A line which makes an acute angle `theta` with the positive direction of the x-axis is drawn through the point `P(3,4)` to meet the line `x=6` at `R` and `y=8` at `Sdot` Then,

Options

A. `PR = 3 "sec"theta`
B. `PS = 4 " cosec"theta`
C. `PR+PS = (2(3"sin" theta + 4 "cos" theta))/("sin" 2theta)`
D. `(9)/(PR)^(2) + (16)/(PS)^(2) = 1`

Answer

A, B, C, D

Explanation

The equation of any line making an acute angle `theta` with the positive direction of the x-axis and passing through P(3, 4) is
`(x-3)/("cos"theta) = (y-4)/("sin" theta) " " (i)`
Where |r| is the distance of any point (x,y) from P. Therefore, `A(r " cos "theta +3, r " sin " theta +4)` is a general point on line (i). If A is R, then
`r " cos "theta +3 = 6 " or " r =(3)/("cos "theta)= 3 "sec"theta`
`"Since "theta "is acute, " cos" theta gt 0`. Therefore,
`PR = r = 3 " sec " theta " "(ii)`
`"If " A-=S, r " sin "theta +4 =8. "Therefore",`
`r = 4 " cosec " theta`
`therefore PS = 4 " cosec " theta`
`"Also, " PR+PS = (3)/("cos"theta) + (4)/("sin" theta)`
` = (2(3"sin" theta + 4"cos" theta))/("sin" 2theta)`
`"and " (9)/((PR)^(2)) + (16)/((PS)^(2)) = "cos"^(2)theta + "sin"^(2)theta = 1`
Therefore, (1),(2), (3), and (4) all are correct.