Solution

Let $P(x, y)$ be the required point. Using the section formula, we obtain

$ \begin{aligned} & x=\dfrac{2 \times 4+3 \times(-1)}{2+3}=\dfrac{8-3}{5}=\dfrac{5}{5}=1 \\ & y=\dfrac{2 \times(-3)+3 \times 7}{2+3}=\dfrac{-6+21}{5}=\dfrac{15}{5}=3 \end{aligned} $

Therefore, the point is $(1,3)$.

Solution

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ are the points of trisection of the line segment joining the given points i.e., $AP=PQ=QB$ Therefore, point $P$ divides $A B$ internally in the ratio 1:2.

$x_1=\dfrac{1 \times(-2)+2 \times 4}{1+2}, \quad y_1=\dfrac{1 \times(-3)+2 \times(-1)}{1+2}$

$x_1=\dfrac{-2+8}{3}=\dfrac{6}{3}=2, \quad y_1=\dfrac{-3-2}{3}=\dfrac{-5}{3}$

Therefore, $P(x_1, y_1)=(2,-\dfrac{5}{3})$

Point $Q$ divides $AB$ internally in the ratio 2:1.

$x_2=\dfrac{2 \times(-2)+1 \times 4}{2+1}, y_2=\dfrac{2 \times(-3)+1 \times(-1)}{2+1}$

$x_2=\dfrac{-4+4}{3}=0, \quad y_2=\dfrac{-6-1}{3}=\dfrac{-7}{3}$

$Q(x_2, y_2)=(0,-\dfrac{7}{3})$

Solution

It can be observed that Niharika posted the green flag at $\dfrac{1}{4}$ of the distance AD i.e., $(\dfrac{1}{4} \times 100) m=25 m$ from the starting point of $2^{\text{nd }}$ line. Therefore, the coordinates of this point $G$ is $(2,25)$.

Similarly, Preet posted red flag at $\dfrac{1}{5}$ of the distance AD i.e., $(\dfrac{1}{5} \times 100) m=20 m$ from the starting point of $8^{\text{th }}$ line. Therefore, the coordinates of this point $R$ are $(8,20)$.

Distance between these flags by using distance formula $=GR$

$=\sqrt{(8-2)^{2}+(25-20)^{2}}=\sqrt{36+25}=\sqrt{61} m$

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be $A(x, y)$.

$x=\dfrac{2+8}{2}, y=\dfrac{25+20}{2}$

$x=\dfrac{10}{2}=5, y=\dfrac{45}{2}=22.5$

Hence, $A(x, y)=(5,22.5)$

Therefore, Rashmi should post her blue flag at $22.5 m$ on $5^{\text{th }}$ line.

Solution

Let the ratio in which the line segment joining $(-3,10)$ and $(6,-8)$ is divided by point $(-1,6)$ be $k: 1$.

Therefore, $-1=\dfrac{6 k-3}{k+1}$

$-k-1=6 k-3$

$7 k=2$

$k=\dfrac{2}{7}$

Therefore, the required ratio is $2: 7$.

Solution

Let the ratio in which the line segment joining $A(1,-5)$ and $B(-4,5)$ is divided by $x$-axisbe $k: 1$.

Therefore, the coordinates of the point of division is $(\dfrac{-4 k+1}{k+1}, \dfrac{5 k-5}{k+1})$.

We know that $y$-coordinate of any point on $x$-axis is 0 .

$\therefore \dfrac{5 k-5}{k+1}=0$

$k=1$

Therefore, $x$-axis divides it in the ratio 1:1.

Division point $=(\dfrac{-4(1)+1}{1+1}, \dfrac{5(1)-5}{1+1})=(\dfrac{-4+1}{2}, \dfrac{5-5}{2})=(\dfrac{-3}{2}, 0)$

Solution

Let $(1,2),(4, y),(x, 6)$, and $(3,5)$ are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point $O$ of diagonal $AC$ and $BD$ also divides these diagonals.

Therefore, $O$ is the mid-point of $AC$ and $BD$.

If $O$ is the mid-point of $AC$, then the coordinates of $O$ are

$(\dfrac{1+x}{2}, \dfrac{2+6}{2}) \Rightarrow(\dfrac{x+1}{2}, 4)$

If $O$ is the mid-point of $B D$, then the coordinates of $O$ are

$(\dfrac{4+3}{2}, \dfrac{5+y}{2}) \Rightarrow(\dfrac{7}{2}, \dfrac{5+y}{2})$

Since both the coordinates are of the same point $O$,

$\therefore \dfrac{x+1}{2}=\dfrac{7}{2}$ and $4=\dfrac{5+y}{2}$

$\Rightarrow x+1=7$ and $5+y=8$

$\Rightarrow x=6$ and $y=3$

Solution

Let the coordinates of point $A$ be $(x, y)$.

Mid-point of $A B$ is $(2,-3)$, which is the center of the circle.

$\therefore(2,-3)=(\dfrac{x+1}{2}, \dfrac{y+4}{2})$

$\Rightarrow \dfrac{x+1}{2}=2$ and $\dfrac{y+4}{2}=-3$

$\Rightarrow x+1=4$ and $y+4=-6$

$\Rightarrow x=3$ and $y=-10$

Therefore, the coordinates of A are $(3,-10)$.

Solution

The coordinates of point $A$ and $B$ are $(-2,-2)$ and $(2,-4)$ respectively.

Since $AP=\dfrac{3}{7} AB$,

Therefore, AP: PB = 3:4

Point $P$ divides the line segment $A B$ in the ratio 3:4.

$ \text{ Coordinates of } \begin{aligned} P & =(\dfrac{3 \times 2+4 \times(-2)}{3+4}, \dfrac{3 \times(-4)+4 \times(-2)}{3+4}) \\ & =(\dfrac{6-8}{7}, \dfrac{-12-8}{7}) \\ & =(-\dfrac{2}{7},-\dfrac{20}{7}) \end{aligned} $

Solution

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

$ \begin{aligned} \text{ Coordinates of } P & =(\dfrac{1 \times 2+3 \times(-2)}{1+3}, \dfrac{1 \times 8+3 \times 2}{1+3}) \\ & =(-1, \dfrac{7}{2}) \end{aligned} $

Coordinates of $Q=(\dfrac{2+(-2)}{2}, \dfrac{2+8}{2})$

$ =(0,5) $

Coordinates of $R=(\dfrac{3 \times 2+1 \times(-2)}{3+1}, \dfrac{3 \times 8+1 \times 2}{3+1})$

$ =(1, \dfrac{13}{2}) $

Solution

Let $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ are the vertices $A, B, C, D$ of a rhombus ABCD.

Length of diagonal $AC=\sqrt{[3-(-1)]^{2}+(0-4)^{2}}$

$ =\sqrt{16+16}=4 \sqrt{2} $

Length of diagonal $BD=\sqrt{[4-(-2)]^{2}+[5-(-1)]^{2}}$

$ =\sqrt{36+36}=6 \sqrt{2} $

Therefore, area of rhombus $ABCD=\dfrac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}$

$ =24 \text{ square units } $