Solution:

Question: Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution:

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Let the points of trisection divide the line segment joining A(4, –1) and B(–2, –3) into three equal parts. Let the points of trisection be P and Q, where P divides the line segment in the ratio 1:2 and Q divides the line segment in the ratio 2:1.

Using the section formula, the coordinates of a point dividing a line segment in the ratio $m:n$ are given by:

$\left( \dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n} \right)$

For point P, dividing the line segment in the ratio 1:2:

Let P be the point dividing AB in the ratio 1:2. So, $m = 1$ and $n = 2$.

The coordinates of P are:

$P = \left( \dfrac{1 \times (-2) + 2 \times 4}{1+2}, \dfrac{1 \times (-3) + 2 \times (-1)}{1+2} \right)$

$P = \left( \dfrac{-2 + 8}{3}, \dfrac{-3 - 2}{3} \right)$

$P = \left( \dfrac{6}{3}, \dfrac{-5}{3} \right)$

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

For point Q, dividing the line segment in the ratio 2:1:

Let Q be the point dividing AB in the ratio 2:1. So, $m = 2$ and $n = 1$.

The coordinates of Q are:

$Q = \left( \dfrac{2 \times (-2) + 1 \times 4}{2+1}, \dfrac{2 \times (-3) + 1 \times (-1)}{2+1} \right)$

$Q = \left( \dfrac{-4 + 4}{3}, \dfrac{-6 - 1}{3} \right)$

$Q = \left( \dfrac{0}{3}, \dfrac{-7}{3} \right)$

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

Solution:

Distance Between Two Flags and Midpoint Calculation

Let the length of side AD of the rectangular school ground ABCD be represented by $l$. The distance between two points along AD is measured in meters, with each line spaced 1 meter apart.

The positions of the flags are determined as follows:

• Niharika posts a green flag on the 2nd line, at $ \frac{1}{4} \times l $ meters along AD.
• Preet posts a red flag on the 8th line, at $ \frac{1}{5} \times l $ meters along AD.

To find the distance between the two flags, we calculate the difference in their positions along the AD line:

Distance between the two flags = $ \left| \frac{1}{4} \times l - \frac{1}{5} \times l \right| $

Now, to calculate Rashmi's position for posting the blue flag, we need to find the midpoint between the two flags. The midpoint is the average of the positions of the green and red flags:

Midpoint = $ \frac{ \left( \frac{1}{4} \times l \right) + \left( \frac{1}{5} \times l \right) }{2} $

Simplifying the expression for the midpoint:

Midpoint = $ \frac{ \left( \frac{5}{20} \times l \right) + \left( \frac{4}{20} \times l \right) }{2} $

Midpoint = $ \frac{ \frac{9}{20} \times l }{2} $

Midpoint = $ \frac{9}{40} \times l $

Rashmi should post the blue flag at a distance of $ \frac{9}{40} \times l $ meters along AD.

Solution:

Question: Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Solution:

Solution:

Question: Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Let the point of division be P(x, 0). The coordinates of A are (1, –5) and the coordinates of B are (–4, 5).

Since the point P lies on the x-axis, its y-coordinate is 0. Let the ratio in which P divides AB be $m : n$.

Using the section formula, the coordinates of P dividing AB in the ratio $m : n$ are given by:

The x-coordinate of P: $x = \frac{m \cdot x_2 + n \cdot x_1}{m + n}$

The y-coordinate of P: $y = \frac{m \cdot y_2 + n \cdot y_1}{m + n}$

Substituting the values for the coordinates of A and B, we have:

For the x-coordinate of P:

$x = \frac{m \cdot (-4) + n \cdot 1}{m + n}$

For the y-coordinate of P:

$0 = \frac{m \cdot 5 + n \cdot (-5)}{m + n}$

Solving the equation for the y-coordinate:

$0 = \frac{5m - 5n}{m + n}$

This simplifies to $5m = 5n$, which gives $m = n$.

Now substitute $m = n$ into the equation for the x-coordinate of P:

$x = \frac{m \cdot (-4) + m \cdot 1}{m + m}$

$x = \frac{-4m + m}{2m}$

$x = \frac{-3m}{2m}$

$x = -\frac{3}{2}$

Therefore, the point P divides the line segment in the ratio $1 : 1$ and the coordinates of the point of division are $\left(-\frac{3}{2}, 0\right)$.

Solution:

Given vertices of a parallelogram: (1, 2), (4, y), (x, 6), (3, 5)

We know that the diagonals of a parallelogram bisect each other. So, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.

Let the vertices be labeled as follows: A(1, 2), B(4, y), C(x, 6), and D(3, 5).

The midpoint of diagonal AC is the midpoint of points A(1, 2) and C(x, 6), and the midpoint of diagonal BD is the midpoint of points B(4, y) and D(3, 5).

Midpoint of AC:
$ \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, 4 \right)$

Midpoint of BD:
$ \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right)$

Since the midpoints of diagonals bisect each other, we equate the corresponding coordinates.

Equating the x-coordinates:
$ \frac{1 + x}{2} = \frac{7}{2} $
$ 1 + x = 7 $
$ x = 6 $

Equating the y-coordinates:
$ 4 = \frac{y + 5}{2} $
$ 8 = y + 5 $
$ y = 3 $

Therefore, the values of x and y are:

$ x = 6 $ and $ y = 3 $

Solution:

Solution:

Solution:

Coordinates of P such that AP = $\dfrac{3}{7}$ AB

Solution:

Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

The coordinates of the points dividing the line segment joining two points in a given ratio can be found using the section formula. The section formula is given by:

For a point dividing the line segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$, the coordinates of the point $(x, y)$ are given by:

$x = \dfrac{m x_2 + n x_1}{m + n}$

$y = \dfrac{m y_2 + n y_1}{m + n}$

Here, the points $A(– 2, 2)$ and $B(2, 8)$ are given. We need to find the coordinates of the points dividing the segment into four equal parts. This means the ratio will be 1:3, 2:2, and 3:1, and the point dividing the segment in each of these ratios will give the required coordinates.

Step 1: Coordinates of the point dividing the line in the ratio 1:3

Using the section formula for the ratio 1:3, we have:

$x = \dfrac{1 \cdot 2 + 3 \cdot (-2)}{1 + 3} = \dfrac{2 - 6}{4} = \dfrac{-4}{4} = -1$

$y = \dfrac{1 \cdot 8 + 3 \cdot 2}{1 + 3} = \dfrac{8 + 6}{4} = \dfrac{14}{4} = 3.5$

So, the coordinates of the point dividing the segment in the ratio 1:3 are $(-1, 3.5)$.

Step 2: Coordinates of the point dividing the line in the ratio 1:1

For the ratio 1:1, the midpoint is found:

$x = \dfrac{1 \cdot 2 + 1 \cdot (-2)}{1 + 1} = \dfrac{2 - 2}{2} = 0$

$y = \dfrac{1 \cdot 8 + 1 \cdot 2}{1 + 1} = \dfrac{8 + 2}{2} = \dfrac{10}{2} = 5$

So, the coordinates of the midpoint are $(0, 5)$.

Step 3: Coordinates of the point dividing the line in the ratio 3:1

Using the section formula for the ratio 3:1, we have:

$x = \dfrac{3 \cdot 2 + 1 \cdot (-2)}{3 + 1} = \dfrac{6 - 2}{4} = \dfrac{4}{4} = 1$

$y = \dfrac{3 \cdot 8 + 1 \cdot 2}{3 + 1} = \dfrac{24 + 2}{4} = \dfrac{26}{4} = 6.5$

So, the coordinates of the point dividing the segment in the ratio 3:1 are $(1, 6.5)$.

Step 4: Coordinates of the point dividing the line in the ratio 1:3 (other direction)

The coordinates of the last point dividing the line in the ratio 3:1 are the same as in Step 3:

So, the coordinates are $(1, 6.5)$.

Solution:

Find the Area of a Rhombus

Given vertices of the rhombus are: (3, 0), (4, 5), (–1, 4), and (–2, –1).

To find the area of a rhombus, we use the formula:

Area = $ \frac{1}{2} \times d_1 \times d_2 $

Where $d_1$ and $d_2$ are the lengths of the diagonals of the rhombus.

First, calculate the length of the diagonals:

The diagonals are the line segments joining opposite vertices.

Diagonal 1: From (3, 0) to (–1, 4)

Length of Diagonal 1, $d_1 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d_1 = \sqrt{((-1) - 3)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Diagonal 2: From (4, 5) to (–2, –1)

Length of Diagonal 2, $d_2 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d_2 = \sqrt{((-2) - 4)^2 + ((-1) - 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$

Now, substitute the values of $d_1$ and $d_2$ into the area formula:

Area = $ \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} $

Area = $ \frac{1}{2} \times 24 \times 2 $

Area = $ \frac{1}{2} \times 48 $

Area = 24 square units