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`        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.`
one year 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.
```
one year ago
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