Tenth Grade Mathematics: Section Formula: Internal and External Division

While the distance formula calculates the total length of a segment joining two points, geometric construction often requires dividing that segment into a prescribed ratio. The Section Formula provides the algebraic coordinates of a dividing point that partitions a line segment internally or externally in a given ratio m:n. Derived using the properties of similar right triangles, the section formula is indispensable for locating centers of mass, calculating midpoints, and performing trisection proofs.

Derivation of the Section Formula for Internal Division

Let A(x1, y1) and B(x2, y2) be two distinct points in the coordinate plane. Let P(x, y) be a point lying on the segment AB that divides it internally in the ratio m1 : m2:

Ratio Condition:
                       AP / PB = m1 / m2

Geometric Construction of Similar Triangles:
                    y
                    |                            B(x2, y2)
                    |                            +
                    |                           /|
                    |              m1          / |
                    |       P(x, y) +---------+ C| (y2 - y)
                    |              /|         |  |
                    |             / | (y - y1)|
                    |    A(x1,y1)+--+ R       |
                    |            |  |         |
                    +----+-------+--+---------+--- x
                         x1      x            x2

From points A, P, and B, drop perpendiculars to the horizontal line parallel to the x-axis to form two right triangles: triangle APR and triangle PBC. The horizontal side of triangle APR is AR = x - x1. The vertical side of triangle APR is PR = y - y1. The horizontal side of triangle PBC is PC = x2 - x. The vertical side of triangle PBC is BC = y2 - y.

Because both triangles are right-angled and their corresponding sides are parallel, triangle APR is similar to triangle PBC (by AA similarity). Therefore, the ratios of their corresponding sides are equal:

AP / PB = AR / PC = PR / BC

Substituting the coordinate expressions and the ratio m1/m2:

m1 / m2 = (x - x1) / (x2 - x)
m1 * (x2 - x) = m2 * (x - x1)
m1*x2 - m1*x = m2*x - m2*x1
(m1 + m2)*x = m1*x2 + m2*x1
x = (m1*x2 + m2*x1) / (m1 + m2)

Similarly, equating the vertical ratio:

m1 / m2 = (y - y1) / (y2 - y)
m1 * (y2 - y) = m2 * (y - y1)
m1*y2 - m1*y = m2*y - m2*y1
(m1 + m2)*y = m1*y2 + m2*y1
y = (m1*y2 + m2*y1) / (m1 + m2)

Theorem (Internal Section Formula):
The coordinates of the point P(x, y) dividing the line segment joining A(x1, y1)
and B(x2, y2) internally in the ratio m1 : m2 are:
               ( m1*x2 + m2*x1     m1*y2 + m2*y1 )
        P(x,y) = ( --------------- , --------------- )
               (    m1 + m2             m1 + m2    )

External Division

If the point P lies outside the line segment AB on the extended ray such that AP : PB = m1 : m2 (with m1 != m2), P is said to divide AB externally.

Theorem (External Section Formula):
The coordinates of the external dividing point are obtained by replacing m2 with -m2:
               ( m1*x2 - m2*x1     m1*y2 - m2*y1 )
        P(x,y) = ( --------------- , --------------- )
               (    m1 - m2             m1 - m2    )

Finding the Ratio in Which a Point or Axis Divides a Segment

When the dividing point (or the dividing line/axis) is known and the ratio is sought, working with two separate variables m1 and m2 is inefficient. Instead, we assume the ratio to be k : 1.

The k : 1 Technique:
Let the ratio m1 : m2 be k : 1. The coordinates of P become:
           P = ( (k*x2 + x1)/(k + 1) ,  (k*y2 + y1)/(k + 1) )
- If divided by the x-axis: set the y-coordinate equal to 0 and solve for k.
- If divided by the y-axis: set the x-coordinate equal to 0 and solve for k.
- If k is positive, division is internal; if k is negative, division is external.

Chapter Exercises

Exercise 1: Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3 internally.

Exercise 2: Find the coordinates of the points of trisection (i.e., points dividing into three equal parts) of the line segment joining (4, -1) and (-2, -3).

Exercise 3: Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

Solutions and Explanations

Solution 1: Let the given points be A(-1, 7) and B(4, -3), with ratio m1 = 2 and m2 = 3. Using the internal section formula: x = (m1x2 + m2x1) / (m1 + m2) = (2(4) + 3(-1)) / (2 + 3) = (8 - 3) / 5 = 5 / 5 = 1. For y: y = (m1y2 + m2y1) / (m1 + m2) = (2(-3) + 3(7)) / (2 + 3) = (-6 + 21) / 5 = 15 / 5 = 3. Thus, the coordinates of the dividing point are (1, 3).

Solution 2: Trisection partitions the segment into three equal parts: AP = PQ = QB. Point P divides AB internally in the ratio 1 : 2. Point Q divides AB internally in the ratio 2 : 1. Given A(4, -1) and B(-2, -3). For point P (ratio 1:2): x_P = (1(-2) + 2(4)) / (1 + 2) = (-2 + 8) / 3 = 6 / 3 = 2; y_P = (1(-3) + 2(-1)) / (1 + 2) = (-3 - 2) / 3 = -5/3. Thus, P has coordinates (2, -5/3). For point Q (ratio 2:1): x_Q = (2(-2) + 1(4)) / (2 + 1) = (-4 + 4) / 3 = 0; y_Q = (2(-3) + 1(-1)) / (2 + 1) = (-6 - 1) / 3 = -7/3. Thus, Q has coordinates (0, -7/3). The two points of trisection are (2, -5/3) and (0, -7/3).

Solution 3: Any point on the y-axis has an x-coordinate of 0. Let the point on the y-axis be P(0, y), and suppose it divides the segment joining A(5, -6) and B(-1, -4) in the ratio k : 1. By the section formula, the x-coordinate of P is: x = (k(-1) + 1(5)) / (k + 1). Since x = 0: (-k + 5) / (k + 1) = 0 ==> -k + 5 = 0 ==> k = 5. Therefore, the y-axis divides the line segment internally in the ratio 5 : 1. Now find the y-coordinate of the point of intersection: y = (k(-4) + 1(-6)) / (k + 1) = (5*(-4) + (-6)) / (5 + 1) = (-20 - 6) / 6 = -26 / 6 = -13 / 3. Thus, the point of intersection on the y-axis is (0, -13/3).