Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

The lines represented by the equation $9{x^2} + 24xy + 16{y^2} + 21x + 28y + 6 = 0$ are
A. Parallel
B. Coincident
C. Perpendicular
D. None of these

Answer
VerifiedVerified
164.4k+ views
Hint: First we will compare given equation $9{x^2} + 24xy + 16{y^2} + 21x + 28y + 6 = 0$ with the general equation $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ and find the values of $a$, $b$, $h$, $g$, and $f$. Then we will check if the lines are perpendicular or not. If the lines are perpendicular we are done if lines are not perpendicular. Then we will check if the lines are parallel.

Formula Used: We know if lines are parallel then,
 $\dfrac{a}{h} = \dfrac{h}{b} = \dfrac{g}{f}$
if lines are perpendicular then,
 $a = - b$

Complete step by step solution: Given Equation is $9{x^2} + 24xy + 16{y^2} + 21x + 28y + 6 = 0$
We know that the general equation is $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
On comparing general equation with given equation, we get
$a = 9$, $b = 16$, $g = \dfrac{{21}}{2}$, $f = 14$ and $h = 12$
Checking if lines are perpendicular
Given lines are not perpendicular that is $a \ne - b$
Now, will check for parallel
$\dfrac{a}{h} = \dfrac{h}{b} = \dfrac{g}{f}$
Cross multiplying, we get
${h^2} = ab$
Putting the value of $a$, $b$ and $h$
${\left( {12} \right)^2} = 9(16)$
After solving, we get
$144 = 144$
Hence, the lines are parallels.

Hence, option A is correct.

Note: Students should correctly find the values of $a$, $b$, $h$, $g$, and $f$ while comparing with the general equation. And should correctly check what kind of these given lines are parallel, perpendicular or coincidence.