 10 level in Coordinate Geometry, with Intermediate difficulty.

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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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