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What are the coordinates of point B on AC such that the ratio of AB to AC is 5:6

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Coordinates of point B are (5x₂+6x₁)/11 and (5y₂+6y₁)/11

What is the section formula in coordinate geometry?

When the line segment is divided internally in the ratio m:n, we use this formula. That is when point C lies somewhere between points A and B. p= (mx₂+nx₁)/(m+n) & q= (my₂+ny₁)/(m+n)

Given here, point B on AC such that the ratio of AB to AC is 5:6

Let the point B(p,q) internally divide the line joining points A(x₁, y₁) and C(x₂, y₂) in the ratio m : n , then,

p= (mx₂+nx₁)/(m+n)

q= (my₂+ny₁)/(m+n)

Here. m=5 and n=6

p=(5x₂+6x₁)/11 , q=(5y₂+6y₁)/11

Hence, the Coordinates of point B are (5x₂+6x₁)/11 and (5y₂+6y₁)/11

I hope this helps <3

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