Question

If the distance between the points (4, p) and (1, 0) is 5 units, then the value of p is

Answer 1

$\underline{ \sf \fcolorbox{red}{pink}{\huge{Solution :)}}}$

Given ,

The given two points are (4,P) & (1,0) and the distance between them is 5 units

We know that , the distance between two points is given by

$\sf \fbox{D = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1})}^{2} } \: \: }$

Substitute the known values , we get

$\sf \mapsto 5 = \sqrt{ {(1 - 4)}^{2} + {(0 - p)}^{2} } \\ \\\sf squaring \: both \: sides \: , \: we \: get \\ \\\sf \mapsto 25 = {( - 3)}^{2} + {(p)}^{2} \\ \\\sf \mapsto 25 = 9 + {(p)}^{2} \\ \\ \sf \mapsto {(p)}^{2} = 16 \\ \\ \sf \mapsto p = ± \: 4$

Hence , the required value of P is ± 4

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Answer 2

Hello !!

To you solve this question is very easy, you just need make use of this formula below.

d = √[(xB - xA)^2 + (yB - yA)^2]

Now that you have knowledge about the formula, you can just put the information of the statement in the formula, develop and find the solution.

d = √[(xB - xA)^2 + (yB - yA)^2]

5 = √[(1 - 4)^2 + (0 - p)^2]

5 = √[(-3)^2 + (0 - p)^2]

5 = √[9 + (0 - p)^2]

5 = √[9 + (0 - p)(0 - p)]

5 = √[9 + p^2]

√[9 + p^2] = 5

[9 + p^2] = 5^2

9 + p^2 = 25

p^2 = 25 - 9

p^2 = 16

p = ±√16

p' = +4

p'' = -4

Therefore, with all this reasoning we have that the (p) has 2 solutions. The first solution is +4 and the second solution is -4.

I hope I have collaborated !