Solution:

Find the distance between the following pairs of points:

(i) Distance between (2,3) and (4,1):

The distance formula is given by:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute the values:
$D = \sqrt{(4 - 2)^2 + (1 - 3)^2}$

$D = \sqrt{2^2 + (-2)^2}$

$D = \sqrt{4 + 4}$

$D = \sqrt{8}$

$D = 2\sqrt{2}$

(ii) Distance between (–5,7) and (–1,3):

The distance formula is given by:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute the values:
$D = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$

$D = \sqrt{(4)^2 + (-4)^2}$

$D = \sqrt{16 + 16}$

$D = \sqrt{32}$

$D = 4\sqrt{2}$

(iii) Distance between (a,b) and (–a,–b):

The distance formula is given by:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute the values:
$D = \sqrt{(-a - a)^2 + (-b - b)^2}$

$D = \sqrt{(-2a)^2 + (-2b)^2}$

$D = \sqrt{4a^2 + 4b^2}$

$D = 2\sqrt{a^2 + b^2}$

Solution:

Find the distance between the points (0, 0) and (36, 15)

To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the distance formula:

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For the points $(0, 0)$ and $(36, 15)$, we have $x_1 = 0$, $y_1 = 0$, $x_2 = 36$, and $y_2 = 15$.

Substitute these values into the distance formula:

Distance = $\sqrt{(36 - 0)^2 + (15 - 0)^2}$

Distance = $\sqrt{36^2 + 15^2}$

Distance = $\sqrt{1296 + 225}$

Distance = $\sqrt{1521}$

Distance = 39

The distance between the points $(0, 0)$ and $(36, 15)$ is 39 units.

Solution:

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear

To check if the points are collinear, we need to find the slope of the line joining each pair of points. If the slopes are equal, then the points are collinear.

Let's first find the slope between the points (1, 5) and (2, 3).

The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For points (1, 5) and (2, 3):

$m_1 = \frac{3 - 5}{2 - 1} = \frac{-2}{1} = -2$

Now, let's find the slope between the points (2, 3) and (–2, –11):

$m_2 = \frac{-11 - 3}{-2 - 2} = \frac{-14}{-4} = 3.5$

Finally, let's find the slope between the points (1, 5) and (–2, –11):

$m_3 = \frac{-11 - 5}{-2 - 1} = \frac{-16}{-3} = \frac{16}{3}$

Since the slopes $m_1$, $m_2$, and $m_3$ are not equal, the points (1, 5), (2, 3) and (–2, –11) are not collinear.

Solution:

Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Let the points be A(5, –2), B(6, 4), and C(7, –2).

To check if the triangle is isosceles, we need to calculate the distances between the points and see if any two sides are equal.

Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

1. Distance between A(5, –2) and B(6, 4):

AB = $\sqrt{(6 - 5)^2 + (4 - (-2))^2}$ = $\sqrt{1^2 + 6^2}$ = $\sqrt{1 + 36}$ = $\sqrt{37}$

2. Distance between B(6, 4) and C(7, –2):

BC = $\sqrt{(7 - 6)^2 + (-2 - 4)^2}$ = $\sqrt{1^2 + (-6)^2}$ = $\sqrt{1 + 36}$ = $\sqrt{37}$

3. Distance between A(5, –2) and C(7, –2):

AC = $\sqrt{(7 - 5)^2 + (-2 - (-2))^2}$ = $\sqrt{2^2 + 0^2}$ = $\sqrt{4}$ = 2

We observe that AB = BC = $\sqrt{37}$, and AC = 2.

Since two sides (AB and BC) are equal, the triangle formed by the points A(5, –2), B(6, 4), and C(7, –2) is an isosceles triangle.

Solution:

Using Distance Formula to Check if ABCD is a Square

Let the coordinates of the points be as follows:

• A(x₁, y₁)
• B(x₂, y₂)
• C(x₃, y₃)
• D(x₄, y₄)

To determine whether ABCD forms a square, we need to calculate the lengths of all four sides and the diagonals. If all four sides are equal and the diagonals are also equal, then ABCD is a square.

Step 1: Calculate the lengths of sides

Use the distance formula:

Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For side AB:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For side BC:

$BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}$

For side CD:

$CD = \sqrt{(x_4 - x_3)^2 + (y_4 - y_3)^2}$

For side DA:

$DA = \sqrt{(x_4 - x_1)^2 + (y_4 - y_1)^2}$

Step 2: Calculate the diagonals

For diagonal AC:

$AC = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$

For diagonal BD:

$BD = \sqrt{(x_4 - x_2)^2 + (y_4 - y_2)^2}$

Step 3: Compare the results

If $AB = BC = CD = DA$ and $AC = BD$, then ABCD is a square.

Solution:

Question (i)

Given points: $(-1, -2), (1, 0), (-1, 2), (-3, 0)$
To determine the type of quadrilateral, we first calculate the lengths of the sides and the slopes of the diagonals. 1. Length of side 1: Distance between points $(-1, -2)$ and $(1, 0)$ is: $$ \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(1 + 1)^2 + (0 + 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} $$ 2. Length of side 2: Distance between points $(1, 0)$ and $(-1, 2)$ is: $$ \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ 3. Length of side 3: Distance between points $(-1, 2)$ and $(-3, 0)$ is: $$ \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ 4. Length of side 4: Distance between points $(-3, 0)$ and $(-1, -2)$ is: $$ \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ As all four sides are equal, it suggests that the quadrilateral may be a rhombus. To confirm, we need to check the diagonals. Diagonal 1: Distance between points $(-1, -2)$ and $(-1, 2)$ is: $$ \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = 4 $$ Diagonal 2: Distance between points $(1, 0)$ and $(-3, 0)$ is: $$ \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2} = 4 $$ Since the diagonals are equal in length, the quadrilateral formed is a **rhombus**.

Question (ii)

Given points: $(-3, 5), (3, 1), (0, 3), (-1, -4)$
We calculate the lengths of the sides and the slopes of the diagonals. 1. Length of side 1: Distance between points $(-3, 5)$ and $(3, 1)$ is: $$ \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{(6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} $$ 2. Length of side 2: Distance between points $(3, 1)$ and $(0, 3)$ is: $$ \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} $$ 3. Length of side 3: Distance between points $(0, 3)$ and $(-1, -4)$ is: $$ \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} $$ 4. Length of side 4: Distance between points $(-1, -4)$ and $(-3, 5)$ is: $$ \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} $$ Since the sides are of different lengths, the quadrilateral formed is not a special quadrilateral (like a rectangle or rhombus). Hence, the quadrilateral is an **irregular quadrilateral**.

Question (iii)

Given points: $(4, 5), (7, 6), (4, 3), (1, 2)$
We calculate the lengths of the sides and the slopes of the diagonals. 1. Length of side 1: Distance between points $(4, 5)$ and $(7, 6)$ is: $$ \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} $$ 2. Length of side 2: Distance between points $(7, 6)$ and $(4, 3)$ is: $$ \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} $$ 3. Length of side 3: Distance between points $(4, 3)$ and $(1, 2)$ is: $$ \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} $$ 4. Length of side 4: Distance between points $(1, 2)$ and $(4, 5)$ is: $$ \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} $$ Since opposite sides are equal in length, this is a **parallelogram**.

Solution:

Question:

Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Let the point on the x-axis be $(x, 0)$. This point is equidistant from $(2, –5)$ and $(–2, 9)$. Therefore, the distances from $(x, 0)$ to $(2, –5)$ and $(x, 0)$ to $(–2, 9)$ must be equal.

Distance between $(x, 0)$ and $(2, –5)$:

$$d_1 = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25}$$

Distance between $(x, 0)$ and $(–2, 9)$:

$$d_2 = \sqrt{(x - (-2))^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81}$$

Since these distances are equal, we set $d_1 = d_2$:

$$\sqrt{(x - 2)^2 + 25} = \sqrt{(x + 2)^2 + 81}$$

Squaring both sides:

$$ (x - 2)^2 + 25 = (x + 2)^2 + 81 $$

Expanding both sides:

$$ (x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81 $$

Simplifying:

$$ x^2 - 4x + 29 = x^2 + 4x + 85 $$

Canceling $x^2$ from both sides:

$$ -4x + 29 = 4x + 85 $$

Solving for $x$:

$$ -4x - 4x = 85 - 29 $$

$$ -8x = 56 $$

$$ x = \frac{56}{-8} $$

$$ x = -7 $$

The point on the x-axis is $(-7, 0)$.

Solution:

Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

We are given two points, P(2, – 3) and Q(10, y), and the distance between them is 10 units. We will use the distance formula to solve for y.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the given points P(2, – 3) and Q(10, y), we have:

$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$

Now, simplify the expression:

$10 = \sqrt{(8)^2 + (y + 3)^2}$

$10 = \sqrt{64 + (y + 3)^2}$

Square both sides:

$100 = 64 + (y + 3)^2$

Subtract 64 from both sides:

$36 = (y + 3)^2$

Take the square root of both sides:

$y + 3 = \pm 6$

Now, solve for y:

Case 1: $y + 3 = 6$

$y = 6 - 3$

$y = 3$

Case 2: $y + 3 = -6$

$y = -6 - 3$

$y = -9$

Therefore, the values of y are $y = 3$ and $y = -9$.

Solution:

Question:

If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution:

Since Q is equidistant from P and R, we use the condition that the distances QP and QR are equal.

Distance QP is given by:

$QP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Here, $Q(0, 1)$ and $P(5, -3)$, so:

$QP = \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$

Distance QR is given by:

$QR = \sqrt{(x - 0)^2 + (6 - 1)^2} = \sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}$

Since QP = QR, we equate the two distances:

$\sqrt{41} = \sqrt{x^2 + 25}$

Squaring both sides:

$41 = x^2 + 25$

Subtracting 25 from both sides:

$x^2 = 16$

Taking the square root of both sides:

$x = \pm 4$

Thus, the values of $x$ are $x = 4$ and $x = -4$.

Now, we calculate the distances QR and PR for both values of x.

For $x = 4$,

Distance PR is:

$PR = \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$

Distance QR is:

$QR = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$

For $x = -4$,

Distance PR is:

$PR = \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$

Distance QR is:

$QR = \sqrt{(-4 - 0)^2 + (6 - 1)^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$

Hence, for both values of $x$, the distances PR and QR are both $\sqrt{41}$.

Solution:

Find a relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (–3,4).

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let the point $(x, y)$ be equidistant from the points $(3, 6)$ and $(-3, 4)$. Therefore, the distance from $(x, y)$ to $(3, 6)$ is equal to the distance from $(x, y)$ to $(-3, 4)$.

The distance from $(x, y)$ to $(3, 6)$ is:

$d_1 = \sqrt{(x - 3)^2 + (y - 6)^2}$

The distance from $(x, y)$ to $(-3, 4)$ is:

$d_2 = \sqrt{(x + 3)^2 + (y - 4)^2}$

Since the point is equidistant from both points, we have:

$d_1 = d_2$

Therefore:

$\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}$

Now square both sides to eliminate the square roots:

$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$

Expand both sides:

$(x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16)$

Simplify both sides:

$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$

Cancel the common terms $x^2$ and $y^2$ from both sides:

$-6x + 9 + 36 - 12y = 6x + 9 + 16 - 8y$

Simplify further:

$-6x + 45 - 12y = 6x + 25 - 8y$

Group like terms:

$-6x - 6x + 45 - 25 = -8y + 12y$

$-12x + 20 = 4y$

Now solve for $y$:

$4y = -12x + 20$

$y = -3x + 5$

The relation between $x$ and $y$ is:

$y = -3x + 5$