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,`P R=3sectheta``P S=4cos e ctheta``P R=+P S=(2(3sintheta+4costheta_)/(sin2theta)``9/((P R)^2)+(16)/((P S)^2)=1`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`

A line which makes an acute angle `theta`with the positive direction o

1.	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,`P R=3sectheta``P S=4cos e ctheta``P R=+P S=(2(3sintheta+4costheta_)/(sin2theta)``9/((P R)^2)+(16)/((P S)^2)=1`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» Correct Answer - A::B::C::D 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.

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