Question

find the distance between the pair of points (-4,6)and (-2,-5).​

Answer 1

Let assume that, The distance between two points is k

So, We have already a formula for the distance

$\sf K^{2} = ( x_2 - x_1 )^{2} + (y_2 -y_1)^{2}$

Where, (According to the question)

$\sf x_1 = -4$

$\sf x_2 = - 2$

$\sf y_1 = 6$

$\sf y_2 = -5$

So, Now we have to put these values in the formula

$\sf K^{2} = ( x_2 - x_1 )^{2} + (y_2 -y_1)^{2}$

$\sf K^{2} = ( ( -2) - (-4) )^{2} + ((-5) - (6))^{2}$

$\sf K^{2} = ( -2 + 4 )^{2} + (-5 - 6)^{2}$

$\sf K^{2} = ( 2 )^{2} + (-11 )^{2}$

$\sf K^{2} = 4 + 121$

$\sf K^{2} = 125$

$\sf K = \sqrt{ 125}$

$\underline{\boxed{\sf K = 5\sqrt{5} \: \: Unit }}$

Hence the distance between two points is 5√5

I hope it helps you ❤️✔️

Answer 2

Solution :-

we know that, the distance between points P(a, b) and Q(c, d) is given by distance formula :-

• D = √[(a - c)² + (b - d)²]

So, comparing given points (-4,6) and (-2,-5) with P(a, b) and Q(c, d) we get,

• a = (-4)
• b = 6
• c = (-2)
• d = (-5) .

then,

→ D = √[(a - c)² + (b - d)²]

putting values we get,

→ D = √[{(-4) - (-2)}² + {6 - (-5)}²]

→ D = √[(-2)² + (11)²]

→ D = √(4 + 121)

→ D = √(125)

→ D = √(5 * 5 * 5)

→ D = 5√5 units (Ans.)

Hence, the distance between the given points is 5√5 units .

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