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Question: Find out the straight line which is represented by the equation \(135{x^2} - 136xy + 33{y^2} = 0\) i...

Find out the straight line which is represented by the equation 135x2136xy+33y2=0135{x^2} - 136xy + 33{y^2} = 0 is equally inclined to the line
A.x2y=7x - 2y = 7
B.x+2y=7x + 2y = 7
C.x2y=4x - 2y = 4
D.3x+2y=43x + 2y = 4

Explanation

Solution

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=0ax + by + c = 0. The equation of bisector of angles between pair of lines is given by x2y2ab=xyh\dfrac{{{x^2} - {y^2}}}{{a - b}} = \dfrac{{xy}}{h}.

Complete step by step solution:
Given pair of line is 135x2136xy+33y2=0135{x^2} - 136xy + 33{y^2} = 0-----(1)
We know that the equation of bisector of angles between pair of lines is given by x2y2ab=xyh\dfrac{{{x^2} - {y^2}}}{{a - b}} = \dfrac{{xy}}{h}.
Using the same formula and substituting a=135,b=33a = 135,b = 33andh=68h = - 68 which we got by (1362)\left( { - \dfrac{{136}}{2}} \right), we get,
x2y213533=xy68 2x2+3xy2y2=0 (x+2y)(2xy)=0  \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=0x + 2y = 0 which is also parallel to the line x+2y=7x + 2y = 7.
Therefore, the line x+2y=7x + 2y = 7 is equally inclined to 135x2136xy+33y2=0135{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 - 1.
3.If a straight line is at a distance kk and is parallel to xx-axis, then the equation of the line is y=±ky = \pm k.
4.If a straight line is at a distance cc and is parallel to yy-axis, then its equation of line is x=±cx = \pm c.