CBSE Class 10 Mathematics Coordinate Geometry Notes Set 08

Section Formula

Formula for Internal Division

The coordinates of the point P which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n are given by

\[ x = \frac{mx_2 + nx_1}{m + n}, \qquad y = \frac{my_2 + ny_1}{m + n} \]

Proof. Let O be the origin and let OX and OY be the X-axis and Y-axis respectively. Let A(x₁, y₁) and B(x₂, y₂) be the given points. Let (x, y) be the coordinates of the point P which divides AB internally in the ratio m : n. Draw AL, BM and PN perpendicular to OX. Also draw AH and PK perpendicular from A and P on PN and BM respectively. Then OL = x₁, ON = x, OM = x₂, AL = y₁, PN = y and BM = y₂.

Therefore AH = LN = ON − OL = x − x₁, PH = PN − HN = PN − AL = y − y₁, PK = NM = OM − ON = x₂ − x and BK = BM − MK = BM − PN = y₂ − y. Clearly, \( \triangle AHP \) and \( \triangle PKB \) are similar. Therefore

\[ \frac{AP}{BP} = \frac{AH}{PK} = \frac{PH}{BK} \;\Rightarrow\; \frac{m}{n} = \frac{x - x_1}{x_2 - x} = \frac{y - y_1}{y_2 - y} \]

Now \( \dfrac{m}{n} = \dfrac{x - x_1}{x_2 - x} \Rightarrow mx_2 - mx = nx - nx_1 \Rightarrow mx + nx = mx_2 + nx_1 \Rightarrow x = \dfrac{mx_2 + nx_1}{m + n} \).

Also \( \dfrac{m}{n} = \dfrac{y - y_1}{y_2 - y} \Rightarrow my_2 - my = ny - ny_1 \Rightarrow my + ny = my_2 + ny_1 \Rightarrow y = \dfrac{my_2 + ny_1}{m + n} \).

Thus, the coordinates of P are \( \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right) \).

Note.
1. Midpoint formula. The midpoint of a line segment divides the line segment in the ratio 1 : 1. Hence, the coordinates of the midpoint of the line segment joining points A(x₁, y₁) and B(x₂, y₂) are \( \left(\dfrac{1 \cdot x_1 + 1 \cdot x_2}{1 + 1}, \dfrac{1 \cdot y_1 + 1 \cdot y_2}{1 + 1}\right) = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \).
2. The general point of division. If we are required to find the ratio in which the point P divides the line segment AB, it is convenient to take the ratio k : 1 instead of m₁ : m₂. Then the coordinates of the point of division will be \( \left(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}\right) \).

Centroid of a Triangle

The coordinates of the centroid of the triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) are \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \), and the medians of a triangle are concurrent.

Proof. Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC whose medians are AD, BE and CF. So D, E and F are respectively the mid-points of BC, CA and AB.

The coordinates of D are \( \left(\dfrac{x_2 + x_3}{2}, \dfrac{y_2 + y_3}{2}\right) \). The coordinates of a point dividing AD in the ratio 2 : 1 are

\[ \left(\frac{1 \cdot x_1 + 2 \cdot \frac{x_2 + x_3}{2}}{1 + 2}, \frac{1 \cdot y_1 + 2 \cdot \frac{y_2 + y_3}{2}}{1 + 2}\right) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]

The coordinates of E are \( \left(\dfrac{x_1 + x_3}{2}, \dfrac{y_1 + y_3}{2}\right) \). The coordinates of a point dividing BE in the ratio 2 : 1 are \( \left(\dfrac{1 \cdot x_2 + 2 \cdot \frac{x_1 + x_3}{2}}{3}, \dfrac{1 \cdot y_2 + 2 \cdot \frac{y_1 + y_3}{2}}{3}\right) = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \).

Similarly, the coordinates of a point dividing CF in the ratio 2 : 1 are \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \). Thus, the point having coordinates \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \) is common to AD, BE and CF and divides them in the ratio 2 : 1. Hence, the medians of a triangle are concurrent and the coordinates of the centroid are \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \).

Solved Examples

Example 9. Find the coordinates of the point which divides the line segment joining (6, 3) and (−4, 5) in the ratio 3 : 2 internally.
Answer:
Let P(x, y) be the required point, with A(6, 3), B(−4, 5), m : n = 3 : 2.
\( x = \dfrac{3(-4) + 2(6)}{3 + 2} = \dfrac{-12 + 12}{5} = 0 \) and \( y = \dfrac{3(5) + 2(3)}{3 + 2} = \dfrac{21}{5} \).
So the coordinates of P are \( \left(0, \dfrac{21}{5}\right) \).

Example 10. In what ratio does the X-axis divide the line segment joining the points (2, −3) and (5, 6)?
Answer:
Let the required ratio be k : 1. The coordinates of the point of division are \( \left(\dfrac{5k + 2}{k + 1}, \dfrac{6k - 3}{k + 1}\right) \).
This is a point on the X-axis, so its y-coordinate is zero: \( \dfrac{6k - 3}{k + 1} = 0 \Rightarrow k = \dfrac{1}{2} \).
The required ratio is \( \dfrac{1}{2} : 1 \), that is 1 : 2.

Example 11. Determine the ratio in which the line \( 3x + y - 9 = 0 \) divides the segment joining the points (1, 3) and (2, 7).
Answer:
Suppose the line divides the segment joining A(1, 3) and B(2, 7) in the ratio k : 1 at the point C. Then \( C = \left(\dfrac{2k + 1}{k + 1}, \dfrac{7k + 3}{k + 1}\right) \).
C lies on \( 3x + y - 9 = 0 \): \( 3\left(\dfrac{2k + 1}{k + 1}\right) + \dfrac{7k + 3}{k + 1} - 9 = 0 \).
\( 6k + 3 + 7k + 3 - 9k - 9 = 0 \Rightarrow 4k - 3 = 0 \Rightarrow k = \dfrac{3}{4} \).
So the required ratio is 3 : 4 internally.

Check Your Progress

Check Your Progress 1. Find the ratio in which the line segment joining (2, −3) and (5, 6) is divided by the x-axis.
Answer:
As in Example 10, the ratio is 1 : 2.

Check Your Progress 2. The centre of a circle is (2a − 1, 3a + 1). Find a if the circle passes through (−3, −1) and the length of its diameter is 20 units.
Answer:
The radius is 10, so \( (2a - 1 + 3)^2 + (3a + 1 + 1)^2 = 100 \Rightarrow (2a + 2)^2 + (3a + 2)^2 = 100 \Rightarrow 13a^2 + 20a - 92 = 0 \).
\( (a - 2)(13a + 46) = 0 \), so a = 2 or a = −46/13.

Check Your Progress 3. Find the ratio in which the Y axis divides the line segment joining the points (1, 3) and (−13, −6).
Answer:
Let the ratio be k : 1. The x-coordinate is \( \dfrac{-13k + 1}{k + 1} = 0 \Rightarrow k = \dfrac{1}{13} \). The ratio is 1 : 13.

Check Your Progress 4. The segment joining (−6, 8) and (8, −6) is divided into four equal parts. Find the co-ordinates of the points of division.
Answer:
The middle point is \( (1, 1) \). The first and third points are the midpoints of the two halves: midpoint of (−6, 8) and (1, 1) is \( \left(-\dfrac{5}{2}, \dfrac{9}{2}\right) \); midpoint of (1, 1) and (8, −6) is \( \left(\dfrac{9}{2}, -\dfrac{5}{2}\right) \).
The points are \( \left(-\dfrac{5}{2}, \dfrac{9}{2}\right) \), (1, 1), \( \left(\dfrac{9}{2}, -\dfrac{5}{2}\right) \).

Check Your Progress 5. Find the vertices of a triangle if the mid points of the sides of the triangle are (2, 3), (4, 1) and (3, 0). Also find the lengths of the sides of the triangle.
Answer:
Let the vertices be A(a₁, a₂), B(b₁, b₂), C(c₁, c₂) and let D(2, 3), E(4, 1), F(3, 0) be the midpoints of BC, CA, AB respectively. Each vertex equals the sum of the two adjacent midpoints minus the opposite midpoint (for example \( A = E + F - D \)).
\( A = (4 + 3 - 2, 1 + 0 - 3) = (5, -2) \), \( B = (2 + 3 - 4, 3 + 0 - 1) = (1, 2) \), \( C = (2 + 4 - 3, 3 + 1 - 0) = (3, 4) \).
Sides: \( AB = \sqrt{16 + 16} = 4\sqrt{2} \), \( BC = \sqrt{4 + 4} = 2\sqrt{2} \), \( CA = \sqrt{4 + 36} = 2\sqrt{10} \).
The vertices are (5, −2), (1, 2), (3, 4) with sides \( 4\sqrt{2}, 2\sqrt{2}, 2\sqrt{10} \).