Question:
The two pairs of straight lines
\[
12x^2+7xy-12y^2=0
\]
and
\[
12x^2+7xy-12y^2-x+7y-1=0
\]
constitute a
The two pairs of straight lines
\[
12x^2+7xy-12y^2=0
\]
and
\[
12x^2+7xy-12y^2-x+7y-1=0
\]
constitute a
Show Hint
To identify the figure formed by two pairs of parallel lines, factorize both equations, check perpendicularity using slopes, and calculate the distance between each pair of parallel lines.
Updated On: Jun 26, 2026
- square of area \(\dfrac{1}{25}\) sq. units
- square of area \(\dfrac{1}{5}\) sq. units
- rectangle of area \(\dfrac{1}{10}\) sq. units
- rectangle of area \(\dfrac{1}{15}\) sq. units
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Factorize the first pair of straight lines.
Given, \[ 12x^2+7xy-12y^2=0 \] We factorize it as \[ (3x+4y)(4x-3y)=0 \] Therefore, the two lines are \[ 3x+4y=0 \] and \[ 4x-3y=0 \]
Step 2: Factorize the second pair of straight lines.
Now consider \[ 12x^2+7xy-12y^2-x+7y-1=0 \] Using the same linear factors, write it as \[ (3x+4y+a)(4x-3y+b)=0 \] Expanding, \[ (3x+4y+a)(4x-3y+b) \] \[ =12x^2+7xy-12y^2+(3b+4a)x+(4b-3a)y+ab \] Comparing with \[ 12x^2+7xy-12y^2-x+7y-1, \] we get \[ 3b+4a=-1 \] \[ 4b-3a=7 \] \[ ab=-1 \] Solving gives \[ a=-1,\quad b=1 \] Hence, the second pair is \[ (3x+4y-1)(4x-3y+1)=0 \] So the two lines are \[ 3x+4y-1=0 \] and \[ 4x-3y+1=0 \]
Step 3: Observe the four lines.
The four lines are \[ 3x+4y=0,\quad 3x+4y-1=0 \] and \[ 4x-3y=0,\quad 4x-3y+1=0 \] The first two lines are parallel.
The last two lines are also parallel.
Step 4: Check the angle between the two directions.
The slopes of \[ 3x+4y=0 \] and \[ 4x-3y=0 \] are respectively \[ -\frac{3}{4} \] and \[ \frac{4}{3} \] Their product is \[ -\frac{3}{4}\cdot \frac{4}{3}=-1 \] Thus, the two directions are perpendicular.
So, the figure formed is a rectangle with perpendicular adjacent sides, that is, a square.
Step 5: Find the distance between the parallel lines.
Distance between \[ 3x+4y=0 \] and \[ 3x+4y-1=0 \] is \[ \frac{|0-(-1)|}{\sqrt{3^2+4^2}} \] \[ =\frac{1}{5} \] Similarly, distance between \[ 4x-3y=0 \] and \[ 4x-3y+1=0 \] is \[ \frac{|0-1|}{\sqrt{4^2+(-3)^2}} \] \[ =\frac{1}{5} \]
Step 6: Find the area.
Since both perpendicular distances are equal, the figure is a square of side \[ \frac{1}{5} \] Therefore, area is \[ \left(\frac{1}{5}\right)^2 \] \[ =\frac{1}{25} \]
Step 7: Final conclusion.
Hence, the two pairs of straight lines constitute a \[ \boxed{\text{square of area } \frac{1}{25}\text{ sq. units}} \]
Given, \[ 12x^2+7xy-12y^2=0 \] We factorize it as \[ (3x+4y)(4x-3y)=0 \] Therefore, the two lines are \[ 3x+4y=0 \] and \[ 4x-3y=0 \]
Step 2: Factorize the second pair of straight lines.
Now consider \[ 12x^2+7xy-12y^2-x+7y-1=0 \] Using the same linear factors, write it as \[ (3x+4y+a)(4x-3y+b)=0 \] Expanding, \[ (3x+4y+a)(4x-3y+b) \] \[ =12x^2+7xy-12y^2+(3b+4a)x+(4b-3a)y+ab \] Comparing with \[ 12x^2+7xy-12y^2-x+7y-1, \] we get \[ 3b+4a=-1 \] \[ 4b-3a=7 \] \[ ab=-1 \] Solving gives \[ a=-1,\quad b=1 \] Hence, the second pair is \[ (3x+4y-1)(4x-3y+1)=0 \] So the two lines are \[ 3x+4y-1=0 \] and \[ 4x-3y+1=0 \]
Step 3: Observe the four lines.
The four lines are \[ 3x+4y=0,\quad 3x+4y-1=0 \] and \[ 4x-3y=0,\quad 4x-3y+1=0 \] The first two lines are parallel.
The last two lines are also parallel.
Step 4: Check the angle between the two directions.
The slopes of \[ 3x+4y=0 \] and \[ 4x-3y=0 \] are respectively \[ -\frac{3}{4} \] and \[ \frac{4}{3} \] Their product is \[ -\frac{3}{4}\cdot \frac{4}{3}=-1 \] Thus, the two directions are perpendicular.
So, the figure formed is a rectangle with perpendicular adjacent sides, that is, a square.
Step 5: Find the distance between the parallel lines.
Distance between \[ 3x+4y=0 \] and \[ 3x+4y-1=0 \] is \[ \frac{|0-(-1)|}{\sqrt{3^2+4^2}} \] \[ =\frac{1}{5} \] Similarly, distance between \[ 4x-3y=0 \] and \[ 4x-3y+1=0 \] is \[ \frac{|0-1|}{\sqrt{4^2+(-3)^2}} \] \[ =\frac{1}{5} \]
Step 6: Find the area.
Since both perpendicular distances are equal, the figure is a square of side \[ \frac{1}{5} \] Therefore, area is \[ \left(\frac{1}{5}\right)^2 \] \[ =\frac{1}{25} \]
Step 7: Final conclusion.
Hence, the two pairs of straight lines constitute a \[ \boxed{\text{square of area } \frac{1}{25}\text{ sq. units}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Family of Lines Questions
- For \( \lambda, \mu \in \mathbb{R} \), the lines \[ (x - 2y - 1) + \lambda (3x + 2y - 11) = 0 \] and \[ (3x + 4y - 11) + \mu (-x + 2y - 3) = 0 \] represent two families of lines. If the equation of the line common to both families is given by \[ ax + by - 5 = 0, \] then \( 2a + b = \) ?
If the pair of lines represented by $$ 3x^2 - 5xy + P y^2 = 0 $$ and $$ 6x^2 - xy - 5y^2 = 0 $$ have one line in common, then the sum of all possible values of \( P \) is:
- If the straight lines $x + 2y - 9 = 0$, $3x + 5y - 5 = 0$ and $ax + by - 1 = 0$ are concurrent, then the straight line $22x - 35y = 1$ passes through the point:
- The equations of the pairs of opposite sides of a parallelogram are \[ x^2-5x+6=0 \] and \[ y^2-6y+5=0. \] Then the equations of its diagonals are
- The family of lines, forming an isosceles triangle with the lines \(3x-4y-2=0\) and \(12x-5y+6=0\), is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions