A line which makes an acute angle θ with the positive direction of x-axis is drawn through the point P(3, 4) to meet the line x = 6 at R and y = 8 at S, then
Equation of any line making an acute angle θ with positive direction of x-axis and passing through P(3, 4) is
x−3cos θ=y−4sin θ=r . . .(i)
where |r| is the distance of any point (x, y) from P.
Therefore, A(r cosθ+3,rsinθ+4) is a general point on line (1).
If A is R, r cos θ+3=6r=3cos θ=3 sec θ
Since θ is acute, cos θ>0
∴ PR=r=3 sec θ . . . (ii)
If A = S, r sin θ+4=8,
∴ r=4 cosec θ,
∴ PS = 4 cosec θ,
Also PR+PS=3cos θ+4sin θ
=2(3 sin θ+4 cos θ)sin 2θ and 9(PR)2+16(PS)2=cos2θ+sin2θ=1
∴ (A), (B), (C) and (D) all are correct.