Question

the coordinates of a point p on the line 2x-y+5=0 such that |PA -PB| is maximum where A is (4-2) B is (2-1) will be​

Answer 1

Answer:

(-2,1)

Step-by-step explanation:

Let us assume that the coordinates of P are (h,k).

Now point P is on the line 2x-y+5=0.

So, we have, 2h-k+5=0....... (1)

Given that A(4,-2) and B(2,-1) are two points such that, IPA-PBI=D(say) is maximum.

Hence, D=$\sqrt{(4-h)^{2}+(-2-k)^{2} } -\sqrt{(2-h)^{2}+(-1-k)^{2} }$

Now, from equation (1), we have

D=$\sqrt{(4-h)^{2}+(-7-2h)^{2} } -\sqrt{(2-h)^{2}+(-6-2h)^{2} }$

So, differentiating the above equation with respect to h, we get

$\frac{dD}{dh}=\frac{2h-8+28+8h}{2\sqrt{(4-h)^{2}+(-7-2h)^{2} }} -\frac{2h-4+24+8h}{2\sqrt{(2-h)^{2}+(-6-2h)^{2} }}$

For D to be maximum  

$\frac{dD}{dh} =0=\frac{2h-8+28+8h}{2\sqrt{(4-h)^{2}+(7+2h)^{2} }} -\frac{2h-4+24+8h}{2\sqrt{(2-h)^{2}+(6+2h)^{2} }}$

⇒$\frac{10h+20}{\sqrt{(4-h)^{2}+(7+2h)^{2} }}=\frac{10h+20}{\sqrt{(2-h)^{2}+(6+2h)^{2} }}$

⇒ $(h+2)[\frac{1}{\sqrt{(4-h)^{2}+(7+2h)^{2} }} -\frac{1}{\sqrt{(2-h)^{2}+(6+2h)^{2} }} ]=0$

Therefore, one of the solution of the above equation is (h+2)=0 or, h=-2.

Hence, from equation (1) k=2h+5=2(-2)+5 =1.

Hence, coordinates of point P are (-2,1). (Answer)