Question

# 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

A
PR=3 sec θ
B
PS=4 cosecθ
C
PR+PS=2(3 sin θ+4 cos θ)sin 2θ
D
9(PR)2+16(PS)2=1

Solution

## The correct options are A PR=3 sec θ B PS=4 cosecθ C PR+PS=2(3 sin θ+4 cos θ)sin 2θ D 9(PR)2+16(PS)2=1Equation 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.

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