Find out the straight line which is represented by the equation $135{x^2} - 136xy + 33{y^2} = 0$ is equally inclined to the line
A.$x - 2y = 7$
B.$x + 2y = 7$
C.$x - 2y = 4$
D.$3x + 2y = 4$
Answer
616.2k+ views
Hint: The given question is related to the concept of straight lines. Straight line is defined as a curve where every point on the line segment joining any two points lies on it. The straight line’s general equation is given by $ax + by + c = 0$. The equation of bisector of angles between pair of lines is given by $\dfrac{{{x^2} - {y^2}}}{{a - b}} = \dfrac{{xy}}{h}$.
Complete step by step solution:
Given pair of line is $135{x^2} - 136xy + 33{y^2} = 0$-----(1)
We know that the equation of bisector of angles between pair of lines is given by $\dfrac{{{x^2} - {y^2}}}{{a - b}} = \dfrac{{xy}}{h}$.
Using the same formula and substituting $a = 135,b = 33$and$h = - 68$ which we got by $\left( { - \dfrac{{136}}{2}} \right)$, we get,
$
\Rightarrow \dfrac{{{x^2} - {y^2}}}{{135 - 33}} = \dfrac{{xy}}{{ - 68}} \\
\Rightarrow 2{x^2} + 3xy - 2{y^2} = 0 \\
\Rightarrow \left( {x + 2y} \right)\left( {2x - y} \right) = 0 \\
$
We got one of the bisectors as $x + 2y = 0$ which is also parallel to the line $x + 2y = 7$.
Therefore, the line $x + 2y = 7$ is equally inclined to $135{x^2} - 136xy + 33{y^2} = 0$.
Hence, the correct option is (B).
Note:
The given question was an easy one. The trick to solve these types of questions without any complexity is to use the formula. Students should be aware of the straight line and related concepts for their own ease. Some facts about straight lines are as follows:
1.If two straight lines are parallel, then their slopes are equal.
2.If two straight lines are perpendicular to each other, then the product of their slopes is $ - 1$.
3.If a straight line is at a distance $k$ and is parallel to $x$-axis, then the equation of the line is $y = \pm k$.
4.If a straight line is at a distance $c$ and is parallel to $y$-axis, then its equation of line is $x = \pm c$.
Complete step by step solution:
Given pair of line is $135{x^2} - 136xy + 33{y^2} = 0$-----(1)
We know that the equation of bisector of angles between pair of lines is given by $\dfrac{{{x^2} - {y^2}}}{{a - b}} = \dfrac{{xy}}{h}$.
Using the same formula and substituting $a = 135,b = 33$and$h = - 68$ which we got by $\left( { - \dfrac{{136}}{2}} \right)$, we get,
$
\Rightarrow \dfrac{{{x^2} - {y^2}}}{{135 - 33}} = \dfrac{{xy}}{{ - 68}} \\
\Rightarrow 2{x^2} + 3xy - 2{y^2} = 0 \\
\Rightarrow \left( {x + 2y} \right)\left( {2x - y} \right) = 0 \\
$
We got one of the bisectors as $x + 2y = 0$ which is also parallel to the line $x + 2y = 7$.
Therefore, the line $x + 2y = 7$ is equally inclined to $135{x^2} - 136xy + 33{y^2} = 0$.
Hence, the correct option is (B).
Note:
The given question was an easy one. The trick to solve these types of questions without any complexity is to use the formula. Students should be aware of the straight line and related concepts for their own ease. Some facts about straight lines are as follows:
1.If two straight lines are parallel, then their slopes are equal.
2.If two straight lines are perpendicular to each other, then the product of their slopes is $ - 1$.
3.If a straight line is at a distance $k$ and is parallel to $x$-axis, then the equation of the line is $y = \pm k$.
4.If a straight line is at a distance $c$ and is parallel to $y$-axis, then its equation of line is $x = \pm c$.
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