Question

Find the equation of the straight lines passing through $(-2,-7)$ & having an intercept of length $3$ between the straight lines $4x+3y=12$ and $4x+3y=3$.

• A. $7x+24y+182=0$
• B. $x=2$
• C. $7x-24y+182=0$
• D. $x=-2$

Answer 1

Answer: (A), (D)

The correct options are
A $7x+24y+182=0$
D $x=-2$

Let $L_1$ and $L_2$ be the given line. They are parallel, since they have equal slopes.
Let $ABC$ be the required line passing through the given point $A(-2,-7)$ and making an intercept $BC=3$ between the given lines.
Let the equation of the line $ABC$ be
$\frac{x+2}{\cos \theta }=\frac{y+7}{\sin \theta }=r$ ...(1)
where $AB=r$
The coordinate of any point on (1) are $(r\cos \theta -2,r\sin \theta -7)$
If they lie on $4x+3y=3$. Then $4(r\cos \theta -2)+3(r\sin \theta -7)=3$ ...(2)
The line (1) will meet $4x+3y=12$ at a distance $AC=(r+3)$
Hence $4[(r+3)\cos \theta -2]+3[(r+3)\sin \theta -7]=12$ ...(3)
Subtrating (2) and (3), we get
$4\cos \theta +3\sin \theta =3$ or $4\cos \theta =3(1-\sin \theta )$
On squaring $16(1-{\sin }^2\theta )=9{(1-\sin \theta )}^2\Rightarrow \sin \theta =1,-\frac{7}{25}$
Hence $\cos \theta =0,\frac{24}{25}$
Putting these values in (1), we get
$x+2=0$ and $7x+24y+182=0$ as the required equation.