Co-ordinate Geometry — Exercise 21(C)

Find the distance between the points :

(i) A(7, 13) and B(10, 9)

(ii) P(-4, 7) and Q(2, -5)

(iii) C(4, -5) and D(12, -11)

(iv) E(-6, -4) and F(9, -12)

Answer

(i) A(7, 13) and B(10, 9)

By distance formula,

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

Substituting values we get :

$AB = \sqrt{(10 - 7)^2 + (9 - 13)^2} \\[1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.$

Hence, AB = 5 units.

(ii) P(-4, 7) and Q(2, -5)

By distance formula,

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

Substituting values we get :

$PQ = \sqrt{(2 - (-4))^2 + (-5 - 7)^2}\\[1em] =\sqrt{(6)^2 + (-12)^2}\\[1em] = \sqrt{36 + 144}\\[1em] = \sqrt{180}\\[1em] = 6 \sqrt{5}$

Hence, PQ = $6\sqrt{5}$ units.

(iii) C(4, -5) and D(12, -11)

By distance formula,

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

Substituting values we get :

$CD = \sqrt{(12 -4)^2 + (-11 + 5)^2}\\[1em] =\sqrt{(8)^2 + (-6)^2}\\[1em] =\sqrt{64 + 36}\\[1em] =\sqrt{100}\\[1em] = 10$

Hence, CD = 10 units.

(iv) E(-6, -4) and F(9, -12)

By distance formula,

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

Substituting values we get :

$EF = \sqrt{(9 - (-6))^2 + (-12 - (-4))^2}\\[1em] = \sqrt{(15)^2 + (-8)^2}\\[1em] = \sqrt{225 + 64}\\[1em] = \sqrt{289}\\[1em] = 17$

Hence, EF = 17 units.

Find the distance of each of the following points from the origin :

(i) A(6, -6)

(ii) B(-5, 5)

(iii) C(4, -6)

Answer

(i) A(6, -6)

Origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$\text{OA } = \sqrt{(6 - 0)^2 + (-6 - 0)^2} \\[1em] = \sqrt{(6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2}.$

Hence, OA = $6\sqrt{2}$ units.

(ii) B(-5, 5)

Let origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$\text{OB} = \sqrt{(-5 - 0)^2 + (5 - 0)^2}\\[1em] = \sqrt{(-5)^2 + (5)^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.$

Hence, OB = $5\sqrt{2}$ units.

(iii) C(4, -6)

Let origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$\text{OC } = \sqrt{(4 - 0)^2 + (-6 - 0)^2}\\[1em] = \sqrt{(4)^2 + (-6)^2}\\[1em] = \sqrt{16 + 36}\\[1em] = \sqrt{52}\\[1em] = 2\sqrt{13}.$

Hence OC = $2\sqrt{13}$ units.

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

Answer

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

As point is equidistant from the points A and B.

Distance of A and P = Distance of P and B

AP = PB

$\Rightarrow \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(- 2 - x)^2 + (9 - 0)^2}$

Squaring both sides, we get :

⇒ (x - 2)² + 5² = (-2 - x)² + 9²

⇒ x² + 4 - 4x + 25 = x² + 4 + 4x + 81

⇒ -4x + 25 = 4x + 81

⇒ -4x - 4x + 25 - 81 = 0

⇒ -8x - 56 = 0

⇒ -8x = 56

⇒ x = -7.

Hence, P(-7, 0) is the point on x-axis which is equidistant from the points A(2, -5) and B(-2, 9).

Find the point on the y-axis, which is equidistant from the points A(-3, 2) and B(5, -2).

Answer

Let the point on y-axis which is equidistant from the points A(-3, 2) and B(5, -2) be P(0, y).

As point is equidistant from the points A and B.

Distance of A and P = Distance of P and B

AP = PB

$\Rightarrow \sqrt{(0 - (-3))^2 + (y - 2)^2} = \sqrt{(5 - 0)^2 + ((-2) - y)^2}$

Squaring both sides, we get :

⇒ 3² + (y - 2)² = 5² + (-2 -y)²

⇒ 9 + y² + 4 - 4y = 25 + y² + 4 + 4y

⇒ y² - 4y + 13 = y² + 4y + 29

⇒ y² - y² + 4y + 4y = 13 - 29

⇒ 8y = -16

⇒ y = -2.

Hence, P(0, -2) is the point on y-axis which is equidistant from the points A(-3, 2) and B(5, -2).

Show that the points A(3, 0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right-angled triangle.

Answer

We will find the length of all sides of the triangle ABC by using distance formula,

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

Substituting values we get :

$\Rightarrow AB = \sqrt{(6 -3)^2 + (4 - 0)^2}\\[1em] = \sqrt{(3)^2 + (4)^2}\\[1em] = \sqrt{9 + 16} = \sqrt{25} \\[1em] = 5\\[1em] \Rightarrow BC = \sqrt{(-1 -6)^2 + (3 - 4)^2}\\[1em] = \sqrt{(-7)^2 + (-1)^2}\\[1em] = \sqrt{49 + 1} = \sqrt{50} \\[1em] = 5\sqrt{2}\\[1em] \Rightarrow AC = \sqrt{(-1 -3)^2 + (3 - 0)^2}\\[1em] = \sqrt{(-4)^2 + (3)^2}\\[1em] = \sqrt{16 + 9} = \sqrt{25} \\[1em] = 5$

⇒ AB = AC

Also,

AB² + AC² = 5² + 5²

= 25 + 25

= 50.

BC² = $(5\sqrt{2})^2$ = 50.

Since,

AB² + AC² = BC²

⇒ △ABC is right angled triangle at A.

As two sides are equal.

⇒ △ABC is an isosceles right angled triangle.

Hence, proved that the points A(3, 0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right-angled triangle.

Show that the points O(0, 0), A(3, $\sqrt{3}$) and B(3, -$\sqrt{3}$) are the vertices of an equilateral triangle.

Answer

Using distance formula,

$\Rightarrow OA = \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2}\\[1em] = \sqrt{(3)^2 + (\sqrt{3})^2}\\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units} \\[1em] \Rightarrow AB = \sqrt{(3 - 3)^2 + (-\sqrt{3} - \sqrt{3})^2}\\[1em] = \sqrt{(0)^2 + (-2\sqrt{3})^2}\\[1em] = \sqrt{0 + 12} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units}\\[1em] \Rightarrow OB = \sqrt{(3 - 0)^2 + (-\sqrt{3} - 0)^2}\\[1em] = \sqrt{(3)^2 + (-\sqrt{3})^2}\\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units}.$

Since OA = AB = OB, all the sides of the triangle are equal.

Hence, the points O(0, 0), A(3, $\sqrt{3}$) and B(3, -$\sqrt{3}$) are the vertices of an equilateral triangle.

Show that the points A(1, 1), B(-1, 5), C(7, 9) and D(9, 5) are the vertices of a rectangle ABCD.

Answer

We will find the length of all sides by using distance formula.

$\Rightarrow AB = \sqrt{(-1 -1)^2 + (5 - 1)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow BC = \sqrt{(7 -(-1))^2 + (9 - 5)^2}\\[1em] = \sqrt{(8)^2 + (4)^2}\\[1em] = \sqrt{64 + 16} = \sqrt{80} \\[1em] = 4\sqrt{5}\\[1em] \Rightarrow CD = \sqrt{(9 - 7)^2 + (5 - 9)^2}\\[1em] = \sqrt{(2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow AD = \sqrt{(9 - 1)^2 + (5 - 1)^2}\\[1em] = \sqrt{(8)^2 + (4)^2}\\[1em] = \sqrt{64 + 16} = \sqrt{80} \\[1em] = 4\sqrt{5}$

⇒ AB = CD and BC = AD

Now,

We will find the length of diagonals AC and BD.

$\Rightarrow AC = \sqrt{(7 -1)^2 + (9 - 1)^2}\\[1em] = \sqrt{(6)^2 + (8)^2}\\[1em] = \sqrt{36 + 64} = \sqrt{100} \\[1em] = 10\\[1em] \Rightarrow BD = \sqrt{(9 -(-1))^2 + (5 - 5)^2}\\[1em] = \sqrt{(9 + 1)^2 + (0)^2}\\[1em] =\sqrt{(10)^2}\\[1em] = 10$

⇒ AC = BD

Since, opposite sides are equal and diagonals are also equal.

⇒ ABCD is a rectangle.

Hence, the points A(1, 1), B(-1, 5), C(7, 9) and D(9, 5) are the vertices of a rectangle ABCD.

Show that the points A(1, 2), B(5, 4), C(3, 8) and D(-1, 6) are the vertices of a square.

Answer

We will find the length of all sides by using distance formula.

$\Rightarrow AB = \sqrt{(5 - 1)^2 + (4 - 2)^2}\\[1em] = \sqrt{(4)^2 + (2)^2}\\[1em] = \sqrt{16 + 4} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow BC = \sqrt{(3 - 5)^2 + (8 - 4)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow CD = \sqrt{(-1 - 3)^2 + (6 - 8)^2}\\[1em] = \sqrt{(-4)^2 + (-2)^2}\\[1em] = \sqrt{16 + 4} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow AD = \sqrt{(-1 -1)^2 + (6 - 2)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}$

⇒ AB = BC = CD = AD

Now,

We will find the length of diagonals AC and BD.

$\Rightarrow AC = \sqrt{(3 -1)^2 + (8 - 2)^2}\\[1em] = \sqrt{(2)^2 + (6)^2}\\[1em] = \sqrt{4 + 36} = \sqrt{40} \\[1em] = 2\sqrt{10}\\[1em] \Rightarrow BD = \sqrt{(-1 - 5)^2 + (6 - 4)^2}\\[1em] =\sqrt{(-6)^2 + (2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] =\sqrt{(40)}\\[1em] = 2\sqrt{10}$

⇒ AC = BD

Since, all sides are equal and diagonals are also equal.

⇒ ABCD is a square.

Hence, the points A(1, 2), B(5, 4), C(3, 8) and D(-1, 6) are the vertices of a square.

Show that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus.

Answer

By distance formula,

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

Calculate the lengths :

$\Rightarrow AB = \sqrt{(3 - 2)^2 + (4 - (-1))^2}\\[1em] = \sqrt{(1)^2 + (5)^2}\\[1em] = \sqrt{1 + 25}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow BC = \sqrt{(-2 -3)^2 + (3 - 4)^2}\\[1em] = \sqrt{(-5)^2 + (-1)^2}\\[1em] = \sqrt{25 + 1}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow CD = \sqrt{(-3 - (-2))^2 + (-2 - 3)^2}\\[1em] = \sqrt{(-1)^2 + (-5)^2}\\[1em] = \sqrt{1 + 25}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow AD = \sqrt{(-3 -2)^2 + (-2 -(-1))^2}\\[1em] = \sqrt{(-5)^2 + (-1)^2}\\[1em] = \sqrt{25 + 1}\\[1em] = \sqrt{26} \text{ units}.$

As, AB = BC = CD = AD

Hence, the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus.