The point P divides the line segment joining the point (1, 2) and (-1, 2) internally in the ratio 1 : 2.

Assertion (A) : The co-ordinates of point P = (1, 6)

Reason (R) : If point P divides the line segment joining the points (x₁, y₁) and (x₂, y₂) in the ratio m₁ : m₂ then :

P = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Point P(x, y) is equidistant from points A(7, 1) and B(3, 5).

Statement 1: x = $\dfrac{7 + 3}{2}$ and y = $\dfrac{1 + 5}{2}$.

Statement 2: (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)².

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

A(20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.

A(-8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5.

Show that : PQ = $\dfrac{3}{8}$BC.