Question

The distance between the point (4, p) and (1,0) is 5 then the value of p is

Answer 1

Step-by-step explanation:

acorrding to distance formula:-

AB= UNDER ROOT (x2-x1)(y2-y1)

5= underroot (1-4)(0-p)

5= underroot (-3)(-p)

5= underroot (3p)

squaring on both side,

25=3p

25/3=p

hope helps...(ans is approx, calculated, can not be sure...)

mark me brainlist...

Answer 2

Answer:

The value of " p = 5 ".

Step-by-step explanation:

$\sf\small\underline\red{★ Given \: that :-}$

A(4, p) ; B(1, 0)

Distance between AB = 5 sq. units.

$\sf\small\underline\red{★ To \: find :-}$

Value of " p ".

$\sf\small\underline\red{★ Formula :-}$

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

★ 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.

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

★ More info :

Some related formulae :-

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

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

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

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