# Find the distance between the point R( 2a+3, 8) and the point P(2a, 4).

Asked on by nadolla

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the points R(2a+3, 8) and the point P(2a, 4).

We need to determine the distance between the point P and the point R. Or we need to find the length of the segment RP.

We will use the distance formula to find the length.

We know that:

Distance = sqrt[ ( xA- xB)^2 + ( yA-yB)^2]

Let us substitute.

==> D (RP) = sqrt[( 2a+3 - 2a)^2 + ( 8-4)^2]

Reduce similar terms.

==> D ( RP ) = sqrt( 3^2) + 4^2) ]

= sqrt( 9 + 16)

= sqrt( 25) = 5

Then, the distance between the point R( 2a+3, 8) and the point P(2a, 4) is 5 units.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'l consider RP the hypothenuse of the right angle triangle RPA,  where A is the intersection of projections of R and P.

We'll use the Pythagorean theorem in a right angle triangle:

RP^2 = RA^2 + AP^2

RA^2 = (2a - 2a - 3)^2

We'll eliminate like terms:

RA^2 = 9 (1)

AP^2 = (8-4)^2

AP^2 = 16 (2)

We'll add (1) and (2) and we'll get:

RP^2 = 9 + 16

RP^2 = 25

RP = 5

Since it is a distance, we'll reject the negative answer.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The distance d between (x1,y1) and (x2,y2) is given by:

d = sqrt{(x2-1)^2+(y2-y1)^2}.

Therefore th distance between the points R(2a+3 , 8) and P(2a , 4) is given by:

d = RP = sqrt((2a - (2a+3))^2+((4-8)^2)} = sqrt{(-3)^2+(_4)^2}

RP = sqrt{9+16} = sqrt25= 5.

RP = 5.

Therefore the distance between R and P is 5.

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