Question

A line which makes an acute angle $\theta$ 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. $PR=3sec\theta$
• B. $PS=4cosec\theta$
• C. $PR+PS=\frac{2(3sin\theta +4cos\theta )}{sin2\theta }$
• D. $\frac{9}{(PR{)}^2}+\frac{16}{(PS{)}^2}=1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $PR=3sec\theta$
B $PS=4cosec\theta$
C $PR+PS=\frac{2(3sin\theta +4cos\theta )}{sin2\theta }$
D $\frac{9}{(PR{)}^2}+\frac{16}{(PS{)}^2}=1$

Equation of any line making an acute angle $\theta$ with positive direction of x-axis and passing through P(3, 4) is

$\frac{x-3}{cos\theta }=\frac{y-4}{sin\theta }=r$ . . .(i)

where |r| is the distance of any point (x, y) from P.

Therefore, $A(rcos\theta +3,rsin\theta +4)$ is a general point on line (1).

If A is R, $rcos\theta +3=6r=\frac{3}{cos\theta }=3sec\theta$

Since $\theta$ is acute, $cos\theta >0$

$\therefore PR=r=3sec\theta$ . . . (ii)

If A = S, r $sin\theta +4=8$,

$\therefore r=4cosec\theta$,

$\therefore$ PS = 4 $cosec\theta$,

Also $PR+PS=\frac{3}{cos\theta }+\frac{4}{sin\theta }$

$=\frac{2(3sin\theta +4cos\theta )}{sin2\theta }$ and $\frac{9}{(PR{)}^2}+\frac{16}{(PS{)}^2}=co{s}^2\theta +si{n}^2\theta =1$

$\therefore$ (A), (B), (C) and (D) all are correct.