Question

If the distance between the points (4,p) & (1,0) is 5, then the value of p is:​

Answer 1

Answer:

The given points are(4,p) and (1,0)

d=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

⇒5=

(4−1)

2

+p

2

⇒25=(4−1)

2

+p

2

⇒25−9=p

2

⇒p

2

=16

⇒p=+4,−4

Answer 2

The value of " p = 5 ".

Step-by-step explanation:

★ Given that :

• A(4, p) ; B(1, 0)
• Distance between AB = 5 sq. units.

★ To find :

• Value of " p ".

★ Formula :

$\boxed{\large{\underline{\tt{\red{ Distance : AB = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}} }}}}}$

★ Let :

• x1 = 4 ; y1 = p
• x2 = 1 ; y2 = 0
• AB = 5

$\sf \implies 5 = \sqrt{(1 - 4)^{2}+(0-p)^{2}}$

$\sf \implies 5 = \sqrt{(-3)^{2}+(-p)^{2}}$

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

• [Squaring on both sides.]

$\sf \implies (5)^{2}=(\sqrt{9+p^{2}})^{2}$

$\sf \implies 25 = 9 + p^{2}$

$\sf \implies 25 - 9 = p^{2}$

$\sf \implies p^{2} = 16$

$\sf \implies p= \sqrt{16}$

$\sf \implies p= 4$

★ Verification :

Substitute all the values in the formula.

$\sf \implies \sqrt{(1-4)^{2}+(0-4)^{2}}$

$\sf \implies \sqrt{(-3)^{2}+(-4)^{2}}$

$\sf \implies \sqrt{9+16}$

$\sf \implies \sqrt{25}$

$\sf \implies 5$

◼ Since, LHS = RHS.

◼ Hence, it was verified.

$\underline{\boxed{\bf{\purple{\therefore The\:value\:of\:p=\pm \: 5.}}}}\:\orange{\bigstar}$

★ More info :

Some related formulae :-

$\boxed{\large{\underline{\rm{\red{ Mid-point : \Bigg( \cfrac{x_{1} + x_{2}}{2} \: \: , \: \: \cfrac{y_{1} + y_{2}}{2} \Bigg) }}}}}$

$\boxed{\large{\underline{\rm{\red{ Section : \Bigg ( \cfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1}+m_{2}} \:\:, \cfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \Bigg) }}}}}$

$\boxed{\large{\underline{\rm{\red{ Slope : \cfrac{y_{2} - y_{1}}{x_{2} - x_{1}}}}}}}$

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