CBSE Class 10 Mathematics Coordinate Geometry Notes Set 06

Section Formula in Coordinate Geometry

The section formula in coordinate geometry is used to find the point that divides a line segment in a given ratio. The coordinates of the point which divides the line segment AB joining the points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) internally in the ratio \( m_1 : m_2 \) are given by\[ \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \]

Proof. Let \( M(x, y) \) be the point that divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) internally in the ratio \( m_1 : m_2 \). Draw AL, MQ and BS perpendicular on the x-axis from the points A, M and B respectively. L, Q and S are the points where these perpendiculars meet the x-axis.Draw \( AN \perp MQ \) and \( MC \perp BS \). As AN and MC are both parallel to the x-axis, \( AN \parallel MC \), and AM is a transversal.
\( \Rightarrow \angle MAN = \angle BMC \) (corresponding angles). Also, in \( \triangle ANM \) and \( \triangle MCB \), \( \angle ANM = \angle MCB = 90^\circ \).
\( \therefore \triangle MAN \sim \triangle BMC \) (by AA similarity criterion) \( \Rightarrow \frac{AN}{MC} = \frac{MN}{BC} = \frac{AM}{MB} \) (corresponding parts of similar triangles).
Since \( \frac{AM}{MB} = \frac{m_1}{m_2} \) (given), we get\[ \frac{AN}{MC} = \frac{m_1}{m_2} \quad \ldots (i) \qquad \text{and} \qquad \frac{MN}{BC} = \frac{m_1}{m_2} \quad \ldots (ii) \]Now \( AN = LQ = OQ - OL = x - x_1 \) and \( MC = QS = OS - OQ = x_2 - x \). Putting these values in (i):\[ \frac{x - x_1}{x_2 - x} = \frac{m_1}{m_2} \Rightarrow m_2(x - x_1) = m_1(x_2 - x) \Rightarrow m_1 x + m_2 x = m_1 x_2 + m_2 x_1 \Rightarrow x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \]Also \( MN = MQ - NQ = MQ - AL = y - y_1 \) and \( BC = BS - CS = BS - MQ = y_2 - y \). Putting these values in (ii):\[ \frac{y - y_1}{y_2 - y} = \frac{m_1}{m_2} \Rightarrow m_2(y - y_1) = m_1(y_2 - y) \Rightarrow y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \]Hence, the coordinates of the point M are \( \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \).

Knowledge Hub

1. If M is the mid-point of PQ, then it divides the line segment PQ in the ratio 1 : 1. Hence its coordinates are \( \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) \).
2. The ratio \( m_1 : m_2 \) can also be written as \( \lambda : 1 \), where \( \lambda = \frac{m_1}{m_2} \). Therefore, the coordinates of the point M which divides the line segment PQ joining the points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) are given by \( \left( \frac{\lambda x_2 + x_1}{\lambda + 1},\ \frac{\lambda y_2 + y_1}{\lambda + 1} \right) \).

Solved Examples

Example 1. Find the coordinates of the point which divides the line segment joining \( (-1, 3) \) and \( (4, -7) \) internally in the ratio \( 3 : 4 \).

Answer:
We have \( A(-1, 3) \) and \( B(4, -7) \). Let a point \( P(x, y) \) divide the line segment joining A and B in the ratio \( 3 : 4 \) internally. By the section formula,
\( P(x, y) = \left( \frac{4(-1) + 3(4)}{3 + 4},\ \frac{4(3) + 3(-7)}{3 + 4} \right) = \left( \frac{8}{7},\ -\frac{9}{7} \right) \).
Therefore, the co-ordinates of point P are \( \left( \frac{8}{7},\ -\frac{9}{7} \right) \).

Example 2. Determine the ratio in which the line segment joining the points \( (1, -7) \) and \( (6, 4) \) is divided by the x-axis.

Answer:
Let the point \( C(x, 0) \) on the x-axis divide the line segment joining \( A(1, -7) \) and \( B(6, 4) \) in the ratio \( k : 1 \) internally. Then \( C(x, 0) = \left( \frac{6k + 1}{k + 1},\ \frac{4k - 7}{k + 1} \right) \).
Equating the y-coordinates: \( \frac{4k - 7}{k + 1} = 0 \Rightarrow 4k - 7 = 0 \Rightarrow k = \frac{7}{4} \).
Hence, the x-axis divides the line segment AB in the ratio \( 7 : 4 \).

Example 3. If \( P(-2, 4),\ Q(0, 0) \) and \( R(4, 2) \) are the vertices of a \( \triangle PQR \), then find the length of the median through the vertex P.

Answer:
Let D be the mid-point of the side QR. Then \( D = \left( \frac{0 + 4}{2},\ \frac{0 + 2}{2} \right) = (2, 1) \).
\( PD = \sqrt{(2 + 2)^2 + (1 - 4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Hence, the length of the median PD is 5 units.

Example 4. Find the ratio in which the line \( 3x + y - 9 = 0 \) divides the line segment joining the points \( (1, 3) \) and \( (2, 7) \).

Answer:
Let the line \( 3x + y - 9 = 0 \) divide the line segment joining \( P(1, 3) \) and \( Q(2, 7) \) in the ratio \( k : 1 \) at the point A. Then \( A = \left( \frac{2k + 1}{k + 1},\ \frac{7k + 3}{k + 1} \right) \).
As A lies on the line \( 3x + y - 9 = 0 \): \( \frac{3(2k + 1)}{k + 1} + \frac{7k + 3}{k + 1} - 9 = 0 \Rightarrow 3(2k + 1) + (7k + 3) - 9(k + 1) = 0 \Rightarrow 4k - 3 = 0 \Rightarrow k = \frac{3}{4} \).
Therefore, the required ratio is \( 3 : 4 \).

Example 5. Find the ratio in which the line segment joining \( P(6, 4) \) and \( Q(-2, -8) \) is divided by the (i) x-axis (ii) y-axis.

Answer:
Let PQ intersect the x-axis at A and the y-axis at B. We have to find \( PA : AQ \) and \( PB : BQ \).
(i) Let \( PA : AQ = m : 1 \). The y-coordinate of A is 0, so \( \frac{m(-8) + 1(4)}{m + 1} = 0 \Rightarrow -8m + 4 = 0 \Rightarrow m = \frac{1}{2} \). Hence the required ratio is \( 1 : 2 \).
(ii) Let \( PB : BQ = m : 1 \). The x-coordinate of B is 0, so \( \frac{m(-2) + 1(6)}{m + 1} = 0 \Rightarrow -2m + 6 = 0 \Rightarrow m = 3 \). Hence the required ratio is \( 3 : 1 \).

Example 6. Find the co-ordinates of the point which divides internally the line joining the point \( (m + n,\ m - n) \) to the point \( (m - n,\ m + n) \) in the ratio \( m : n \).
Answer:
The co-ordinates of the point dividing two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m : n \) internally are \( \left( \frac{m x_2 + n x_1}{m + n},\ \frac{m y_2 + n y_1}{m + n} \right) \).
Here \( (x_1, y_1) = (m + n,\ m - n) \) and \( (x_2, y_2) = (m - n,\ m + n) \).
\( x = \frac{m(m - n) + n(m + n)}{m + n} = \frac{m^2 + n^2}{m + n} \) and \( y = \frac{m(m + n) + n(m - n)}{m + n} = \frac{m^2 - n^2 + 2mn}{m + n} \).
Hence the point is \( \left( \frac{m^2 + n^2}{m + n},\ \frac{m^2 - n^2 + 2mn}{m + n} \right) \).

Collinearity of Three Points

Let A, B and C be three given points. The points A, B and C will be collinear if the sum of the lengths of any two segments is equal to the length of the third line segment. Three points A, B and C are collinear if and only if (i) \( AB + BC = AC \) or (ii) \( AB + AC = BC \) or (iii) \( AC + BC = AB \).

Example 7. Show that the points \( (1, -1),\ (5, 2) \) and \( (9, 5) \) are collinear.
Answer:
Let \( A(1, -1),\ B(5, 2) \) and \( C(9, 5) \). By the distance formula:
\( AB = \sqrt{(5 - 1)^2 + (2 + 1)^2} = 5 \) units, \( BC = \sqrt{(9 - 5)^2 + (5 - 2)^2} = 5 \) units, \( AC = \sqrt{(9 - 1)^2 + (5 + 1)^2} = 10 \) units.
\( AC = AB + BC \), hence A, B, C are collinear points.

Summary