# Consider a point A(0,1) and B(2,0).Let P is a point on line 4x+3y+9=0. Find co-ordinates of P such that |PA-PB| is maximum.

Grade:11

## 1 Answers

Arun
25757 Points
4 years ago
Solving the equation given for y:

4x + 3y + 9 = 0
==> 3y = -4x - 9
==> y = (-4/3)x - 3.

So any point on this line takes the form of [x, (-4/3)x - 3].

The distance between A(0, 1) and P[x, (-4/3)x - 3] is:

d(P, A) = √{[(-4/3)x - 3 - 1]^2 + (x - 0)^2]} = √[(25/9)x^2 + (32/3)x + 16].

The distance between B(2, 0) and P[x, (-4/3)x - 3] is:

d(P, B) = √{[(-4/3)x - 3]^2 + (x - 2)^2} = √[(25/9)x^2 + 4x + 13].

Then:

d(P, A) - d(P, B) = √[(25/9)x^2 + (32/3)x + 16] - √[(25/9)x^2 + 4x + 13].

Taking derivatives of √[(25/9)x^2 + (32/3)x + 16] - √[(25/9)x^2 + 4x + 13], we see that:

d/dx √[(25/9)x^2 + (32/3)x + 16] - √[(25/9)x^2 + 4x + 13]
= [(50/9)x + 32/3]/{2√[(25/9)x^2 + (32/3)x + 16]} - [(50/9)x^2 + 4]/{2√[(25/9)x^2 + 4x + 13]}.

Setting this expression equal to zero yields x = -24/5. This is the x-coordinate of P. Then, the corresponding y-coordinate is (-4/3)(-24/5) - 3 = 17/5. Therefore, the required point is (-24/5, 17/5) as required.

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