In what direction should a line be drawn through the point (1, 2)

Solution:

Let the given line $x+y=4$ and the required line ' $T$ ' intersect at $B(a, b)$ Slope of line ' $T$ ' is

$\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{b}-2}{\mathrm{a}-1}$

And we also know that, $m=\tan \theta$

$\therefore \tan \theta=\frac{\mathrm{b}-2}{\mathrm{a}-1}$ $\ldots$ (i)

Given that $A B=\sqrt{6} / 3$

So, by distance formula for point $A(1,2)$ and $B(a, b)$, we get

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\Rightarrow \frac{\sqrt{6}}{3}=\sqrt{(a-1)^{2}+(b-2)^{2}}$

On squaring both the sides, we get

$\Rightarrow \frac{6}{9}=(a-1)^{2}+(b-2)^{2}$

Using $(a-b)^{2}$ formula we get

$\Rightarrow \frac{2}{3}=a^{2}+1-2 a+b^{2}+4-4 b$

On cross multiplication we get

$\Rightarrow 2=3 a^{2}+3-6 a+3 b^{2}+12-12 b$

$\Rightarrow 2=3 a^{2}+3 b^{2}-6 a-12 b+15$

$\Rightarrow 3 a^{2}+3 b^{2}-6 a-12 b+13=0$ $\ldots$ (ii)

Point $B(a, b)$ also satisfies the equation $x+y=4$

$\therefore a+b=4$

$\Rightarrow \mathrm{b}=4-\mathrm{a} \ldots$ (iii)

Putting the value of $b$ in equation (ii),

we get $3 a^{2}+3(4-a)^{2}-6 a-12(4-a)+13=0$

$3 a^{2}+3(4-a)^{2}-6 a-12(4-a)+13=0$

Computing and simplifying we get

$\Rightarrow 3 a^{2}+3\left(16+a^{2}-8 a\right)-6 a-48+12 a+13=0$

$\Rightarrow 3 a^{2}+48+3 a^{2}-24 a-6 a-48+12 a+13=0$

$\Rightarrow 6 a^{2}-18 a+13=0$

Now, we solve the above equation by using this formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$a=\frac{-(-18) \pm \sqrt{(-18)^{2}-4 \times 6 \times 13}}{2 \times 6}$

$a=\frac{18 \pm \sqrt{324-312}}{12}$

$a=\frac{18 \pm \sqrt{12}}{12}$

$a=\frac{9 \pm \sqrt{3}}{6}$

$a=\frac{\sqrt{3}(3 \sqrt{3} \pm 1)}{\sqrt{3}(2 \sqrt{3})}$

$\Rightarrow a=\frac{3 \sqrt{3} \pm 1}{2 \sqrt{3}}$

$\Rightarrow a=\frac{3 \sqrt{3}+1}{2 \sqrt{3}}$ or $a=\frac{3 \sqrt{3}-1}{2 \sqrt{3}}$

Putting the value of a in equation (iii), we get

$\mathrm{b}=4-\frac{3 \sqrt{3} \pm 1}{2 \sqrt{3}}$

Taking LCM and simplifying we get

$\Rightarrow \mathrm{b}=\frac{8 \sqrt{3}-3 \sqrt{3} \pm 1}{2 \sqrt{3}}$

$\Rightarrow \mathrm{b}=\frac{5 \sqrt{3} \pm 1}{2 \sqrt{3}}$

$\Rightarrow \mathrm{b}=\frac{5 \sqrt{3}+1}{2 \sqrt{3}}$ or $\mathrm{b}=\frac{5 \sqrt{3}-1}{2 \sqrt{3}}$

Now, putting the value of a and b in equation (i), we get

$\tan \theta=\frac{b-2}{a-1}$

Now, putting the value of $a$ and $b$ in equation (i), we get

$\tan \theta=\frac{b-2}{a-1}$

$\Rightarrow \tan \theta=\frac{\frac{5 \sqrt{3} \pm 1}{2 \sqrt{3}}-2}{\frac{3 \sqrt{3} \pm 1}{2 \sqrt{3}}-1}$

Taking LCM and simplifying we get

$\Rightarrow \tan \theta=\frac{\frac{5 \sqrt{3} \pm 1-4 \sqrt{3}}{2 \sqrt{3}}}{\frac{3 \sqrt{3} \pm 1-2 \sqrt{3}}{2 \sqrt{3}}}$

$\Rightarrow \tan \theta=\frac{\sqrt{3} \pm 1}{\sqrt{3} \pm 1}$

$\Rightarrow \tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ ........1

$\tan \theta=\frac{\sqrt{3}-1}{\sqrt{3}+1}$...........2

We solve the equation 1 to get the value of $\theta$, we get We know that,

$\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

$\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

if $x=\sqrt{3}$ and $y=1$

$=\tan ^{-1}\left(\frac{\sqrt{3}-1}{1+(\sqrt{3})(1)}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{3}-1}{1+\sqrt{3}}\right)$

We have,

$\theta=\tan ^{-1}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$

$\theta=\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)$

$\theta=\tan ^{-1}\left(\tan 60^{\circ}\right)-\tan ^{-1}\left(\tan 45^{\circ}\right)$

$\theta=60^{\circ}-45^{\circ}$

$\theta=15^{\circ}$

$\theta=15^{\circ}$

Now, we solve the equation 2

We know that,

$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

if $x=\sqrt{3}$ and $y=1$

$=\tan ^{-1}\left(\frac{\sqrt{3}+1}{1-(\sqrt{3})(1)}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{3}+1}{1-\sqrt{3}}\right)$

We have,

$\theta=\tan ^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$

$\theta=\tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)$

$\theta=\tan ^{-1}\left(\tan 60^{\circ}\right)+\tan ^{-1}\left(\tan 45^{\circ}\right)$

$\theta=60^{\circ}+45^{\circ}$

$\theta=105^{\circ}$