Coordinate Geometry — Exercise 18.4

Find the distance between the following pair of points :

(2, 3), (4, 1)

Answer

By distance formula,

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

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (2, 3)

Co-ordinates of Q = (4, 1)

$PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} \\[1em] = \sqrt{2^2 + (-2)^2} \\[1em] = \sqrt{4 + 4}\\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.$

Hence, distance between (2, 3) and (4, 1) = $2\sqrt{2}$ units.

Find the distance between the following pair of points :

(0, 0), (36, 15)

Answer

By distance formula,

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

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (0, 0)

Co-ordinates of Q = (36, 15)

$PQ = \sqrt{(36 - 0)^2 + (15 - 0)^2} \\[1em] = \sqrt{36^2 + 15^2} \\[1em] = \sqrt{1296 + 225}\\[1em] = \sqrt{1521} \\[1em] = 39.$

Hence, distance between (0, 0) and (36, 15) = 39 units.

Find the distance between the following pair of points :

(a, b), (-a, -b)

Answer

By distance formula,

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

Let P(x₁, y₁) and Q(x₂, y₂) be the given points

Co-ordinates of P = (a, b)

Co-ordinates of Q = (-a, -b)

$PQ = \sqrt{(-a - a)^2 + (-b - b)^2} \\[1em] = \sqrt{(-2a)^2 + (-2b)^2} \\[1em] = \sqrt{4a^2 + 4b^2}\\[1em] = \sqrt{4(a^2 + b^2)} \\[1em] = 2\sqrt{a^2 + b^2}.$

Hence, distance between (a, b) and (-a, -b) = $2\sqrt{a^2 + b^2}$ units.

A is a point on y-axis whose ordinate is 4 and B is a point on x-axis whose abscissa is -3. Find the length of the line segment AB.

Answer

Since, point A is on y-axis so its x-coordinate = 0.

Given,

Ordinate of A = 4.

∴ A = (0, 4).

Since, point B is on x-axis so its y-coordinate = 0.

Given,

Abscissa of B = -3.

∴ B = (-3, 0).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(-3 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.$

Hence, AB = 5 units.

Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.

Answer

By distance formula,

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

$\Rightarrow AB = \sqrt{[a - (-3)]^2 + [(-5) - (-14)]^2} \\[1em] \Rightarrow 9 = \sqrt{(a + 3)^2 + (-5 + 14)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 9 + 6a + (9)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 6a + 9 + 81}$

On squaring both sides,

$\Rightarrow a^2 + 6a + 90 = (9)^2 \\[1em] \Rightarrow a^2 + 6a + 90 = 81 \\[1em] \Rightarrow a^2 + 6a + 90 - 81 = 0 \\[1em] \Rightarrow a^2 + 6a + 9 = 0 \\[1em] \Rightarrow a^2 + 3a + 3a + 9 = 0 \\[1em] \Rightarrow a(a + 3) + 3(a + 3) = 0 \\[1em] \Rightarrow (a + 3)(a + 3) = 0 \\[1em] \Rightarrow a + 3 = 0 \\[1em] \Rightarrow a = -3.$

Hence, value of a = -3.

Find points on the x-axis which are at a distance of 5 units from the point (5, -4).

Answer

We know that,

y-coordinate of any point on x-axis = 0.

Let point on x-axis which is at a distance of 5 units from (5, -4) be P(x, 0).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 5 = \sqrt{(x - 5)^2 + [0 - (-4)]^2} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 25 - 10x + 16} \\[1em] \Rightarrow (5)^2 = x^2 - 10x + 41$

On squaring both sides,

$\Rightarrow 25 = x^2 - 10x + 41 \\[1em] \Rightarrow x^2 - 10x + 41 - 25 = 0 \\[1em] \Rightarrow x^2 - 10x + 16 = 0 \\[1em] \Rightarrow x^2 - 8x - 2x + 16 = 0 \\[1em] \Rightarrow x(x - 8) - 2(x - 8) = 0 \\[1em] \Rightarrow (x - 2)(x - 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = 8.$

∴ P = (x, 0) = (2, 0) or (8, 0).

Hence, points on the x-axis which are at a distance of 5 units from the point (5, -4) are (2, 0) or (8, 0).

Find points on the y-axis which are at a distance of 10 units from the point (8, 8).

Answer

We know that,

x-coordinate of any point on y-axis = 0.

Let point on y-axis which is at a distance of 10 units from (8, 8) be P(0, y).

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow 10 = \sqrt{(0 - 8)^2 + (y - 8)^2} \\[1em] \Rightarrow 10 = \sqrt{(-8)^2 + y^2 + 8^2 - 16y} \\[1em] \Rightarrow 10 = \sqrt{64 + y^2 + 64 - 16y} \\[1em]$

On squaring both sides,

$\Rightarrow 10^2 = 64 + y^2 + 64 - 16y \\[1em] \Rightarrow 100 = y^2 + 128 - 16y \\[1em] \Rightarrow y^2 - 16y + 128 - 100 = 0 \\[1em] \Rightarrow y^2 - 16y + 28 = 0 \\[1em] \Rightarrow y^2 - 14y - 2y + 28 = 0 \\[1em] \Rightarrow y(y - 14) - 2(y - 14) = 0 \\[1em] \Rightarrow (y - 2)(y - 14) = 0 \\[1em] \Rightarrow y - 2 = 0 \text{ or } y - 14 = 0 \\[1em] \Rightarrow y = 2 \text{ or } y = 14.$

∴ P = (0, y) = (0, 2) or (0, 14).

Hence, points on the y-axis which are at a distance of 10 units from the point (8, 8) are (0, 2) or (0, 14).

Find point (or points) which are at distance of $\sqrt{10}$ units from the point (4, 3) given that the ordinate of the point (or points) is twice the abscissa.

Answer

Given,

Ordinate of the point is twice the abscissa.

Let abscissa of point be k, then ordinate = 2k

Let the required point be P.

∴ P = (k, 2k).

By distance formula,

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

Given,

Distance between (4, 3) and (k, 2k) = $\sqrt{10}$ units.

$\Rightarrow \sqrt{10} = \sqrt{(k - 4)^2 + (2k - 3)^2} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + (4)^2 - 8k + (2k)^2 + (3)^2 - 12k} \\[1em] \Rightarrow \sqrt{10} = \sqrt{k^2 + 16 - 8k + 4k^2 + 9 - 12k}$

On squaring both sides

$\Rightarrow 10 = k^2 + 16 - 8k + 4k^2 + 9 - 12k \Rightarrow 10 = 5k^2 - 20k + 25 \\[1em] \Rightarrow 10 = 5(k^2 - 4k + 5) \\[1em] \Rightarrow k^2 - 4k + 5 = 2 \\[1em] \Rightarrow k^2 - 4k + 5 - 2 = 0 \\[1em] \Rightarrow k^2 - 4k + 3 = 0 \\[1em] \Rightarrow k^2 - 3k - k + 3 = 0 \\[1em] \Rightarrow k(k - 3) - 1(k - 3) = 0 \\[1em] \Rightarrow (k - 1)(k - 3) = 0 \\[1em] \Rightarrow k - 1 = 0 \text{ or } k - 3 = 0 \\[1em] \Rightarrow k = 1 \text{ or } k = 3.$

When k = 1,

P = (k, 2k) = (1, 2).

When k = 3,

P = (k, 2k) = (3, 6).

Hence, required points are (1, 2) or (3, 6).

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

Answer

We know that,

y-coordinate of any point on x-axis = 0.

By distance formula,

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

Let point on x-axis which is equidistant from the points (2, -5) and (-2, 9) be P(x, 0).

∴ Distance between (2, -5) and (x, 0) = Distance between (-2, 9) and (x, 0).

$\Rightarrow \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \\[1em] \Rightarrow \sqrt{x^2 + 4 - 4x + 5^2} = \sqrt{(x + 2)^2 + (-9)^2} \\[1em] \Rightarrow \sqrt{x^2 - 4x + 25 + 4} = \sqrt{x^2 + 4 + 4x + 81} \\[1em]$

On squaring both sides,

$\Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \\[1em] \Rightarrow x^2 - x^2 + 4x + 4x = 29 - 85 \\[1em] \Rightarrow 8x = -56 \\[1em] \Rightarrow x = -7.$

P = (x, 0) = (-7, 0).

Hence, required point is (-7, 0).

Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.

Answer

By distance formula,

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

Given, PQ = QR.

∴ Distance between (6, -1) and (1, 3) = Distance between (1, 3) and (x, 8).

$\Rightarrow \sqrt{(1 - 6)^2 + [3 - (-1)]^2} = \sqrt{(x - 1)^2 + (8 - 3)^2} \\[1em] \Rightarrow \sqrt{(-5)^2 + [4]^2} = \sqrt{x^2 + 1 - 2x + (5)^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 - 2x + 1 + 25} \\[1em] \Rightarrow \sqrt{41} = \sqrt{x^2 - 2x + 26}$

On squaring both sides,

$\Rightarrow 41 = x^2 - 2x + 26 \\[1em] \Rightarrow x^2 - 2x + 26 - 41 = 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3(x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 5.$

Hence, x = -3 or 5.

If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.

Answer

By distance formula,

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

Given,

∴ Distance between Q(0, 1) and P(5, -3) = Distance between Q(0, 1) and R(x, 6).

$\Rightarrow \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \\[1em] \Rightarrow \sqrt{5^2 + (-4)^2} = \sqrt{x^2 + 5^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 + 25}$

On squaring both sides,

$\Rightarrow 25 + 16 = x^2 + 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4.$

Hence, x = ±4.

Find a relation between x and y such that point (x, y) is equidistant from the points (7, 1) and (3, 5).

Answer

By distance formula,

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

Given,

∴ Distance between (x, y) and (7, 1) = Distance between (x, y) and (3, 5).

$\therefore \sqrt{(7 - x)^2 + (1 - y)^2} = \sqrt{(3 - x)^2 + (5 - y)^2} \\[1em] \Rightarrow \sqrt{49 + x^2 - 14x + 1 + y^2 - 2y} = \sqrt{9 + x^2 - 6x + 25 + y^2 - 10y}$

On squaring both sides,

$\Rightarrow 49 + x^2 - 14x + 1 + y^2 - 2y = 9 + x^2 - 6x + 25 + y^2 - 10y \\[1em] \Rightarrow x^2 + y^2 - 14x - 2y + 50 = x^2 + y^2 - 6x - 10y + 34 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 - 6x + 14x = 10y - 2y + 50 - 34 \\[1em] \Rightarrow 8x = 8y + 16 \\[1em] \Rightarrow 8x = 8(y + 2) \\[1em] \Rightarrow x = y + 2 \\[1em] \Rightarrow x - y = 2.$

Hence, relation between x and y is x - y = 2.

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q(2, -5) and R(-3, 6), then find the coordinates of P.

Answer

Let y-coordinate of point P be k.

∴ x-coordinate = 2k.

P = (2k, k)

By distance formula,

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

Given,

P is equidistant from the points Q(2, -5) and R(-3, 6).

$\Rightarrow \sqrt{(2 - 2k)^2 + (-5 - k)^2} = \sqrt{(-3 - 2k)^2 + (6 - k)^2} \\[1em] \Rightarrow \sqrt{4 + 4k^2 - 8k + 25 + k^2 + 10k} = \sqrt{9 + 4k^2 + 12k + 36 + k^2 - 12k} \\[1em] \Rightarrow \sqrt{5k^2 + 2k + 29} = \sqrt{5k^2 + 45}$

On squaring both sides,

$\Rightarrow 5k^2 + 2k + 29 = 5k^2 + 45 \\[1em] \Rightarrow 5k^2 - 5k^2 + 2k = 45 - 29 \\[1em] \Rightarrow 2k = 16 \\[1em] \Rightarrow k = \dfrac{16}{2} \\[1em] \Rightarrow k = 8.$

P = (2k, k) = (16, 8).

Hence, coordinates of P = (16, 8).

If the points A(4, 3) and B(x, 5) are on a circle with center C(2, 3), find the values of x.

Answer

Given,

A(4, 3) and B(x, 5) are on a circle with center C(2, 3).

∴ AC = BC [Radius of same circle]

By distance formula,

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

Substituting values in above equation, we get :

$\Rightarrow \sqrt{(2 - 4)^2 + (3 - 3)^2} = \sqrt{(2 - x)^2 + (3 - 5)^2} \\[1em] \Rightarrow \sqrt{(-2)^2} = \sqrt{4 + x^2 - 4x + (-2)^2} \\[1em] \Rightarrow \sqrt{4} = \sqrt{4 + x^2 - 4x + 4}$

On squaring both sides,

$\Rightarrow 4 = x^2 - 4x + 8 \\[1em] \Rightarrow x^2 - 4x + 4 = 0 \\[1em] \Rightarrow x^2 - 2x - 2x + 4 = 0 \\[1em] \Rightarrow x(x - 2) - 2(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(x - 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \\[1em] \Rightarrow x = 2.$

Hence, value of x = 2.

If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.

Answer

By distance formula,

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

Given,

AB = AC.

$\Rightarrow \sqrt{(3 - 0)^2 + (p - 2)^2} = \sqrt{(p - 0)^2 + (5 - 2)^2} \\[1em] \Rightarrow \sqrt{3^2 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2} \\[1em] \Rightarrow \sqrt{9 + p^2 + 4 - 4p} = \sqrt{p^2 + 3^2}$

On squaring both sides,

$\Rightarrow p^2 + 13 - 4p = p^2 + 9 \\[1em] \Rightarrow p^2 - p^2 + 4p = 13 - 9 \\[1em] \Rightarrow 4p = 4 \\[1em] \Rightarrow p = 1.$

Hence, p = 1.

Using distance formula, show that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

Answer

Given,

Center (O) = (3, 3)

Let, A = (6, 2), B = (0, 4) and C = (4, 6).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AO = \sqrt{(6 - 3)^2 + (2 - 3)^2} \\[1em] = \sqrt{(3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow BO = \sqrt{(0 - 3)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-3)^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10}. \\[1em] \Rightarrow CO = \sqrt{(4 - 3)^2 + (6 - 3)^2} \\[1em] = \sqrt{1^2 + 3^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10}. \\[1em]$

Since, AO = BO = CO.

Hence, proved that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).

The center of a circle is C(2α - 1, 3α + 1) and it passes through the point A(-3, -1). If a diameter of the circle is of length 20 units, find the value(s) of α.

Answer

Given,

Diameter = 20 units

Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{20}{2}$ = 10 units.

∴ AC = 10 units

$\Rightarrow \sqrt{[-3 - (2α - 1)]^2 + [-1 - (3α + 1)]^2} = 10 \\[1em] \Rightarrow \sqrt{(-3 - 2α + 1)^2 + (-1 - 3α - 1)^2} = 10 \\[1em] \Rightarrow \sqrt{(-2 - 2α)^2 + (-3α - 2)^2} = 10 \\[1em] \Rightarrow \sqrt{4 + 4α^2 + 8α + 9α^2 + 4 + 12α} = 10 \\[1em] \Rightarrow \sqrt{13α^2 + 20α + 8} = 10$

On squaring both sides,

$\Rightarrow 13α^2 + 20α + 8 = 10^2 \\[1em] \Rightarrow 13α^2 + 20α + 8 = 100 \\[1em] \Rightarrow 13α^2 + 20α + 8 - 100 = 0 \\[1em] \Rightarrow 13α^2 + 20α - 92 = 0 \\[1em] \Rightarrow 13α^2 + 46α - 26α - 92 =0 \\[1em] \Rightarrow α(13α + 46) - 2(13α + 46) = 0 \\[1em] \Rightarrow (13α + 46)(α - 2) = 0 \\[1em] \Rightarrow 13α + 46 = 0 \text{ or } α - 2 = 0 \\[1em] \Rightarrow 13α = -46 \text{ or } α = 2 \\[1em] \Rightarrow α = -\dfrac{46}{13} \text{ or } α = 2.$

Hence, α = $-\dfrac{46}{13}$ or α = 2.

Using distance formula, show that the points A(3, 1), B(6, 4) and C(8, 6) are collinear.

Answer

By distance formula,

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

$AB = \sqrt{(6 - 3)^2 + (4 - 1)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}. \\[1em] BC = \sqrt{(8 - 6)^2 + (6 - 4)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{(8 - 3)^2 + (6 - 1)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}. \\[1em] \Rightarrow AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC.$

Thus, AC = AB + BC.

Hence, proved that A, B and C are collinear.

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

Answer

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

By distance formula,

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

$AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} \\[1em] = \sqrt{1^2 + 6^2} \\[1em] = \sqrt{1 + 36} \\[1em] = \sqrt{37}. \\[1em] BC = \sqrt{(7 - 6)^2 + [-2 - 4]^2} \\[1em] = \sqrt{1^2 + (-6)^2} \\[1em] = \sqrt{37}.$

Since, AB = BC.

Hence, proved that (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Name the type of triangle formed by the points A(-5, 6), B(-4, -2) and C(7, 5).

Answer

By distance formula,

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

$AB = \sqrt{[-4 - (-5)]^2 + [-2 - 6]^2} \\[1em] = \sqrt{(-4 + 5)^2 + (-8)^2} \\[1em] = \sqrt{1^2 + 64} \\[1em] = \sqrt{65}. \\[1em] BC = \sqrt{[7 - (-4)]^2 + [5 - (-2)]^2} \\[1em] = \sqrt{11^2 + 7^2} \\[1em] = \sqrt{121 + 49} \\[1em] = \sqrt{170}. \\[1em] AC = \sqrt{[7 - (-5)]^2 + (5 - 6)^2} \\[1em] = \sqrt{12^2 + (-1)^2} \\[1em] = \sqrt{144 + 1} \\[1em] = \sqrt{145}.$

Since, AB ≠ BC ≠ AC.

Hence, ABC is a scalene triangle.

Show that the points A(1, 1), B(-1, -1) and C(-$\sqrt{3}, \sqrt{3}$) form an equilateral triangle.

Answer

By distance formula,

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

$AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] BC = \sqrt{[-\sqrt{3} - (-1)]^2 + [\sqrt{3} - (-1)]^2} \\[1em] = \sqrt{[-\sqrt{3} + 1]^2 + [\sqrt{3} + 1]^2} \\[1em] = \sqrt{3 + 1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{[-\sqrt{3} - 1]^2 + [\sqrt{3} - 1]^2} \\[1em] = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em]$

Since, AB = BC = AC.

Hence, ABC is an equilateral triangle.

Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.

Answer

Let points be A(7, 10), B(-2, 5) and C(3, -4).

By distance formula,

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

$AB = \sqrt{(-2 - 7)^2 + (5 - 10)^2} \\[1em] = \sqrt{(-9)^2 + (-5)^2} \\[1em] = \sqrt{81 + 25} \\[1em] = \sqrt{106} \text{ units}. \\[1em] BC = \sqrt{[3 - (-2)]^2 + [-4 - 5]^2} \\[1em] = \sqrt{5^2 + (-9)^2} \\[1em] = \sqrt{25 + 81} \\[1em] = \sqrt{106} \text{ units}. \\[1em] AC = \sqrt{(3 - 7)^2 + (-4 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-14)^2} \\[1em] = \sqrt{16 + 196} \\[1em] = \sqrt{212} \text{ units}.$

⇒ AB² + BC² = $(\sqrt{106})^2 + (\sqrt{106})^2$

= 106 + 106

= 212

= AC².

Since, AB = BC and AB² + BC² = AC².

∴ ABC is an isosceles right angle triangle, right angled at B.

Hence, proved that (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles triangle.

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

Answer

By distance formula,

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

$AB = \sqrt{(-2 - 0)^2 + (a - 3)^2} \\[1em] = \sqrt{(-2)^2 + a^2 + 9 - 6a} \\[1em] = \sqrt{4 + a^2 + 9 - 6a} \\[1em] = \sqrt{a^2 - 6a + 13}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + (4 - a)^2} \\[1em] = \sqrt{(-1 + 2)^2 + (4 - a)^2} \\[1em] = \sqrt{(1)^2 + 16 + a^2 - 8a} \\[1em] = \sqrt{1 + 16 + a^2 - 8a} \\[1em] = \sqrt{a^2 - 8a + 17}. \\[1em] AC = \sqrt{(-1 - 0)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-1)^2 + (1)^2} \\[1em] = \sqrt{1 + 1} \\[1em] = \sqrt{2}.$

By pythagoras theorem,

AB² + AC² = BC²

$\Rightarrow \Big(\sqrt{a^2 - 6a + 13}\Big)^2 + \Big(\sqrt{2}\Big)^2 = \Big(\sqrt{a^2 - 8a + 17}\Big)^2 \\[1em] \Rightarrow a^2 - 6a + 13 + 2 = a^2 - 8a + 17 \\[1em] \Rightarrow a^2 - a^2 - 6a + 8a = 17 - 15 \\[1em] \Rightarrow 2a = 2 \\[1em] \Rightarrow a = 1.$

Hence, value of a = 1.

Show that the points (0, -1), (-2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area.

Answer

Let A(0, -1), B(-2, 3), C(6, 7) and D(8, 3).

By distance formula,

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

$AB = \sqrt{(-2 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{(-2)^2 + [3 + 1]^2}\\[1em] = \sqrt{4 + 4^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20}. \\[1em] BC = \sqrt{[6 - (-2)]^2 + (7 - 3)^2} \\[1em] = \sqrt{[6 + 2]^2 + (4)^2} \\[1em] = \sqrt{8^2 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}. \\[1em] CD = \sqrt{(8 - 6)^2 + (3 - 7)^2} \\[1em] = \sqrt{2^2 + (-4)^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20} \\[1em] AD = \sqrt{(8 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{8^2 + [3 + 1]^2} \\[1em] = \sqrt{64 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}.$

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Area of rectangle ABCD = AB × BC

$= \sqrt{20} \times \sqrt{80} \\[1em] = 2\sqrt{5} \times 4\sqrt{5} \\[1em] = 8 \times 5 \\[1em] = 40 \text{ sq. units}.$

Hence, area of rectangle = 40 sq. units.

If P(2, -1), Q(3, 4), R(-2, 3) and S(-3, -2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

Answer

By distance formula,

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

$PQ = \sqrt{(3 - 2)^2 + [4 - (-1)]^2} \\[1em] = \sqrt{1^2 + [4 + 1]^2} \\[1em] = \sqrt{1 + 5^2} \\[1em] = \sqrt{1 + 25} \\[1em] = \sqrt{26}. \\[1em] QR = \sqrt{(-2 - 3)^2 + (3 - 4)^2} \\[1em] = \sqrt{(-5)^2 + (-1)^2} \\[1em] = \sqrt{25 + 1} \\[1em] = \sqrt{26}. \\[1em] RS = \sqrt{[-3 - (-2)]^2 + [-2 - 3]^2} \\[1em] = \sqrt{[-3 + 2]^2 + [-5]^2} \\[1em] = \sqrt{[-1]^2 + 25} \\[1em] = \sqrt{26}. \\[1em] PS = \sqrt{[-3 - 2]^2 + [-2 - (-1)]^2} \\[1em] = \sqrt{[-5]^2 + [-2 + 1]^2} \\[1em] = \sqrt{25 + [-1]^2} \\[1em] = \sqrt{25 + 1} \\[1em] = \sqrt{26}.$

Here, PQ = QR = RS = PS.

So, PQRS can be a square or rhombus.

Calculating diagonal,

$PR = \sqrt{(-2 - 2)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{(-4)^2 + [3 + 1]^2} \\[1em] = \sqrt{16 + 4^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = 4\sqrt{2}. \\[1em] QS = \sqrt{(-3 - 3)^2 + (-2 - 4)^2} \\[1em] = \sqrt{(-6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2}.$

PR ≠ QS.

Since, diagonal of rhombus are not equal,

∴ PQRS is a rhombus

Area of rhombus = $\dfrac{1}{2} \times d_1 \times d_2$

= $\dfrac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2}$

= $\dfrac{48}{2}$ = 24 sq. units.

Hence, PQRS is a rhombus and area = 24 sq. units.

Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD.

Answer

By distance formula,

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

$AB = \sqrt{(-2 - 2)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-4)^2 + (-1)^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + [-2 - 2]^2} \\[1em] = \sqrt{[-1 + 2]^2 + [-4]^2} \\[1em] = \sqrt{1^2 + 16} \\[1em] = \sqrt{17}. \\[1em] CD = \sqrt{[3 - (-1)]^2 + [-1 - (-2)]^2} \\[1em] = \sqrt{[3 + 1]^2 + [-1 + 2]^2} \\[1em] = \sqrt{4^2 + 1^2} \\[1em] = \sqrt{16 + 1} \\[1em] = \sqrt{17}. \\[1em] AD = \sqrt{(3 - 2)^2 + (-1 - 3)^2} \\[1em] = \sqrt{1^2 + (-4)^2} \\[1em] = \sqrt{1 + 16} \\[1em] = \sqrt{17}.$

Calculating diagonals,

$AC = \sqrt{(-1 - 2)^2 + (-2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-5)^2} \\[1em] = \sqrt{9 + 25} \\[1em] = \sqrt{34}. \\[1em] BD = \sqrt{[3 - (-2)]^2 + (-1 - 2)^2} \\[1em] = \sqrt{[3 + 2]^2 + (-3)^2} \\[1em] = \sqrt{5^2 + 9} \\[1em] = \sqrt{25 + 9} \\[1em] = \sqrt{34}.$

Since, AB = BC =CD = AD and AC = BD.

Hence, proved that A, B, C and D are the vertices of a square.

Name the type of quadrilateral formed by the following points and give reasons for your answer :

(-1, -2), (1, 0), (-1, 2), (-3, 0).

Answer

Let coordinates be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).

$AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[1 + 1]^2 + [0 + 2]^2}\\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] CD = \sqrt{[-3 - (-1)]^2 + [0 - 2]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [-2]^2} \\[1em] = \sqrt{[-2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}. \\[1em] AD = \sqrt{[-3 - (-1)]^2 + [0 - (-2)]^2} \\[1em] = \sqrt{[-3 + 1]^2 + [2]^2} \\[1em] =\sqrt{[-2]^2 + [2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8}.$

Since, AB = BC = CD = AD.

So, ABCD can be square or rhombus.

Calculating diagonal,

$AC = \sqrt{[-1 - (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[-1 + 1]^2 + [2 + 2]^2} \\[1em] = \sqrt{0 + 4^2} \\[1em] = \sqrt{16} \\[1em] = 4. \\[1em] BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4.$

Since, diagonals are also equal.

Hence, (-1, -2), (1, 0), (-1, 2), (-3, 0) are the vertices of a square.

Name the type of quadrilateral formed by the following points and give reasons for your answer :

(4, 5), (7, 6), (4, 3), (1, 2)

Answer

Let A(4, 5), B(7, 6), C(4, 3), D(1, 2) are the given points.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} \\[1em] = \sqrt{3^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}. \\[1em] CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} \\[1em] = \sqrt{3^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ units}.$

Calculating diagonals :

$AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} \\[1em] = \sqrt{0 + (-2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}. \\[1em] BD = \sqrt{(1 - 7)^2 + (2 - 6)^2}\\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units}.$

Since, opposite sides are equal and diagonals are not equal.

Hence, (4, 5), (7, 6), (4, 3), (1, 2) are the vertices of a parallelogram.

Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also, find its circumradius.

Answer

Let O(x, y) be the circumcentre of the circle.

Let A(8, 6), B(8, -2) and C(2, -2) be the vertices of the triangle.

OB = OC [Radii of same circle]

By distance formula,

$\Rightarrow \sqrt{(x - 8)^2 + [y - (-2)]^2} = \sqrt{(x - 2)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 64 - 16x + [y + 2]^2} = \sqrt{x^2 + 4 - 4x + [y + 2]^2}$

Squaring both sides we get,

$\Rightarrow x^2 + 64 - 16x + [y + 2]^2 = x^2 + 4 - 4x + [y + 2]^2 \\[1em] \Rightarrow x^2 - x^2 + [y + 2]^2 - [y + 2]^2 + 64 - 4 = 16x - 4x \\[1em] \Rightarrow 12x = 60 \\[1em] \Rightarrow x = \dfrac{60}{12} \\[1em] \Rightarrow x = 5.$

Also,

OA = OB [Radii of same circle]

By distance formula,

$\Rightarrow \sqrt{(x - 8)^2 + (y - 6)^2} = \sqrt{(x - 8)^2 + [y - (-2)]^2} \\[1em] \Rightarrow \sqrt{(x - 8)^2 + y^2 + 36 - 12y} = \sqrt{(x - 8)^2 + [y + 2]^2}$

On squaring both sides,

$\Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + [y + 2]^2 \\[1em] \Rightarrow (x - 8)^2 + y^2 + 36 - 12y = (x - 8)^2 + y^2 + 4 + 4y \\[1em] \Rightarrow (x - 8)^2 - (x - 8)^2 + y^2 - y^2 + 4y + 12y = 36 - 4 \\[1em] \Rightarrow 16y = 32 \\[1em] \Rightarrow y = \dfrac{32}{16} \\[1em] \Rightarrow y = 2.$

O = (x, y) = (5, 2).

Radius = OA.

$OA = \sqrt{(8 - 5)^2 + (6 - 2)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, circumcenter = (5, 2) and circumradius = 5 units.