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Q: In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance √6/3 from the given point?

Answers (1)

Let the given line x+y=4 and the required line'l'intersect at B (a,b)

Slope of line 'l' is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{b-2}{a-1}$

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

Given that $AB=\frac{\sqrt{6}}{3}$

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

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

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

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

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

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

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

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

a+b=4; b=4-a

Putting the value of b in equation (i )we get 3a²+3(4-a)²-6a-12(4-a)+13=0

3a²+48+3a²-24a-6a-48+12a+13=0

6a²-18a+13=0

Using the formula $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$a=\frac{-\left ( -18 \right )\pm \sqrt{\left ( -18 \right )^{2}-4*6*13}}{2*6}=\frac{18\pm \sqrt{324-312}}{12}$

$=\frac{18\pm \sqrt{12}}{12}=\frac{9\pm \sqrt{3}}{6}$

Putting the value of a in the equation we get

$b=4-\frac{\left (9\pm \sqrt{3} \right )}{6} =\frac{15\pm \sqrt{3}}{6}$

Now putting the value of a and b in equation $\tan \theta =\frac{b-2}{a-1}$

$=\frac{\frac{\left (15 \pm \sqrt{3}\right )}{6}-2} {\frac{9\pm \sqrt{3} }{6}-1}=\frac{3 \pm \sqrt{3}}{3\pm \sqrt{3}}$

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

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

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

θ= 60⁰-45⁰=15⁰

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

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

θ= 60⁰+45⁰=105⁰

Posted by

infoexpert22

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