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.
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,
$\theta=\tan ^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$
$\theta=\tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)$
$\theta=\tan ^{-1}\left(\tan 60^{\circ}\right)+\tan ^{-1}\left(\tan 45^{\circ}\right)$
$\theta=60^{\circ}+45^{\circ}$
$\theta=105^{\circ}$
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All Study Material
- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Nuclei
- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,