The line `y=(3x)/4` meets the lines `x-y=0` and `2x-y=0` at points `Aa n dB` , respectively. If `P` on the line `y=(3x)/4` satisfies the condition `P

Correct Answer - 3
Point P which lies on the line y=3x/4 can be chosen as P(h, 3h/4).
If `theta` is the angle that the line y=3x/4 makes with the positive direction of the x-axis, then
`"tan " theta = (3)/(4) " or cos " theta =(4)/(5) " and sin" theta = (3)/(5)`
Now, the coordinates of points A and B which lie on the line y=3x/4 can be chosen as
`A-=(h+(4r_(1))/(5), (3h)/(4) + (3r_(1))/(5)) " and "B-=(h+(4r_(2))/(5), (3h)/(4) + (3r_(2))/(5))`
Since A lies on the line x-y+1 =0, we have
`(h+(4r_(1))/(5)) -((3h)/(4) + (3r_(1))/(5)) +1 =0`
`"or " r_(1) = (-5)/(4)(h+4)`
Since B lies on the line 2x-y-5=0, we have
`2(h+(4r_(2))/(5))- ((3h)/(4) + (3r_(2))/(5))-5=0`
`"or " r_(2) = (-5)/(4)(h-4)`
According to the given condition, we have
`PA * PB =25`
`i.e., |r_(1)| * |r_(2)| = 25`
`i.e., (25)/(16)(h^(2) -16) = +- 25`
`i.e., h^(2) = 16+-16 = 32, 0`
`i.e., h=+-4sqrt(2), 0`
Hence, the required points are (0, 0), `(4sqrt(2), 3sqrt(2)), " and " (-4sqrt(2), -3sqrt(2)).`