Question

Find distance between the points P(-1, 1) and Q(5.-7).​

Answer 1

Answer:

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Answer 2

Given

⇒P(-1,1) and Q(5,-7) are point

To Find

⇒Distance Between PQ

Formula

⇒PQ² = (x₂ - x₁)² + (y₂ - y₁)²

Now we have

⇒x₁ = -1 , y₁= 1 , x₂ = 5 and y₂ = -7

⇒PQ² = (5 - (-1))² + (-7-1)²

⇒PQ² = (5+1)² + (-8)²

⇒PQ² = (6)² + 64

⇒PQ² = 36 + 64

⇒PQ² = 100

⇒PQ = √(100)

⇒PQ = 10 Units

Answer

⇒Distance Between PQ =  10 Units

More Formula

Distance Formula

⇒D² = (x₂ - x₁)² + (y₂ - y₁)²

Section Formula

⇒P(x,y) = [(mx₂ + nx₁)/(m+n) , (my₂ + ny₁)/(m+n)]

Area of Triangle

⇒ΔABC = 1/2{x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)}