Question

If the distance between the
points (4,p)and (1,6) is 5 units
Then find the value of p

Answer 1

❏Question :-

If the distance between the points (4,p) and (1,6) is 5 units ,then find the value of p.

❏Answer :-

→ p = 2 or 10 .

❏To Find :-

→ Value of p

❏Formula used :-

$\sf \: distance \: between \: the \: points \: (x_1, y_1) \\ \sf and \: (x_2 \: ,y_2) \: is \: - \\ \boxed{ \sf D = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 -y_1) }^{2} } }$

❏Solution :-

Given that the distance between point (4,p) and (1,6) is 5 units ,

$\sf (4,p) \implies x_1 = 4, \: y_1= p \\ \\ \sf (1,6) \implies \: x_2 = 1 \:, \: y_2 = 6 \: \\ \\ \sf D = 5 \: Units \: \\ \\ \bf \red{Now \: ,apply \: the \: given \: formula } \\ \\ \to \sf 5 = \sqrt{ {(4 - 1)}^{2} + {(6 - p)}^{2} } \\ \\ \bf \green{Squaring \: on \: both \: sides } \\ \\ \to \: {(5)}^{2} = { \bigg \{\sqrt{ {(4 - 1)}^{2} + {(6 - p)}^{2} } \bigg \} \: }^{2} \\ \\ \to \sf \: 25 = 9 + {(6 - p)}^{2} \\ \\ \to \sf \: 25 - 9 = {(6 - p)}^{2} \\ \\ \to \sf 16 = {(6 - p)}^{2} \\ \because \bf {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ \to \sf 16 = 36 + {p}^{2} - 12p \\ \\ \to \sf \: {p}^{2} - 12p + 36 - 16 = 0 \\ \\ \to \sf {p}^{2} - 12p + 20 = 0 \\ \\ \to \sf {p}^{2} - 10p - 2p + 20 = 0 \\ \\ \to \sf \: p(p - 10) - 2(p - 10) = 0 \\ \\ \to \sf (p - 10)(p - 2) = 0 \\ \\ \bf \: case \: (1) \\ \\ \implies \sf p - 10 = 0 \\ \implies \boxed{ \sf p = 10 } \\ \\ \bf case \: (2) \\ \\ \implies \sf p - 2 = 0 \\ \implies \boxed {\sf p = 2}$

Answer 2

Answer:

p can be 10 or 2

Step-by-step explanation:

We Have :-

Points ( 4 , p ) and ( 1 , 6 )

Distance between them is 5 units

To Find :-

p

Formula Used :-

Distance² = ( x₂ - x₁ )² + ( y₂ - y₁ )²

Solution :-

( 5 )² = ( 1 - 4 )² + ( 6 - p )²

25 = 9 + 36 + p² - 12p

p² - 12p + 20 = 0

p² - 10p - 2p + 20 = 0

p ( p - 10 ) - 2 ( p - 10 ) = 0

( p - 10 ) ( p - 2 ) = 0

so , p = 10 or 2

p can be 10 or 2