Two straight lines pass through the point $(-2,5)$ such that one of them makes an angle of $\arctan \frac 34$
with the given line $x-y+5=0$
and the given line makes an angle of $\arctan \frac 23$
with the other line . Find the equations of the two lines.
$\bf{Try} :$
Let the equation of the line be $y=mx+c$.$\cdots (1)$
If $(1)$ makes angle $\theta$ with the given line then $\tan\theta =| \frac {m-1}{1+m}|$
So by the given condition we have
$| \frac {m-1}{1+m}|=\frac 34$
Solving we get $m=7,\frac 17$
Putting these values of $m$ and using the condition that the line passes through $(-2,5)$ we get two lines as
$y=7x+19 \cdots (2)$
$7y=x+37\cdots (3)$
Similarly $| \frac {m-1}{1+m}|=\frac 23$
gives $m=5,\frac 15$ and we get the equations of the lines as
$y=5x+15\cdots (4)$
and $5y=x+27\cdots (5)$
But the answer is given as the lines $(2)$ and $(5)$. Why the lines $(3)$ and $(4)$ are excluded?
Best Answer
In coordinate geometry angles are usually measured in counter-clockwise direction from positive $x-$ axis. Your book is probably following that convention. You need to find two lines $L_1$ and $L_2$, where $L_1$ is the line that makes angle of $\arctan \frac34$ with the given line.
The given line makes an angle of $\frac \pi 4$ with positive $x-$ axis. So the angle that $L_1$ makes with positive $x-$ axis $=\frac \pi 4+\arctan \frac 34$ and so slope (gradient) of $L_1=\tan (\pi/4+\arctan \frac 34)=7$ and since this line passes through $(-2,5)$, its equation is: $y-5=7(x+2)$
Similarly, you may get equation of $L_2$.