Question

Find the distance and midpoint of the line joining points are (5,-3)(3,0)​

Answer 1

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Answer 2

Step-by-step explanation:

Given :-

points are (5,-3),(3,0)

To find :-

Find the distance and midpoint of the line joining points are (5,-3),(3,0) ?

Solution :-

Given points are : (5,-3) and (3,0)

Let (x1, y1)=(5,-3)=> x1 = 5 and y1 = -3

Let (x2, y2)=(3,0)=>x2 = 3 and y2 = 0

Finding the distance :-

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

=> The distance between two points

=> √[(3-5)²+(0-(-3))²]

=> √[(-2)²+(0+3)²]

=>√[(-2)²+(3)²]

=> √(4+9)

=>√13 units

The distance= √13 units

Finding the mid point :-

Given points are : (5,-3) and (3,0)

Let (x1, y1)=(5,-3)=> x1 = 5 and y1 = -3

Let (x2, y2)=(3,0)=>x2 = 3 and y2 =0

We know that

The Mid point of the line segment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2,{y1+y2}/2)

On Substituting these values in the above formula then

=> ( {5+3}/2 , {-3+0}/2 )

=> (8/2 , -3/2)

=> (4 , -3/2)

Mid point = (4 , -3/2)

Answer:-

i) The distance between the given two points is √13 units

ii) The mid point of the linesegment joining the given two points is (4,-3/2)

Used formulae :-

Distance formula :-

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

Mid Point formula :-

The Mid point of the line segment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2,{y1+y2}/2)