Find equations of the lines passing through the point &nbs

Find equations of the lines passing through the point $\mathrm{(2, 3)}$ and making intercept of length $\mathrm{2}$ units between the lines $\mathrm{y + 2x = 3}$ and $\mathrm{ y + 2x = 5. }$

Option: 1
$\mathrm{3x+2y-9=0 }$ and $\mathrm{x-2=0}$

Option: 2
$\mathrm{3x+4y-18=0 }$ and $\mathrm{x-2=0}$ $\mathrm{x-2=0}$

Option: 3
$\mathrm{4x-3y-9=0}$ and $\mathrm{x+2y+5=0}$

Option: 4
$\mathrm{3x+4y-18=0}$ and $\mathrm{x+2y+5=0}$

Answers (1)

The given lines are $\mathrm{y + 2x = 3}$ & $\mathrm{y + 2x = 5}$ which are parallel.
Equation of any line through $\mathrm{P}$ $\mathrm{(2, 3)}$ is $\mathrm{\frac{x-2}{cos\Theta }=\frac{y-2}{sin\Theta }=r}$
Since this line makes an intercept of $\mathrm{2}$ between the given lines,

If $\mathrm{(2\ +\ r\ cos\ \Theta ,3\ +r\ sin\Theta )}$ lies on y + 2x = 3
then $\mathrm{(2 + (r + 2) cos \Theta , 3 + (r + 2) sin\ \Theta )}$ line on $\mathrm{ y + 2 x = 5. }$
$\mathrm{\Rightarrow }$ $\mathrm{3 + r\ sin \Theta + 2\ ( 2 + r\ cos \Theta ) = 3 }$
and $\mathrm{3 + (r + 2)\ sin\ \Theta + 2\(2 + ( r + 2) cos\ \Theta ) = 5 }$
Subtracting, we get $\mathrm{2\ sin\ \Theta + 4\ cos\ \Theta = 2 }$
$\mathrm{\Rightarrow }$ $\mathrm{sin\ \Theta + 2\ cos\ \Theta = 1 \Rightarrow 4\ cos\2\ \Theta = (1 -sin\ \Theta )^{2} }$
$\Rightarrow$ $\mathrm{(1 - sin \Theta ) [4 + 4 sin \Theta - 1 + sin\Theta ] = 0 }$
$\mathrm{\Rightarrow sin \Theta = 1 \ or\ sin \Theta = - 3/5 }$
Now if $\mathrm{ sin \Theta = -3/5, }$ $\mathrm{ cos \Theta = 4/5 }$ and the equation of the required line is $\mathrm{\frac{5\left ( x-2 \right )}{4}=\frac{5\left ( y-3 \right )}{-3} }$
$\mathrm{\Rightarrow 3x + 4y = 18 }$
and if $\mathrm{sin \Theta = 1,}$ $\mathrm{cos \Theta = 0 }$ and the required line is $\mathrm{ x - 2 = 0.}$