Distance Formula — Test Yourself

The distance of point (-8, 6) from x-axis is:

1. 8

2. 6

3. 10

4. none of these

Answer

Plot the point P(-8, 6) on the graph paper.

Draw a perpendicular line from the point P to the x-axis.

From graph, it is clear that the distance between the point (-8, 6) and the x-axis is 6 units.

Hence, option 2 is the correct option.

The distance of point (-4, -3) from the origin is:

1. -10 unit

2. 10 unit

3. $\sqrt{4^2 - 3^2}$

4. none of these

Answer

Origin (O) = (0, 0) and P = (-4, -3).

By distance formula,

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

$OP = \sqrt{(-4 - 0)^2 + (-3 - 0)^2}\\[1em] = \sqrt{(-4)^2 + (-3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = \text{5 units}.$

Hence, option 4 is the correct option.

The co-ordinates of point P are:

1. (0, 5)

2. (5, 0)

3. (4, 3)

4. (3, 4)

Answer

Since, OPQ is right angled triangle. Using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ OP² = OQ² + QP²

⇒ 5² = 3² + QP²

⇒ 25 = 9 + QP²

⇒ QP² = 25 - 9

⇒ QP² = 16

⇒ QP = $\sqrt{16}$

⇒ QP = 4 units

Since, OQ = 3 units and QP = 4 units.

The co-ordinates of point P are (3, 4).

Hence, option 4 is the correct option.

AB (= 10 unit) is diameter of a circle with center at point P = (x, 0) and point B = (0, y). The relation between x and y is:

1. x + y = 10

2. x + y = 25

3. x² + y² = 5

4. x² + y² = 25

Answer

Given, AB is diameter of a circle with center at point P = (x, 0) and point B = (0, y).

By distance formula,

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

From figure,

AB (Diameter) = 2 x PB (Radius)

$\Rightarrow 10 = 2 \times \sqrt{(x - 0)^2 + (0 - y)^2}\\[1em] \Rightarrow \dfrac{10}{2} = \sqrt{x^2 + (-y)^2}\\[1em] \Rightarrow 5 = \sqrt{x^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.$

Hence, option 4 is the correct option.

Statement 1: For the point P, x = -4 and y = 3, the distance of P from origin is 3 + 4 = 7.

Statement 2: P = (-4, 3) and its distance from origin = $\sqrt{(-4)^2 + (3)^2}$.

1. Both the statements are true.

2. Both the statements are false.

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

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

Answer

Given, point P, x = -4 and y = 3.

Using distance formula,

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

Distance of P(-4, 3) from origin (0, 0)

$= \sqrt{(0 - (-4))^2 + (0 - 3)^2}\\[1em] = \sqrt{4^2 + (-3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 \text{ units}.$

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

Hence, option 4 is the correct option.

Statement 1: The point P(x, y) is at a distance of 6 unit from origin, then P lies in the first quadrant.

Statement 2: Point P can lie in any quadrant.

1. Both the statements are true.

2. Both the statements are false.

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

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

Answer

Given, the point P(x, y) is at a distance of 6 unit from origin.

By distance formula,

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

$\Rightarrow 6 = \sqrt{(x - 0)^2 + (y - 0)^2}\\[1em] \Rightarrow 6 = \sqrt{(x)^2 + (y)^2}\\[1em] \Rightarrow 6^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 36.$

The above equation defines a circle centered at the origin with radius 6. That circle covers all four quadrants, so P can lie anywhere on that circle, not necessarily in the first quadrant.

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

Hence, option 4 is the correct option.

Assertion (A): If A = (2x, y), B(x, 2y) and AB = 5 unit, then x + y = 5.

Reason (R): $\sqrt{(x - 2x)^2 + (2y - y)^2}$ = 5

⇒ x² + y² = 25

1. A is true, but R is false.

2. A is false, but R is true.

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

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

Answer

Given, A = (2x, y), B = (x, 2y).

By distance formula,

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

Since, AB = 5 unit.

Substituting the values, we get :

$\Rightarrow 5 = \sqrt{(x - 2x)^2 + (2y - y)^2}\\[1em] \Rightarrow 5 = \sqrt{(-x)^2 + y^2}\\[1em] \Rightarrow 5^2 = x^2 + y^2\\[1em] \Rightarrow x^2 + y^2 = 25.$

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Assertion (A): The distance between the points A(x, 2x) and B(x, 0) is 4 unit, the point B is (2, 0).

Reason (R): $\sqrt{(x - x)^2 + (2x - 0)^2}$ = 4

⇒ x² = 4 and x = ± 2

1. A is true, but R is false.

2. A is false, but R is true.

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

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

Answer

Given, the distance between the points A(x, 2x) and B(x, 0) is 4 unit.

By distance formula,

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

Substituting the values, we get :

$\Rightarrow 4 = \sqrt{(x - x)^2 + (2x - 0)^2}\\[1em] \Rightarrow 4 = \sqrt{0^2 + (2x)^2}\\[1em] \Rightarrow 4 = \sqrt{4x^2}\\[1em] \Rightarrow 4^2 = 4x^2\\[1em] \Rightarrow 16 = 4x^2\\[1em] \Rightarrow x^2 = \dfrac{16}{4}\\[1em] \Rightarrow x^2 = 4\\[1em] \Rightarrow x = \sqrt{4}\\[1em] \Rightarrow x = \pm 2.$

The coordinates of B = (x, 0) = (2, 0) or (-2, 0)

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Find the points on the y-axis which are at a distance of $2{\sqrt5}$ units from the point (-4, 7).

Answer

Let the point on y-axis be (0, y).

Let (0, y) = (x₁, y₁) and (-4, 7) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4 - 0)^2 + (7 - y)^2}\\[1em] ⇒ 2 \sqrt5 = \sqrt{(-4)^2 + (7 - y)^2}\\[1em] ⇒ (2 \sqrt5)^2 = (-4)^2 + (7 - y)^2\\[1em] ⇒ 20 = 16 + 49 + y^2 - 14y\\[1em] ⇒ 20 = 65 + y^2 - 14y\\[1em] ⇒ 65 + y^2 - 14y - 20 = 0\\[1em] ⇒ y^2 - 14y + 45 = 0\\[1em] ⇒ y^2 - (9y + 5y) + 45 = 0\\[1em] ⇒ y^2 - 9y - 5y + 45 = 0\\[1em] ⇒ (y^2 - 9y) - (5y - 45) = 0\\[1em] ⇒ y(y - 9) - 5(y - 9) = 0\\[1em] ⇒ (y - 9)(y - 5) = 0\\[1em] ⇒ y = 9 \text{ or } 5$

Hence, the the points on the y-axis are (0, 9) or (0, 5).

Find the value of k, if the points (5, k) and (k, 7) are equidistant from point (2, 4).

Answer

Given (2, 4) is equidistant from (5, k) and (k, 7).

i.e. distance between (2, 4) and (5, k) = distance between (2, 4) and (k, 7)

$\sqrt{(2 - 5)^2 + (4 - k)^2} = \sqrt{(2 - k)^2 + (4 - 7)^2}\\[1em] ⇒ (2 - 5)^2 + (4 - k)^2 = (2 - k)^2 + (4 - 7)^2\\[1em] ⇒ (- 3)^2 + (4 - k)^2 = (2 - k)^2 + (- 3)^2\\[1em] ⇒ 9 + 16 + k^2 - 8k = 4 + k^2 - 4k + 9\\[1em] ⇒ 25 + k^2 - 8k = 13 + k^2 - 4k\\[1em] ⇒ 13 + k^2 - 4k - 25 - k^2 + 8k = 0\\[1em] ⇒ - 12 + 4k = 0\\[1em] ⇒ 4k = 12\\[1em] ⇒ k = \dfrac{12}{4}\\[1em] ⇒ k = 3$

Hence, the value of k = 3.

The centre of a circle is (2a, a - 7). Find the value (values) of a, if the circle passes through the point (11, -9) and has diameter $10{\sqrt2}$ units.

Answer

Radius of circle = $\dfrac{\text{Diameter}}{2}$

= $\dfrac{10 \sqrt2}{2}$

= $5 \sqrt2$

⇒ Distance between the given points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Radius of the circle = The distance between the center (2a, a - 7) and the point (11, -9), which lies on the circle

$5 \sqrt2 = \sqrt{(2a - 11)^2 + ((a - 7) - (-9))^2}\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a - 7 + 9)^2\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a + 2)^2\\[1em] ⇒ 50 = 4a^2 + 121 - 44a + a^2 + 4 + 4a\\[1em] ⇒ 50 = 5a^2 + 125 - 40a\\[1em] ⇒ 5a^2 + 125 - 40a - 50 = 0\\[1em] ⇒ 5a^2 - 40a + 75 = 0\\[1em] ⇒ a^2 - 8a + 15 = 0\\[1em] ⇒ a^2 - (3a + 5a) + 15 = 0\\[1em] ⇒ a^2 - 3a - 5a + 15 = 0\\[1em] ⇒ a(a - 3) - 5(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 5) = 0\\[1em] ⇒ a = 3 \text{ or } 5$

Hence, the value of a = 3 or a = 5.

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Answer

The points A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) are the vertices of a quadrilateral. We need to show that this quadrilateral is a rhombus.

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

Distance between points A(-3, 2) and B(-5, -5):

$= \sqrt{((-5) - (-3))^2 + ((-5) - 2)^2}\\[1em] = \sqrt{(-2)^2 + (-7)^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

Distance between points B(-5, -5) and C(2, -3):

$= \sqrt{(2 - (-5))^2 + ((-3) - (-5))^2}\\[1em] = \sqrt{7^2 + 2^2}\\[1em] = \sqrt{49 + 4}\\[1em] = \sqrt{53}$

Distance between points C(2, -3) and D(4, 4):

$= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

Distance between points D(4, 4) and A(-3, 2):

$= \sqrt{(4 - 2)^2 + (4 - (-3))^2}\\[1em] = \sqrt{2^2 + 7^2}\\[1em] = \sqrt{4 + 49}\\[1em] = \sqrt{53}$

AB = BC = CD = DA = $\sqrt{53}$

Since all sides are equal, the quadrilateral is a rhombus.

Hence, the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Points A (-3, -2), B (-6, a), C (-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.

Answer

Given:

Points A(-3, -2), B(-6, a), C(-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD and AB = CD.

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

Distance between points A(-3, -2) and B(-6, a):

$= \sqrt{((-6) - (-3))^2 + (a - (-2))^2}\\[1em] = \sqrt{(-3)^2 + (a - (-2))^2}\\[1em] = \sqrt{9 + a^2 + 4 + 4a}\\[1em] = \sqrt{13 + a^2 + 4a}$

Distance between points C (-3, -4) and D(0, -1):

$= \sqrt{(0 - (-3))^2 + ((-1) - (-4))^2}\\[1em] = \sqrt{3^2 + 3^2}\\[1em] = \sqrt{9 + 9}\\[1em] = \sqrt{18}$

Since AB = CD,

$⇒ \sqrt{13 + a^2 + 4a} = \sqrt{18}\\[1em] ⇒ 13 + a^2 + 4a = 18\\[1em] ⇒ 13 + a^2 + 4a - 18 = 0\\[1em] ⇒ a^2 + 4a - 5 = 0\\[1em] ⇒ a^2 + 5a - 1a - 5 = 0\\[1em] ⇒ (a^2 + 5a) - (1a + 5) = 0\\[1em] ⇒ a(a + 5) - 1(a + 5) = 0\\[1em] ⇒ (a + 5)(a - 1) = 0\\[1em] ⇒ a = -5 \text{ or } 1$

Since it is given that a is negative, we select a = -5.

Hence, the value of a = -5.

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Answer

Let the circumcentre of the triangle be P(x, y).

The circumcentre is equidistant from all three vertices of the triangle.

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

The distance from P to A is equal to the distance from P to B, so:

$⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 1)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 1 - 2y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 122 - 22x + y^2 - 2y\\[1em] ⇒ 26 - 10x = 122 - 22x\\[1em] ⇒ 22x - 10x = 122 - 26\\[1em] ⇒ 12x = 96\\[1em] ⇒ x = \dfrac{96}{12}\\[1em] ⇒ x = 8$

The distance from P to A is equal to the distance from P to C, so:

$⇒ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x - 11)^2 + (y - 9)^2}\\[1em] ⇒ (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2\\[1em] ⇒ x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 121 - 22x + y^2 + 81 - 18y\\[1em] ⇒ x^2 + 26 - 10x + y^2 - 2y = x^2 + 202 - 22x + y^2 - 18y\\[1em] ⇒ 26 - 10x - 2y = 202 - 22x - 18y\\[1em] ⇒ 26 - 10x - 2y - 202 + 22x + 18y = 0\\[1em] ⇒ 12x + 16y = 176\\[1em] ⇒ 3x + 4y = 44\\[1em]$

Putting the value of x = 8 in the above equation,

$⇒ 3 \times 8 + 4y = 44\\[1em] ⇒ 24 + 4y = 44\\[1em] ⇒ 4y = 44 - 24\\[1em] ⇒ 4y = 20\\[1em] ⇒ y = \dfrac{20}{4}\\[1em] ⇒ y = 5$

Hence, the co-ordinates of the circumcentre of the triangle is (8, 5).

Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.

Answer

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

Distance between A(3, 1) and B(0, y - 1):

$⇒ \sqrt{(0 - 3)^2 + ((y - 1) - 1)^2} = 5\\[1em] ⇒ (-3)^2 + (y - 1 - 1)^2 = 5^2\\[1em] ⇒ 9 + (y - 2)^2 = 25\\[1em] ⇒ 9 + y^2 + 4 - 4y = 25\\[1em] ⇒ y^2 - 4y + 13 = 25\\[1em] ⇒ y^2 - 4y - 25 + 13 = 0\\[1em] ⇒ y^2 - 4y - 12 = 0\\[1em] ⇒ y^2 - 6y + 2y - 12 = 0\\[1em] ⇒ (y^2 - 6y) + (2y - 12) = 0\\[1em] ⇒ y(y - 6) + 2(y - 6) = 0\\[1em] ⇒ (y - 6)(y + 2) = 0\\[1em] ⇒ y = 6 \text{ or } -2$

Hence, the values of y are 6 and -2.

Given A = (x + 2, -2) and B = (11, 6). Find x if AB = 17.

Answer

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

Distance between A(x + 2, -2) and B(11, 6):

$⇒ \sqrt{(11 - (x + 2))^2 + (6 - (-2))^2} = 17\\[1em] ⇒ (11 - x - 2)^2 + (6 + 2)^2 = 17^2\\[1em] ⇒ (9 - x)^2 + 8^2 = 289\\[1em] ⇒ 81 + x^2 - 18x + 64 = 289\\[1em] ⇒ x^2 - 18x + 145 = 289\\[1em] ⇒ x^2 - 18x + 145 - 289 = 0\\[1em] ⇒ x^2 - 18x - 144 = 0\\[1em] ⇒ x^2 - 24x + 6x - 144 = 0\\[1em] ⇒ (x^2 - 24x) + (6x - 144) = 0\\[1em] ⇒ x(x - 24) + 6(x - 24) = 0\\[1em] ⇒ (x - 24)(x + 6) = 0\\[1em] ⇒ x = 24 \text{ and } -6$

Hence, the values of x are 24 and -6.

The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 units.

Answer

The diameter of the circle is given as 20 units, so the radius is 10 units.

Distance between the centre A (2x - 1, 3x + 1) and point B (-3, -1) = Radius of circle AB = 10

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

⇒ AB² = 100

Distance between B(-3, -1) and A(2x - 1, 3x + 1):

$⇒ ((2x - 1) - (-3))^2 + ((3x + 1) - (-1))^2 = 100\\[1em] ⇒ (2x - 1 + 3)^2 + (3x + 1 + 1)^2 = 100\\[1em] ⇒ (2x + 2)^2 + (3x + 2)^2 = 100\\[1em] ⇒ 4x^2 + 4 + 8x + 9x^2 + 4 + 12x = 100\\[1em] ⇒ 13x^2 + 8 + 20x = 100\\[1em] ⇒ 13x^2 + 8 + 20x - 100 = 0\\[1em] ⇒ 13x^2 + 20x - 92 = 0\\[1em] ⇒ x = \dfrac{-20 + \sqrt{400 + 4784}}{26} \text { or } \dfrac{-20 - \sqrt{400 + 4784}}{26}\\[1em] ⇒ x = \dfrac{-20 + \sqrt{5184}}{26} \text { or } \dfrac{-20 - \sqrt{5184}}{26}\\[1em] ⇒ x = \dfrac{-20 + 72}{26} \text { or } \dfrac{-20 - 72}{26}\\[1em] ⇒ x = \dfrac{52}{26} \text { or } \dfrac{-92}{26}\\[1em] ⇒ x = 2 \text { or } \dfrac{-46}{13}\\[1em]$

Hence, the values of x are 2 and $\dfrac{-46}{13}$.

The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

Answer

Let the co-ordinates of point Q be (10, y).

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

Distance between P(2, -3) and Q(10, y):

$⇒ \sqrt{(10 - 2)^2 + (y - (-3))^2} = 10\\[1em] ⇒ (10 - 2)^2 + (y - (-3))^2 = 100\\[1em] ⇒ 8^2 + (y + 3)^2 = 100\\[1em] ⇒ 64 + y^2 + 9 + 6y = 100\\[1em] ⇒ y^2 + 6y + 73 = 100\\[1em] ⇒ y^2 + 6y + 73 - 100 = 0\\[1em] ⇒ y^2 + 6y - 27 = 0\\[1em] ⇒ y^2 + 9y - 3y - 27 = 0\\[1em] ⇒ (y^2 + 9y) - (3y + 27) = 0\\[1em] ⇒ y(y + 9) - 3(y + 9) = 0\\[1em] ⇒ (y + 9)(y - 3) = 0\\[1em] ⇒ y = -9 \text{ or } 3$

Hence, the required co-ordinates of the point Q are (10, -9) and (10, 3).

Point P (2, -7) is the centre of a circle with radius 13 units, PT is perpendicular to chord AB and T = (-2, -4); Calculate the length of :

(i) AT

(ii) AB.

Answer

(i) Given:

Radius = PA = PB = 13 units

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

Distance between P(2, -7) and T(-2, -4):

$= \sqrt{(-2 - 2)^2 + (-4 + 7)^2}\\[1em] = \sqrt{(-4)^2 + (3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5$

Using Pythagoras theorem in triangle PAT,

PA² = PT² + AT²

⇒ AT² = PA² - PT²

⇒ AT² = 13² - 5²

⇒ AT² = 169 - 25

⇒ AT² = 144

⇒ AT = $\sqrt{144}$

⇒ AT = 12 units

Hence, the value of AT = 12 units.

(ii) We know that the perpendicular from the center of a circle to a chord bisects the chord.

AB = 2AT

= 2 x 12 units

= 24 units

Hence, the length of AB = 24 units.

Calculate the distance between the points P(2, 2) and Q(5, 4) correct to three significant figures.

Answer

∵ Distance between the given points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between P(2, 2) and Q(5, 4):

$= \sqrt{(5 - 2)^2 + (4 - 2)^2}\\[1em] = \sqrt{3^2 + 2^2}\\[1em] = \sqrt{9 + 4}\\[1em] = \sqrt{13}\\[1em] = 3.605 \text{ units}$

Hence, the distance between the points P(2, 2) and Q(5, 4) is 3.605 units.

Calculate the distance between A(7, 3) and B on the x-axis whose abscissa is 11.

Answer

We know that any point on x-axis has co-ordinates of the form (x, 0).

The abscissa of point B is 11.

So, the point B is (11, 0).

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

Distance between A(7, 3) and B(11, 0):

$= \sqrt{(11 - 7)^2 + (0 - 3)^2}\\[1em] = \sqrt{4^2 + (- 3)^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5 \text{ units}$

Hence, the distance between A(7, 3) and B(11, 0) is 5 units.

Calculate the distance between A(5, -3) and B on the y-axis whose ordinate is 9.

Answer

We know that any point on the y-axis has co-ordinates of the form (0, y).

The ordinate of point B is 9.

So, the point B is (0, 9).

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

Distance between A(5, -3) and B(0, 9):

$= \sqrt{(0 - 5)^2 + (9 - (-3))^2}\\[1em] = \sqrt{(-5)^2 + 12^2}\\[1em] = \sqrt{25 + 144}\\[1em] = \sqrt{169}\\[1em] = 13$

Hence, the distance between A(5, -3) and B(0, 9) is 13 units.

Find the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

Answer

Let the point on y-axis be P(0, y).

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

Distance between A(6, 7) and P(0, y):

$= \sqrt{(0 - 6)^2 + (y - 7)^2}\\[1em] = \sqrt{(- 6)^2 + (y - 7)^2}\\[1em] = \sqrt{36 + y^2 + 49 - 14y}\\[1em] = \sqrt{y^2 - 14y + 85}\\[1em]$

Distance between B(4, -3) and P(0, y):

$= \sqrt{(0 - 4)^2 + (y - (-3))^2}\\[1em] = \sqrt{(- 4)^2 + (y + 3)^2}\\[1em] = \sqrt{16 + y^2 + 9 + 6y}\\[1em] = \sqrt{y^2 + 6y + 25}\\[1em]$

It is given that the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.

$⇒\dfrac{PA}{PB} = \dfrac{1}{2}\\[1em] ⇒\dfrac{\sqrt{y^2 - 14y + 85}}{\sqrt{y^2 + 6y + 25}} = \dfrac{1}{2}\\[1em] ⇒\dfrac{y^2 - 14y + 85}{y^2 + 6y + 25} = \dfrac{1}{4}\\[1em] ⇒4(y^2 - 14y + 85) = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 = y^2 + 6y + 25\\[1em] ⇒ 4y^2 - 56y + 340 - y^2 - 6y - 25 = 0\\[1em] ⇒ 3y^2 - 62y + 315 = 0\\[1em] ⇒ y = \dfrac{62 + \sqrt{3844 - 3780}}{6} \text { or } \dfrac{62 - \sqrt{3844 - 3780}}{6}\\[1em] ⇒ y = \dfrac{62 + \sqrt{64}}{6} \text { or } \dfrac{62 - \sqrt{64}}{6}\\[1em] ⇒ y = \dfrac{62 + 8}{6} \text { or } \dfrac{62 - 8}{6}\\[1em] ⇒ y = \dfrac{70}{6} \text { or } \dfrac{54}{6}\\[1em] ⇒ y = \dfrac{35}{3} \text { or } 9 \\[1em]$

Hence, the required points on y-axis are (0, 9) and $\Big(0, \dfrac{35}{3}\Big)$.

The distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.

Show that : 5x² + 5y² - 34x + 70y + 58 = 0.

Answer

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

Distance between A(1, -3) and P(x, y):

$= \sqrt{(x - 1)^2 + (y - (-3))^2}\\[1em] = \sqrt{(x - 1)^2 + (y + 3)^2}\\[1em] = \sqrt{x^2 + 1 - 2x + y^2 + 9 + 6y}\\[1em] = \sqrt{x^2 - 2x + y^2 + 10 + 6y}\\[1em]$

Distance between B(-2, 2) and P(x, y):

$= \sqrt{(x - (-2))^2 + (y - 2)^2}\\[1em] = \sqrt{(x + 2)^2 + (y - 2)^2}\\[1em] = \sqrt{x^2 + 4 + 4x + y^2 + 4 - 4y}\\[1em] = \sqrt{x^2 + 4x + y^2 + 8 - 4y}\\[1em]$

It is given that the distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.

$⇒\dfrac{PA}{PB} = \dfrac{2}{3}\\[1em] ⇒\dfrac{\sqrt{x^2 - 2x + y^2 + 10 + 6y}}{\sqrt{x^2 + 4x + y^2 + 8 - 4y}} = \dfrac{2}{3}\\[1em] ⇒\dfrac{x^2 - 2x + y^2 + 10 + 6y}{x^2 + 4x + y^2 + 8 - 4y} = \dfrac{4}{9}\\[1em] ⇒9(x^2 - 2x + y^2 + 10 + 6y) = 4(x^2 + 4x + y^2 + 8 - 4y)\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y = 4x^2 + 16x + 4y^2 + 32 - 16y\\[1em] ⇒9x^2 - 18x + 9y^2 + 90 + 54y - 4x^2 - 16x - 4y^2 - 32 + 16y = 0\\[1em] ⇒ 5x^2 + 5y^2 - 34x + 70y + 58 = 0\\[1em]$

Hence, proved :- 5x² + 5y² - 34x + 70y + 58 = 0.

The points A(3, 0), B(a, -2) and C(4, -1) are the vertices of triangle ABC right-angled at vertex A. Find the value of a.

Answer

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

The length of AC:

$= \sqrt{(4 - 3)^2 + ((-1) - 0)^2}\\[1em] = \sqrt{1^2 + (-1)^2}\\[1em] = \sqrt{1 + 1}\\[1em] = \sqrt{2}$

The length of BC:

$= \sqrt{(a - 4)^2 + ((-2) - (-1))^2}\\[1em] = \sqrt{(a - 4)^2 + (-1)^2}\\[1em] = \sqrt{a^2 + 16 - 8a + 1}\\[1em] = \sqrt{a^2 - 8a + 17}$

The length of AB:

$= \sqrt{(a - 3)^2 + ((-2) - 0)^2}\\[1em] = \sqrt{(a - 3)^2 + (-2)^2}\\[1em] = \sqrt{a^2 + 9 - 6a + 4}\\[1em] = \sqrt{a^2 - 6a + 13}$

Using Pythagoras theorem in triangle ABC,

BC² = AB² + AC²

$⇒ (\sqrt2)^2 + (\sqrt{a^2 - 6a + 13})^2 = (\sqrt{a^2 - 8a + 17})^2\\[1em] ⇒ 2 + a^2 - 6a + 13 = a^2 - 8a + 17\\[1em] ⇒ a^2 - 6a + 15 = a^2 - 8a + 17 \\[1em] ⇒ - 6a + 15 + 8a - 17 = 0\\[1em] ⇒ 2a - 2 = 0\\[1em] ⇒ 2a = 2 \\[1em] ⇒ a = \dfrac{2}{2} \\[1em] ⇒ a = 1$

Hence, the value of a = 1.

If two vertices of an equilateral triangle be (0,0) and (3, $\sqrt{3}$), find the third vertex using distance formula.

Answer

Given,

A = (0, 0)

B =(3, $\sqrt{3}$)

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

Length of AB:

$= \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2} \\[1em] = \sqrt{(3)^2 + (\sqrt{3})^2} \\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12}$

Let the third vertex be C(x, y)

Since the triangle is equilateral, the distance from C to A and C to B must also be $\sqrt{12}$.

Length of AC = Length of AB

$\Rightarrow \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow x^2 + y^2 = 12 ........(1)$

Length of BC = Length of AB

$\Rightarrow \sqrt{(x - 3)^2 + (y - \sqrt{3})^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow (x - 3)^2 + (y - \sqrt{3})^2 = 12 \\[1em] \Rightarrow x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y + 12 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y = 0 ..........(2) \\[1em] \text{Substitute equation (1) in (2)} \\[1em] \Rightarrow 12 - 6x - 2\sqrt{3}y = 0 \\[1em] \text{Divide by 2} \\[1em] \Rightarrow 6 - 3x - \sqrt{3}y = 0 \\[1em] \Rightarrow \sqrt{3}y = 6 - 3x \\[1em] \Rightarrow y = \dfrac{6 - 3x}{\sqrt{3}} = 2\sqrt{3} - x\sqrt{3} \\[1em] \Rightarrow y = \sqrt{3}(2 - x)$

Substitute the value of y in equation (1),

x² + ($\sqrt{3}$(2 - x))² = 12

x² + 3(4 - 4x + x²) = 12

x² + 12 - 12x + 3x² = 12

4x² - 12x = 0

4x(x - 3) = 0

∴ x = 0 or x = 3

If x = 0, then

y = $\sqrt{3}(2 - 0) = 2\sqrt{3}$

Vertex C = (0, $2\sqrt{3}$)

If x = 3, then

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

Vertex C = (3, $-\sqrt{3}$).

Hence, third vertex = (0, $2\sqrt{3}$) and (3, $-\sqrt{3}$).