Q: 20    If the sum of the perpendicular distances of a variable point  \small P(x,y)  from the lines  \small x+y-5=0  and  \small 3x-2y+7=0  is always  \small 10.  Show that  \small P  must move on a line.     

Answers (1)

Given the equation of line are
x+y-5=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
3x-2y+7=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
Now, perpendicular distances of a variable point  \small P(x,y)  from the lines are

d_1=\left | \frac{1.x+1.y-5}{\sqrt{1^2+1^2}} \right |                                      d_2=\left | \frac{3.x-2.y+7}{\sqrt{3^2+2^2}} \right |
d_1=\left | \frac{x+y-5}{\sqrt2} \right |                                               d_2=\left | \frac{3x-2y+7}{\sqrt{13}} \right |
Now, it is given that
d_1+d_2= 10
Therefore,
\frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10
                                                                 (assuming \ x+y-5 > 0 \ and \ 3x-2y+7 >0)
(x+y-5)\sqrt{13}+(3x-2y+7)\sqrt2=10\sqrt{26}

x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt2

 Which is the equation of the line 
Hence proved

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