Question

Find the equation of the lines passes through the origin and having slopes \[3\] and \[ - \dfrac{1}{3}\] .
A. \[{{3}}{{{y}}^{{2}}} + {{8xy}}-{{3}}{{{x}}^{{2}}} = {{0}}\]
B. \[{{3}}{x^{{2}}} + {{8xy}}-{{3}}{y^{{2}}} = {{0}}\]
C. \[{{3}}{{{y}}^{{2}}} - {{8xy}} + {{3}}{{{x}}^{{2}}} = {{0}}\]
D. \[{{3}}{x^{{2}}} + {{8xy}} + {{3}}{y^{{2}}} = {{0}}\]

Answer 1

Hint: Firstly, we need to find the equations of straight lines after comparing them with the general equation of the straight line in slope form. After that multiply the linear equations to get the required solution.

Formula used
General equation of straight line in slope form, \[y = mx + c\] , where \[m,c\] are real numbers.

Complete step by step solution
First, we find the equation which passes through the origin and has a slope \[3\].
Equation of the line is \[y = 3x\]
\[ \Rightarrow y - 3x = 0\] …………………(1)
Now, we find the equation which passes through the origin and has a slope \[ - \dfrac{1}{3}\].
Equation of the line is \[y = - \dfrac{1}{3}x\]
\[ \Rightarrow - 3y = x\]
\[ \Rightarrow 3y + x = 0\] ………………..(2)
Now, multiplying both the equations (1) and (2), we get
\[\left( {y - 3x} \right)\left( {3y + x} \right) = 0 \times 0\]
\[ \Rightarrow 3{y^2} - 9xy + xy - 3{x^2} = 0\]
\[ \Rightarrow 3{y^2} - 8xy - 3{x^2} = 0\] ………………………(3)
Taking \[( - 1)\] from both sides of the equation (3) and we get
\[ \Rightarrow - \left( {3{x^2} + 8xy - 3{y^2}} \right) = 0 \times \left( { - 1} \right)\]
\[ \Rightarrow 3{x^2} + 8xy - 3{y^2} = 0\]
Therefore, the required solution is \[3{x^2} + 8xy - 3{y^2} = 0\].

Hence option B is correct.

Note: Students got confused in taking common value or sign from the equation. When we take common in this time it should be maintained that we take the value from each term of the equation.