Question

Question related to CO-ORDINATE GEOMETRY

Explain the concept of distance between two points very briefly and find distance of point (-5,0) from origin

Answer 1

Answer:

$\huge\orange{\boxed{\purple{\mathbb{\overbrace{\underbrace{\fcolorbox{orange}{aqua}{\underline{\red{☠ Answer ☠}}}}}}}}}$ As a special case of the distance formula, suppose we want to know the distance of a point (x,y) to the origin. According to the distance formula, this is √(x−0)2+(y−0)2=√x2+y2. A point (x,y) is at a distance r from the origin if and only if √x2+y2=r, or, if we square both sides: x2+y2=r2.

Answer 2

$\Large\boxed{\bf{\textsc{Concept Of Distance Between Two Points :}}}$

Consider two points A (x₁ , y₁) and B (x₂ , y₂). Draw perpendicular AL and BM

from A and B to x - axis. AN is the perpendicular drawn from A on to BM.

From Right angled Trinagle ABN :

AB = $\sf{\sqrt{AN^2\:\:+\:\:BN^2}}$

Here, AN = x₂ - x₁ and BN = y₂ - y₁

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

Hence, the distance between two points  A (x₁ , y₁) and B (x₂ , y₂) is :

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

Note :

The distance of a point A (x₁ , y₁) and from origin O is OA = $\sf{\sqrt{x_1\:^2+y_1\:^2}}}$

• Co-ordinates of origin is (0, 0)

Example : Find the distance between the points (-4 ,5) and (2, -3)

The given points are (-4 ,5) and (2, -3)

Here :

x₁ = -4

y₁ = 5

x₂ = 2

y₂ = -3

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

AB = $\sf{\sqrt{(2-(-4))^2\:\:+\:\:(-3-5)^2}}$

AB = $\sf{\sqrt{(2+4)^2\:\:+\:\:(-3-5)^2}}$

AB = $\sf{\sqrt{(6)^2\:\:+\:\:(-8)^2}}$

AB = $\sf{\sqrt{36\:\:+\:\:64}}$

AB = $\sf{\sqrt{100}}$

AB = 10 UNITS

Hence distance between the points (-4 ,5) and (2, -3) is 10 units .

( Refer to Attachment 2 for Image related of concept )

Important Points to remember :

• Co-ordinates of the origin is (0, 0).

• The y-coordinates of every point on X-axis is zero

• The x-coordinates of every point on Y-axis is zero.

• Any point on X-axis is of the form (x, 0).

• Any point on Y-axis is of the form (0, y).

• A line which is parallel to the X-axis is called a horizontal line. On the horizontal line. The y-coordinates of all the point are same.

• A line which is parallel to the Y-axis is called a vertical line. On vertical line, the x-coordinates of all points are same.

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$\Large\boxed{\bf{\textsc{Coming\:To Question :}}}$

Given :-

Points :  (-5 ,0) and origin

To Find :-

Distance between the points

Note :

• Co-ordinates of origin is (0, 0)

Solution :

The distance of a point A (x₁ , y₁) and from origin O is OA = $\sf{\sqrt{x_1\:^2+y_1\:^2}}}$

Here :

x₁ = -5

y₁ = 0

OA = $\sf{\sqrt{x_1\:^2+y_1\:^2}}}$

OA = $\sf{\sqrt{(-5)\:^2+0\:^2}}}$

OA = $\sf{\sqrt{25+0}}}$

OA = $\sf{\sqrt{25}}}$

OA = 5 UNITS

Hence distance between the points (-5, 0) and origin is 5 units .