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Question: Consider the points \(A(0,1)\) and \(B(2,0)\) and \(P\) be a point on the line \(4x + 3y + 9 = 0\). ...

Consider the points A(0,1)A(0,1) and B(2,0)B(2,0) and PP be a point on the line 4x+3y+9=04x + 3y + 9 = 0. The coordinates of PP such that PAPB|PA - PB| is maximum are
A. (125,15)(\dfrac{{12}}{5},\dfrac{{ - 1}}{5})
B. (845,135)(\dfrac{{ - 84}}{5},\dfrac{{13}}{5})
C. (245,175)(\dfrac{{ - 24}}{5},\dfrac{{17}}{5})
D. (0,3)(0, - 3)

Explanation

Solution

The given question shall be solved with the help of diagrams. Also, any expression containing mod (| |) actually expresses the distance. So here PAPB|PA - PB| expresses distance.

Complete step by step solution:
The given two points are A(0,1)A(0,1) and B(2,0)B(2,0). Since PP is a point which lies on the line 4x+3y+9=04x + 3y + 9 = 0, soPAPA and PBPB can be shown as given below:

(Source: Geogebra Application)

As PP is a point on the line 4x+3y+9=04x + 3y + 9 = 0, we can show this line as given below:

From the figure, it is evident that PAPB|PA - PB| is actually the distance between point AA and point BB. Let the coordinates of PP be (x,y)(x,y). So when we move PP upwards the line, we stop at a point where P(x,y)P(x,y), A(0,1)A(0,1) and B(2,0)B(2,0) becomes collinear. This means that PP, AAand BB lies on the same line as shown below:

Hence, this is where P(x,y)P(x,y) must be located on the line 4x+3y+9=04x + 3y + 9 = 0 so as to maximize PAPB|PA - PB|. Now, to find the coordinates of PP, we will solve the equation for both the lines, the first one is on which PP lies, i.e. 4x+3y+9=04x + 3y + 9 = 0 and the second one is on which AA and BB lies.
But for the second line, we will have to find the equation of that line on which A(0,1)A(0,1) and B(2,0)B(2,0) lies. So we will use the two-point method, such that
(x1,y1)=(0,1)\Rightarrow ({x_1},{y_1}) = (0,1)
x1=0\Rightarrow {x_1} = 0 and y1=1 \Rightarrow {y_1} = 1
And, (x2,y2)=(2,0) \Rightarrow ({x_2},{y_2}) = (2,0)
x2=2\Rightarrow {x_2} = 2 and y2=0 \Rightarrow {y_2} = 0
On substituting these values in the equation for the two-point method below, we will get
(yy1)=y2y1x2x1(xx1)\Rightarrow (y - {y_1}) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})
(y1)=0120(x0)\Rightarrow (y - 1) = \dfrac{{0 - 1}}{{2 - 0}}(x - 0)
On simplifying, we will get
(y1)=12x\Rightarrow (y - 1) = \dfrac{{ - 1}}{2}x
On multiplying both sides by 22, we will get
2(y1)=x\Rightarrow 2(y - 1) = - x
2y2=x\Rightarrow 2y - 2 = - x
On rearranging variables and constants on different sides, we will get
2y+x=2\Rightarrow 2y + x = 2
x+2y2=0\Rightarrow x + 2y - 2 = 0
Hence, the equation of the line on which AA and BBlies is x+2y2=0x + 2y - 2 = 0.
Now we will solve 4x+3y+9=04x + 3y + 9 = 0 and x+2y2=0x + 2y - 2 = 0, so as to obtain the coordinates of point PP. For this we will find the value of xx from x+2y2=0x + 2y - 2 = 0, such that
x=22y\Rightarrow x = 2 - 2y
We will now substitute this value of xx into the equation 4x+3y+9=04x + 3y + 9 = 0, such that
4(22y)+3y+9=0\Rightarrow 4(2 - 2y) + 3y + 9 = 0
88y+3y+9=0\Rightarrow 8 - 8y + 3y + 9 = 0
On arranging the variables and constants on the left hand side and the right hand side respectively, we will get
3y8y=98\Rightarrow 3y - 8y = - 9 - 8
5y=17\Rightarrow - 5y = - 17
Dividing both sides by 5 - 5, we get
y=175\Rightarrow y = \dfrac{{17}}{5}
Now, on putting this value of yy in the equation x+2y2=0x + 2y - 2 = 0, we will get
2×175+x2=0\Rightarrow 2 \times \dfrac{{17}}{5} + x - 2 = 0
345+x=2\Rightarrow \dfrac{{34}}{5} + x = 2
On subtracting 345\dfrac{{34}}{5} from both the sides, we will get
x=2345\Rightarrow x = 2 - \dfrac{{34}}{5}
x=10345\Rightarrow x = \dfrac{{10 - 34}}{5}
Further solving, we will get
x=245\Rightarrow x = \dfrac{{ - 24}}{5}
Hence the coordinates of the point PP for which PAPB|PA - PB| is maximum is (245,175)(\dfrac{{ - 24}}{5},\dfrac{{17}}{5}).

Hence, the correct answer is option C.

Note:
It will seem appropriate at first to apply the distance formula here so as to calculate PAPB|PA - PB| but doing so will not lead us to any answer. Therefore in such questions, we must avoid using the distance formula and try to solve the question with the help of graphs.