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Distance Between Two Factors



Let P(x1, y1) and Q(x2, y2) be two factors within the Cartesian airplane (or xy –airplane), at a distance ‘d’ aside. 

That’s

d  =  PQ

By the definition of coordinates, 

OM  =  x1

ON  =  x2

MP  =  y1

NQ  =  y2

Step 1 :

Now, (PR  NQ)

PR  =  MN

(Reverse sides of the rectangle MNRP)

Step 2 :

Discover the size of MN. 

MN  =  ON – OM

MN  =  x2 – x1

Step 3 :

Discover the size of RQ. 

RQ  =  NQ – NR

RQ  =  y– y1

Step 4 : 

Triangle PQR is correct angled at R. (PR  NQ)

By Pythagorean Theorem, 

PQ2  =  PR2 + RQ

d2  =  (x2 – x1)2 + (y2 – y1)2

By taking optimistic sq. root, 

d  =  √[(x2 – x1)2 + (y2 – y1)2]

Given two factors (x1, y1) and (x2, y2), the gap between these factors is given by the formulation

√[(x2 – x1)2 + (y2 – y1)2]

Instance 1 :

Discover the gap between the 2 factors given beneath. 

(-12, 3) and (2, 5)

Answer :

Components for the gap between the 2 factors is

√[(x2 – x1)2 + (y2 – y1)2]

Substitute (x1, y1)  =  (-12, 3) and (x2, y2)  =  (2, 5).


√[(2 + 12)2 + (5 – 3)2]

 √[142 + 22]

=  
√[196 + 4]

= 
√200

=  
√(2 ⋅ 10 ⋅ 10)

=   10√2

So, the gap between the given factors is 10√2 items. 

Instance 2 :

Discover the gap between the 2 factors given beneath. 

(-2, -3) and (6, -5)

Answer :

Components for the gap between the 2 factors is

√[(x2 – x1)2 + (y2 – y1)2]

Substitute (x1, y1)  =  (-2, -3) and (x2, y2)  =  (6, -5).

 √[(6 + 2)2 + (-5 + 3)2]

 √[82 + (-2)2]

=   √[64 + 4]

=  √68

=   √(2 ⋅ 2 ⋅ 17)

=   2√17

So, the gap between the given factors is 2√17 items. 

Instance 3 :

If the gap between the 2 factors given beneath is 2√29, then discover the worth of ok, on condition that ok > 0.  

(-7, 2) and (3, ok)

Answer :

Distance between the above two factors  =  2√29

√[(x2 – x1)2 + (y2 – y1)2]  =  2√29

Substitute (x1, y1)  =  (-7, 2) and (x2, y2)  =  (3, ok).

√[(3 + 7)2 + (k – 2)2]  =  2√29

√[102 + (k – 2)2]  =  2√29

√[100 + (k – 2)2]  =  2√29

Sq. each side. 

100 + (ok – 2)2  =  (2√29)2

100 + ok2 – 2(ok)(2) +  22  =  22(√29)2

100 + ok2 – 4k +  4  =  4(29)

ok2 – 4k + 104  =  116

Subtract 116 from all sides.

ok2 – 4k – 12  =  0

(ok – 6)(ok + 2)  =  0

ok – 6  =  0  or  ok + 2  =  0

ok  =  6  or  ok  =  -2

As a result of ok > 0, we have now

ok  =  6

Instance 4 :

Discover the gap between the factors A and B within the xy-pane proven beneath. 

Answer :

Establish the factors A and B within the xy-plane above. 

Components for the gap between the 2 factors is

√[(x2 – x1)2 + (y2 – y1)2]

To search out the gap between the factors A and B, substitute (x1, y1)  =  (2, -3) and (x2, y2)  =  (5, 5).  

AB  =  √[(5 – 2)2 + (5 + 3)2]

AB  =  √[32 + 82]

AB  =  √(9 + 64)

AB  =  √73 items

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