Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

P and Q are the points of trisection of AB,
therefore, AP=PQ=QB
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.

C.

7.5

From the figure, the coordinates of A, B, and C are (1,3), (-1, 0) and (4, 0)

respectively.

$$\begin{gathered} \mathrm{Area} \mathrm{of}∆\mathrm{ABC} \\ = \frac{1}{2} \left|1(0-0) + (-1)(0-3) +(3-0)\right| \\ =\frac{1}{2} \left|0+3+12\right| \\ =\frac{1}{2} \left|15\right| \\ = 7.5 \mathrm{sq}. \mathrm{units} \end{gathered}$$

B.

( 3, 5 )

It is given that the point P divides AB in the ratio  2 : 1.

Using the section formula, the coordinates of the point P are

$$\begin{gathered} \left( \frac{\left(1\mathrm{x}1\right) + \left(2\mathrm{x}4\right)}{2 + 1} , \frac{\left(1\mathrm{x}3\right) + \left(2\mathrm{x}6\right)}{2 + 1}\right) \\ = \left(\frac{1 + 8}{3}, \frac{3 + 12}{3}\right) = (3,5) \end{gathered}$$

Hence, the coordinates of the point P are (3,5).

A.

( -6, 7)

Let the coordinates of the other end of the diameter be (x,y).

We know that the centre is the midpoint of the diameter. So, O(-2, 5 )

is the midpoint of the diameter AB. The coordinates of the point  A  and  B  are (2,3) and  (x,y) respectively.

Using midpoint formula, we have,

$$\begin{gathered} -2 = \frac{2 + \mathrm{x}}{2} \Rightarrow -4 = 2+\mathrm{x} \Rightarrow \mathrm{x}= -6 \\ 5 = \frac{3 + \mathrm{y}}{2} \Rightarrow 10 = 3 + \mathrm{y} \Rightarrow \mathrm{y} = 7 \end{gathered}$$

Hence, the coordinates of the other end of the diameter are ( -6,7).

The area of the triangle, whose vertises are (x₁, y₁),  (x₂, y₂)  and (x_3, y₃) is

$\mathrm{Area} \mathrm{of} \mathrm{a} ∆ = \frac{1}{2}$[ x₁ (y₂ - y₃) + x₂ ( y₃ -  y₁ ) + x₃ ( y₁ - y₂ ) ]

Substituting the given coordinates

$\mathrm{Area} \mathrm{of} ∆ = \frac{1}{2}\left[{\mathrm{x}}_1 ({\mathrm{y}}_2 - {\mathrm{y}}_3 ) + {\mathrm{x}}_2 ({\mathrm{y}}_3 - {\mathrm{y}}_1) + {\mathrm{x}}_3 ( {\mathrm{y}}_1 - {\mathrm{y}}_2)\right]$

$\Rightarrow \frac{1}{2}\left[\left(\mathrm{p} - 7\right) + 40 + 27 + 9\mathrm{p}\right] = 15$

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> 10p\; +\; 60$  = $\pm$30

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> 10p\; =\; 30 \; or\; 10p\; =\; -90$

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> p\; =\; -3 \; \; or \; \; p\; =\; -9$