Question

The distance between (2,4) and (p,3) is 5 .Then p=.

Answer 1

Answer:

The value of p is 6.9 or - 2.9 approximately.

Step-by-step-explanation:

Let the two points be A & B.

• A ≡ ( 2, 4 ) ≡ ( x₁, y₁ )
• B ≡ ( p, 3 ) ≡ ( x₂, y₂ )

We have given that,

The distance between given points is 5 units.

We have to find the value of p.

Now, by distance formula,

d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]

⇒ 5 = √[ ( 2 - p )² + ( 4 - 3 )² ]

By squaring both sides, we get,

⇒ 25 = ( 2 - p )² + ( 1 )²

⇒ 25 = 2² - 2 * 2 * p + p² + 1

⇒ 4 - 4p + p² + 1 = 25

⇒ p² - 4p + 4 + 1 - 25 = 0

⇒ p² - 4p + 5 - 25 = 0

⇒ p² - 4p - 20 = 0

Comparing with ax² + bx + c = 0, we get,

• a = 1
• b = - 4
• c = - 20

Now, by quadratic formula,

$\displaystyle{\boxed{\pink{\sf\:p\:=\:\dfrac{-\:b\:\pm\:\sqrt{b^2\:-\:4ac}}{2a}}}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{-\:(\:-\:4\:)\:\pm\:\sqrt{(\:-\:4\:)^2\:-\:4\:\times\:1\:\times\:(\:-\:20\:)}}{2\:\times\:1}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{4\:\pm\:\sqrt{16\:-\:(\:-\:80\:)}}{2}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{4\:\pm\:\sqrt{16\:+\:80}}{2}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{4\:\pm\:\sqrt{96}}{2}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{4\:\pm\:\sqrt{16\:\times\:6}}{2}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{4\:\pm\:4\:\sqrt{6}}{2}}$

$\displaystyle{\implies\sf\:p\:=\:\dfrac{\cancel{4}\:(\:1\:\pm\:\sqrt{6}\:)}{\cancel{2}}}$

$\displaystyle{\implies\sf\:p\:=\:2\:(\:1\:\pm\:\sqrt{6}\:)}$

$\displaystyle{\implies\sf\:p\:=\:2\:\pm\:2\:\sqrt{6}}$

$\displaystyle{\implies\sf\:p\:=\:2\:+\:2\:\sqrt{6}\:\quad\:OR\:\quad\:p\:=\:2\:-\:2\:\sqrt{6}}$

$\displaystyle{\implies\sf\:p\:=\:2\:+\:2\:\times\:2.449\:\quad\:OR\:\quad\:p\:=\:2\:-\:2\:\times\:2.449}$

$\displaystyle{\implies\sf\:p\:=\:2\:+\:4.898\:\quad\:OR\:\quad\:p\:=\:2\:-\:4.898}$

$\displaystyle{\implies\sf\:p\:=\:6.898\:\quad\:OR\:\quad\:p\:=\:-\:2.898}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:p\:\approx\:6.9\:}}}\sf\:\quad\:OR\:\quad\:\underline{\boxed{\red{\sf\:p\:\approx\:-\:2.9\:}}}}$

∴ The value of p is 6.9 or - 2.9 approximately.

Answer 2

Given Data :

let i consider the points be X & Y

X = (2,4) = (x₁, y₁)

Y = (p,3) = (x₂, y₂)

and, Distance b/w X & Y is 5 units .

To Find :

Value of p .

Solution :

by using Distance Formula,

(XY)² = (x₂ - x₁)² + (y₂ - y₁)²

Substituting the values,

⇒ (5)² = (p - 2)² + (3 - 4)²

⇒ 25 = (p - 2)² + (-1)²

✿ Squaring on both sides,

⇒ 5 - 1 = (p - 2)

⇒ p - 2 = 4

⇒ p = 4 + 2

⇒ p = 6

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.°. The Value of p is 6 .