Question

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

Answer 1

given , distance between points (4,p ) and (1,0) is 5 units.

then p = ?

√(1-4)²+(0-p)² = 5

9+p²= 25

p²= 25-9

p = √16

p = 4 units

Answer 2

The value of p is either -4 and 4.

Step-by-step explanation:

Distance formula:

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

It is given that the distance between the points (4,p) and (1,0) is 5. We need to find the value of p.

Using distance formula the distance between (4,p) and (1,0) is

$D=\sqrt{(1-4)^2+(0-p)^2}$

$D=\sqrt{(-3)^2+(-p)^2}$

$D=\sqrt{9+p^2}$

Substitute D=5 in the above equation.

$5=\sqrt{9+p^2}$

Taking square on both sides.

$25=9+p^2$

Subtract 9 from both sides.

$25-9=9+p^2-9$

$16=p^2$

Taking square root on both sides.

$\pm \sqrt{16}=p$

$\pm 4=p$

Therefore, the value of p is either -4 and 4.

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