The area of the quadrilateral formed by the lines 4 x -3 y - a -0-3 x -4 y - a -0-4 x -3 y -3 a -0 and 3 x -4 y -2 a -0 isA- 2 a 2-11 sq- units11B- 2 a 2- g 2 sq- unitsC- 4-7 sq- unitsD- 2 a 2-7 sq- units

The correct option is D 2a^27 sq. unitsGiven lines nbsp; L_1:4x 3y a = 0 L_2:3x 4y+a = 0 L_3:4x 3y 3a =0 L_4:3x 4y+2a =0 We know the distance between two ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Foot of Perpendicular from a Point on a Line

The area of t...

Question

The area of the quadrilateral formed by the lines $4 x - 3 y - a = 0, 3 x - 4 y + a = 0$ , $4 x - 3 y - 3 a = 0$ and $3 x - 4 y + 2 a = 0$ is

\frac{2 a^{2}}{11} sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 a^{2}}{9} sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a^{2}}{7} sq. units

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 a^{2}}{7} sq. units

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{2 a^{2}}{7} sq. units$
Given lines
$L_{1} : 4 x - 3 y - a = 0 L_{2} : 3 x - 4 y + a = 0 L_{3} : 4 x - 3 y - 3 a = 0 L_{4} : 3 x - 4 y + 2 a = 0$
We know the distance between two parallel lines is
$p_{1} = ∣ ∣ ∣ ∣ \frac{- 3 a - (- a)}{\sqrt{4^{2} + 3^{2}}} ∣ ∣ ∣ ∣ = \frac{2 a}{5} units p_{2} = ∣ ∣ ∣ ∣ \frac{a - 2 a}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = \frac{a}{5} units$
Slope of the lines are
$m_{1} = \frac{3}{4}, m_{2} = \frac{4}{3}$
Now,
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣ \Rightarrow tan θ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{3}{4} - \frac{4}{3}}{1 + 1} ∣ ∣ ∣ ∣ ∣ ∣ = \frac{7}{24} \Rightarrow sin θ = \frac{7}{25}$

Now, the area of the parallelogram
$= \frac{p_{1} p_{2}}{sin θ} = \frac{25 \times 2 a^{2}}{7 \times 25} = \frac{2 a^{2}}{7} sq. units$

Suggest Corrections

Similar questions

Prove that the area of the parallelogram formed by the lines $3 x - 4 y + a = 0$ , $3 x - 4 y + 3 a = 0$ , $4 x - 3 y - a = 0$ and $4 x - 3 y - 2 a = 0$ is $\frac{2 a^{2}}{7}$ sq. units.