Chapter – 04 : Co-ordinate Geometry Distance Formula | Chapter Solution Class 9

Ganit Prakash 2023 Co-ordinate Geometry Distance Formula Solution Class 9 Maths. Chapter 4 – Co-ordinate Geometry Distance Formula is provided here with simple step-by-step explanations. These solutions for Co-ordinate Geometry Distance Formula are extremely popular among class 9 students. The Co-ordinate Geometry Distance Formula solution is handy for quickly completing your homework and preparing for exams.

Here, we use the formula to measure the length of the straight line

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

(i) Distance = 18 – 8 =10 units

(ii) Distance = 15 – 4 =11

(iii) Distance = -2-(-7)

= -2 + 7 =5 units

(iv) Distance = -3 – (-10)

= -3 + 10

= 7 units

(v) Distance = 6 – (-2)

= 6 + 2

= 8 units

(vi) Distance = 9 -(-5)

= 9 + 5

= 14 units

(vii) Distance = \sqrt{(5-0)^2+(0-10)^2}

= \sqrt{25+100}

= \sqrt{125} units

(viii) Distance = \sqrt{(3-0)^2+(0-4)^2}

= \sqrt{(3)^2+(-4)^2}

= \sqrt{9+16}

= \sqrt{25}

= 5 units

(ix) Distance = \sqrt{(4-2)^2+(3-1)^2}

= \sqrt{(2)^2+(2)^2}

= \sqrt{4+4}

= \sqrt{8} units

(x) Distance = \sqrt{(-2-2)^2+(-2-2)^2}

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

= \sqrt{16+16}

= \sqrt{32} units

Distance = \sqrt{(7-0)^2+(-24-0)^2}

= \sqrt{(7)^2+(-24)^2}

= \sqrt{49+576}

= \sqrt{625} = 25 units

Distance = \sqrt{(3-0)^2+(-4-0)^2} = \sqrt{(3)^2+(-4)^2}

= \sqrt{9+16}

= \sqrt{25} = 5 units

Distance = \sqrt{(\mathrm{a}+\mathrm{b}-0)^2+(\mathrm{a}-\mathrm{b}-0)^2}

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

= \sqrt{\mathrm{a}^2+ \mathrm{2ab} + \mathrm{b}^2+\mathrm{a}^2- \mathrm{2ab} + \mathrm{b}^2}

= \sqrt{2\left(\mathrm{a}^2+\mathrm{b}^2\right)} units

Distance = \sqrt{(5-8)^2+(7-3)^2}

= \sqrt{(-3)^2+(4)^2}

= \sqrt{9+16}

= \sqrt{25} = 5 units

Distance = \sqrt{(7-2)^2+\{0-(-12)\}^2}

= \sqrt{(5)^2+(0+12)^2}

= \sqrt{25+144}

= \sqrt{169}=13 units

Distance = \sqrt{\left(-\frac{3}{2}-0\right)^2+\{0-(-2)\}^2}

= \sqrt{\left(-\frac{3}{2}\right)^2+(2)^2}

= \sqrt{\frac{9}{4}+4}

= \sqrt{\frac{9+16}{4}}

= \sqrt{\frac{25}{4}}

= \sqrt{5} units

Distance = \sqrt{\{3-(-2)\}^2+\{6-(-6)\}^2}

= \sqrt{(3+2)^2+(6+6)^2}

= \sqrt{(5)^2+(12)^2}

= \sqrt{25+144}

= \sqrt{169} = 13 units

Distance = \sqrt{(1-8)^2+(-3-3)^2}

= \sqrt{(-7)^2+(-6)^2}

= \sqrt{49+36}

= \sqrt{85} units

Distance = \sqrt{(5-8)^2+(7-3)^2}

= \sqrt{(-3)^2+(4)^2}

= \sqrt{9+16}

= \sqrt{25} = 5 units

Let the coordinates of the points A, B & C be (-2, -11),(-3, 7) and (4, 6) respectively.

AB = \sqrt{\left.\{(-2)-(-3)\}^2+\{(-11)-7)\right\}^2}

= \sqrt{(-2+3)^2+(-11-7)^2}

= \sqrt{(1)^2+(-18)^2}

= \sqrt{1+324}=\sqrt{325} units

AC = \sqrt{\{4-(-2)\}^2+(-11-6)^2}

= \sqrt{(4+2)^2+(-17)^2}

= \sqrt{(6)^2+(-17)^2}

= \sqrt{36+289}= \sqrt{325}

∴ AB = AC

Hence, point A is equidistant from points B & C

Length of AB = \sqrt{(7-3)^{2}+\{9-(-7)\}^{2}}

Length of BC = \sqrt{\{3-(-3)\}^{2}+(-7-3)^{2}}

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

= \sqrt{(6)^{2}+(100)^{2}}

= \sqrt{36+100}

= \sqrt{136} units

∴ BC² = 136

Length of AC = \sqrt{(-3-7)^{2}+(3-9)^{2}}

= \sqrt{(-10)^{2}+(-6)^{2}}

= \sqrt{100+36}

= \sqrt{136} units

∴ AC² = 136

Now, AC² + BC² = 272 = AB²

This shows that A, B & C i.e. the given points are the vertices of a right-angle triangle.

Let A (1, 4), B (4, 1) and C (8, 8)

AB = \sqrt{(1-4)^{2}+(4-1)^{2}}

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

= \sqrt{9+9}

= \sqrt{18} units

AB = \sqrt{(8-3)^{2}+(-2-3)^{2}}

= \sqrt{(5)^{2}+(-5)^{2}}

= \sqrt{25+25}

= \sqrt{50} units

∴ AB² = 50

BC = \sqrt{\{8-(-2)\}^{2}+\{-2-(-2)\}^{2}}

= \sqrt{(8+2)^{2}+(-2+2)^{2}}

= \sqrt{(10)^{2}+0}

= \sqrt{100}

∴ BC² = 100

CA = \sqrt{(-2-3)^{2}+(-2-3)^{2}}

= \sqrt{(-5)^{2}+(-5)^{2}}

= \sqrt{25+25}

= \sqrt{50} units

= CA² = 50

∴ \quad \mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{CA}^{2}

This shows that the three points A (3, 3), B (8, -2) and C (-2, -2) are the vertices of a right-angled triangle.

Length of the hypotenuse of Δ ABC = BC = \sqrt{100} units = 10

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

AB = \sqrt{(2-0)^{2}+(1-0)^{2}}

= \sqrt{(2)^{2}+(1)^{2}}

= \sqrt{4+1}

= \sqrt{5} units

BC = \sqrt{(-1-0)^{2}+(2-0)^{2}}

= \sqrt{(-1)^{2}+(2)^{2}}

= \sqrt{1+4}

= \sqrt{5} units

CD = \sqrt{(-1-1)^{2}+(2-3)^{2}}

= \sqrt{(-2)^{2}+(-1)^{2}}

= \sqrt{4+1}

= \sqrt{5} units

AD = \sqrt{(2-1)^{2}+(1-3)^{2}}

= \sqrt{(1)^{2}+(-2)^{2}}

= \sqrt{1+4}

= \sqrt{5} units

∴ \quad \mathrm{AB} =\mathrm{BC} =\mathrm {CD} = \mathrm{AD}

This shows that the four points A, B, C and D i.e. the given points are the vertices of a square.

By the question,

\sqrt{(2-10)^{2}+(-9-y)^{2}} = 10

or, \sqrt{(-8)^{2}+(-9-y)^{2}} = 10

Squaring both sides,

or, (- 8)² + (- 9 - y)² = 10²

or, 64 + (- 9 - y)² = 100

or, (- 9 - y)² = 100 - 64

or, (- 9 - y)² = 36

or, (- 9 - y)² = 6²

or, - 9 - y = ±6

Taking positive,

or, - 9 - y = 6

or, - y = - 6 + 9

or, - y = 15y = - 15

Taking negative

or, - 9 - y = 6

or, y = 6 + 9

or, y = 3

∴ y = - 3

∴ y = - 3, - 15

Let the points on the x-axis be (x, 0) which is equidistant from the two points (3, 4) & (1, 3)

∴ \sqrt{(x-3)^{2}+(5-0)^{2}}=\sqrt{(x-1)^{2}+(3-0)^{2}}

or, (x-3)^{2}+(5)^{2}=(x-1)^{2}+(3)^{2}

or, (x)^2-2.x.3 + (3)^2 + (5)^2 = (x)^2 - 2.x.1 + (1)^2 + (3)^2

or, x^2-6x +9 +25= x^2 - 2x + 1 + 9

or, -6x+25= -2x + 1

or, -6x+2x= 1-25

or, -4x=-24

x = {\frac{-24}{-4}= 6}

or, x = 6

∴ The required point is (6.0)

OA = \sqrt{(4-0)^{2}+(3-0)^{2}}

= \sqrt{4^{2}+3^{2}}

= \sqrt{16+9}

= \sqrt{25} = 5

AB = \sqrt{(8-4)^{2}+(6-3)^{2}}

= \sqrt{4^{2}+3^{2}}

= \sqrt{16+9}

= \sqrt{25}=5

\mathrm{OB} =\sqrt{(8-0)^{2}+(6-0)^{2}}

= \sqrt{8^{2}+6^{2}}

= \sqrt{64+36}

= \sqrt{100} = 10 units

∴ \quad \mathrm{OA}+\mathrm{AB}=\mathrm{OB}

So, the points O (0, 0), A (4, 3) and B (8, 6) are collinear. (Proved)

AB = \sqrt{(-2-2)^{2}+(-2-2)^{2}}

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

= \sqrt{16+16}

= \sqrt{32} units

BC = \sqrt{(-2+2 \sqrt{3})^{2}+(-2-2 \sqrt{3})^{2}}

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

= \sqrt{2\left\{(2 \sqrt{ } 3)^{2}+(2)^{2}\right\}}

= \sqrt{2(12+4)}

= \sqrt{2 \times 16}

= \sqrt{32} units

CA = \sqrt{(2+2 \sqrt{3})^{2}+(2-2 \sqrt{3})^{2}}

= \sqrt{2(2)^{2}+(2\sqrt{3})^{2}}

= \sqrt{2 .(4+12)}=\sqrt{2 \times 16}

= \sqrt{32 \text { units }}

∴ AB = BC = CA

This shows that the points A, B and C i.e. the given points are the vertices of an equilateral triangle.

Let A (-7,2), B (19,18), C (15,-6) & D (-11,-22)

AB = \sqrt{(19+7)^{2}+(18-2)^{2}}

= \sqrt{(26)^{2}+(16)^{2}}

= \sqrt{676+256}

= \sqrt{932} units

BC = \sqrt{(19-15)^{2}+(18+6)^{2}}

= \sqrt{(4)^{2}+(24)^{2}}

= \sqrt{16+476}=\sqrt{592} units

CD = \sqrt{(15+11)^{2}+(-6+22)^{2}}

= \sqrt{(26)^{2}+(6)^{2}}

= \sqrt{676+6}=\sqrt{932} units

AD = \sqrt{(-7+11)^{2}+(2+22)^{2}}

= \sqrt{(4)^{2}+(24)^{2}}

= \sqrt{16+426}=\sqrt{592} units

Since AB = CD and BC = AD

This shows that the points A, B, C and D the given points are the vertices of a parallelogram (Proved).

Let A (2,-2), B (8,4), C (5,7) & D (-1,1)

AB = \sqrt{(2-8)^{2}+(-2-4)^{2}}

= \sqrt{(-6)^{2}+(-6)^{2}}

= \sqrt{36+36}=\sqrt{72 } units

BC = \sqrt{(8-5)^{2}+(4-7)^{2}}

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

= \sqrt{9+9}=\sqrt{18 \text { units }}

\mathrm{CD}=\sqrt{(-1-5)^{2}+(1-7)^{2}}

= \sqrt{(-6)^{2}+(-6)^{2}}

= \sqrt{36+36}=\sqrt{72} units

DA = =\sqrt{(-1-2)^{2}+(1+2)^{2}}

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

= \sqrt{9+9}=\sqrt{18} units

Since AB = CD and BC = AD

This shows that the points A, B, C and D  i.e. the given points are the vertices of a rectangle (Proved).

Let A (2,5), B (5,9), C (9,12) & D (6,8)

AB = \sqrt{(5-2)^{2}+(9-5)^{2}}

= \sqrt{(3)^{2}+(4)^{2}}

= \sqrt{9+16}=\sqrt{25}

= 5 units

BC = \sqrt{(9-5)^{2}+(12-9)^{2}}

= \sqrt{(4)^{2}+(3)^{2}}

= \sqrt{16+9}=\sqrt{25}

= 5 units

CD = \sqrt{(9-6)^{2}+(12-8)^{2}}

= \sqrt{(3)^{2}+(4)^{2}}

= \sqrt{9+16}=\sqrt{25}

= 5 units

DA = \sqrt{(6-2)^{2}+(8-5)^{2}}

= \sqrt{(4)^{2}+(3)^{2}}

= \sqrt{16+9} =\sqrt{25}

= 5 units

∴ AB = BC = CD = DA

This shows that the points A, B, C  and D  i.e. the given points are the vertices of a Rhombus (Proved).

(i) (b) 2 \sqrt{b^{2}+d^{2}}

Explanation

= \sqrt{\{(a+b)-(a-b)\}^{2} + \{(c-d)-(c+d)\}^{2}}

= \sqrt{(a+b-a+b)^{2}+(c-d-c-d)^{2}}

= \sqrt{(2 a)^{2}+(-2 d)^{2}}

= \sqrt{4} \cdot \sqrt{b^{2}+d^{2}}

= 2 \sqrt{b^{2}+d^{2}} units

(a) 0 or 6Explanation

Given, Distance = 5 units

or, \sqrt{(x-3)^{2}+(-7+3)^{2}} = 5

or, \sqrt{(x-3)^{2}+(-4)^{2}} = 5

or, (x-3)^{2}+16 = 25

or, (x-3)^{2} = 25-16 = 9

or, (x-3)^{2}=3^{2}

∴ x-3 = \pm 3

Taking positive,

x - 3 = 3

or, x = 3 + 3

or, x = 6

∴ x = 6

Taking negative

x - 3 = -3

or, x = -3 + 3

or, x = 0

∴ x = 0

∴ x = 0 or 6

Distance = 5 units

or, \sqrt{(x-0)^{2}+(4-0)^{2}} = 5²

or, x² + 4² = 5²

or, x² + 16 = 25

or, x² = 25 - 16 = 9

∴ x = √9 = ± 3

(b) isosceles

Explanation

Let A (3, 0), B(-3, 0) and C (0, 3)

AB = \sqrt{(-3-3)^{2}+(0-0)^{2}}

= \sqrt{(-6)^{2}+(0)^{2}}

= \sqrt{36} = 6 units

BC = \sqrt{(-3-0)^{2}+(0-3)^{2}}

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

= \sqrt{9+9}=\sqrt{18} units

AC = \sqrt{(0-3)^{2}+(3-0)^{2}}=\sqrt{9+9}

= \sqrt{18} units

∴ AC = BC ≠ AB

This shows that A, B and C are the vertices of an isosceles triangle.

= \sqrt{(3-0)^{2}+(4-0)^{2}}

= \sqrt{3^{2}+4^{2}}

= \sqrt{9+16}

= \sqrt{25}

= 5 units

Distance = 5 units.

or, \sqrt{(-4-0)^{2}+(y-0)^{2}}=5

or, (-4)^{2}+y^{2}=5^{2}

or, 16+y² =25

or, y² =25-16

or, y ²=9

or, y = √9 = ± 3

Let A (2, 3) and B (-1, 2)

Let a point on y-axis be C (0, y)

AC = BC

or, AC² = BC²

or, (2 - 0)² + (3 - y)² = (-1 - 0)² + (2 - y)²

or, (2)² + (3)² - 2 × 3 × y + y² = (-1)² + (2)² - 2 × 2 × y + y²

or, 4 + 9 - 6y + y² = 1 + 4 - 4y + y²

or, 13 - 6y = 5 - 4y

or, -6y + 4y = 5 - 13

or, -2y = -8

or, y = -8 / -2 = 4

The point (0, 4) on the y-axis is equidistant from two points (2, 3) and (-1, 2)

(i) (0, 5) and (0, -5)

(ii) (0, 10) and (0, -10)