Question

The equation of the line which is parallel to the lines common to the pair of lines given by \[6{x^2} - xy - 12{y^2} = 0\] and \[15{x^2} + 14xy - 8{y^2} = 0\] and 5 units away from the origin and having positive \[x\] and \[y\] intercepts is:
A) \[4x + 3y = 12\]
B) \[3x + 4y = 25\]
C) \[12x + 5y = 65\]
D) \[15x + 8y = 85\]

Answer 1

Hint:
We will find the line that is common to \[6{x^2} - xy - 12{y^2} = 0\] and \[15{x^2} + 14xy - 8{y^2} = 0\]. We will find the line’s slope. The slope of the required line will also be the same. We will assume that the line has constant \[k\]. We will find the line’s distance from the origin and equate it with 5. We will find the value of \[k\] using this equation. We will check if the line has positive intercepts.

Formulas used:
1. Quadratic equation \[a{x^2} + b + c = 0\] has the roots \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
2. Distance of a line \[ax + by + c = 0\] from a point \[\left( {x,y} \right)\] is given by \[\dfrac{{\left| {ax + by + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\].
3. Slope of a line \[ax + by + c = 0\] is given by \[m = \dfrac{{ - a}}{b}\].
4. The intercept form of a line is given by \[\dfrac{x}{a} + \dfrac{y}{b} = 1\] where \[a\] and \[b\] are the \[x\] and \[y\] intercepts respectively.

Complete step by step solution:
We will find the lines represented by the pair of lines \[6{x^2} - xy - 12{y^2} = 0\]. We will substitute 6 for \[a\], \[ - y\] for \[b\] and \[ - 12{y^2}\] for \[c\] in the first formula:
\[x = \dfrac{{ - \left( { - y} \right) \pm \sqrt {{{\left( { - y} \right)}^2} - 4 \cdot 6\left( { - 12{y^2}} \right)} }}{{2 \cdot 6}}\]
We will simplify the above equation:
\[\begin{array}{l}x = \dfrac{{y \pm \sqrt {{y^2} + 288{y^2}} }}{{12}}\\ \Rightarrow x = \dfrac{{y \pm 17y}}{{12}}\\ \Rightarrow x = \dfrac{{18y}}{{12}},\dfrac{{ - 16y}}{{12}}\\ \Rightarrow x = \dfrac{{3y}}{2}, - \dfrac{{4y}}{3}\end{array}\]
\[\begin{array}{l}x = \dfrac{{3y}}{2}\\ \Rightarrow 2x = 3y\\ \Rightarrow 2x - 3y = 0\end{array}\]
\[\begin{array}{l}x = - \dfrac{{4y}}{3}\\ \Rightarrow 3x = - 4y\\ \Rightarrow 3x + 4y = 0\end{array}\]
\[\therefore \] The pair of lines \[6{x^2} - xy - 12{y^2} = 0\] represents the line \[2x - 3y = 0\] and the line \[3x + 4y = 0\].
We will find the lines represented by the pair of lines \[15{x^2} + 14xy - 8{y^2} = 0\]. We will substitute 15 for \[a\], \[14y\] for \[b\] and \[ - 8{y^2}\] for \[c\] in the first formula:
\[x = \dfrac{{ - \left( {14y} \right) \pm \sqrt {{{\left( { - 14y} \right)}^2} - 4 \cdot 15\left( { - 8{y^2}} \right)} }}{{2 \cdot 15}}\]
We will simplify the above equation:
\[\begin{array}{l}x = \dfrac{{ - 14y \pm \sqrt {196{y^2} + 480{y^2}} }}{{30}}\\ \Rightarrow x = \dfrac{{ - 14y \pm 26y}}{{30}}\\ \Rightarrow x = \dfrac{{ - 14y + 26y}}{{30}},\dfrac{{ - 14y - 26y}}{{30}}\\ \Rightarrow x = \dfrac{{12y}}{{30}},\dfrac{{ - 40y}}{{30}}\end{array}\]
\[\begin{array}{l}x = \dfrac{{12y}}{{30}}\\ \Rightarrow x = \dfrac{{2y}}{5}\\ \Rightarrow 5x = 2y\\ \Rightarrow 5x - 2y = 0\end{array}\]
\[\begin{array}{l}x = - \dfrac{{40y}}{{30}}\\ \Rightarrow x = - \dfrac{{4y}}{3}\\ \Rightarrow 3x = - 4y\\ \Rightarrow 3x + 4y = 0\end{array}\]
\[\therefore \] The pair of lines \[15{x^2} + 14xy - 8{y^2} = 0\] represents the line \[5x - 2y = 0\] and the line \[3x + 4y = 0\].
We can see that the line common to both the pairs is \[3x + 4y = 0\].
We will find the slope of the line. We will substitute 3 for \[a\] and 4 for \[b\] in the third formula:
\[m = - \dfrac{3}{4}\]
We know that the slope of the required line will also be \[ - \dfrac{3}{4}\].
We will assume that the line is \[ax + by + c = 0\]. As the slope of the line is \[ - \dfrac{3}{4}\], we can assume that the equation of the line will be \[3x + 4y + k = 0\].
We will find the distance of origin from this line and equate it with 5. We will substitute 0 for \[x\], \[y\]; \[ - 3\] for \[a\] and 4 for \[b\] in the 2nd formula:
\[\begin{array}{l}{\rm{ }}5 = \dfrac{{\left| {3\left( 0 \right) + 4\left( 0 \right) + k} \right|}}{{\sqrt {{{\left( { - 3} \right)}^2} + {4^2}} }}\\ \Rightarrow {\rm{ }}5 = \dfrac{{\left| k \right|}}{{\sqrt {25} }}\\ \Rightarrow {\rm{ }}5 \times 5 = \left| k \right|\\ \Rightarrow {\rm{ }}\left| k \right| = 25\\
 \Rightarrow k = 25, - 25\end{array}\]
The 2 possibilities for the equation of line are:
\[\begin{array}{l}{\rm{ }}3x + 4y + 25 = 0\\ \Rightarrow \Rightarrow {\rm{ }} - 3x - 4y = 25\end{array}\]
and
\[\begin{array}{l}3x + 4y - 25 = 0\\ \Rightarrow \Rightarrow {\rm{ }}3x + 4y = 25\end{array}\]
We will divide the equations of both the lines by 25 to convert them into the intercept form.
\[\begin{array}{l}{\rm{ }} - \dfrac{{3x}}{{25}} - \dfrac{{4x}}{{25}} = \dfrac{{25}}{{25}}\\
 \Rightarrow \dfrac{x}{{ - 3/25}} + \dfrac{y}{{ - 4/25}} = 1\end{array}\]
\[\begin{array}{l}{\rm{ }}\dfrac{{3x}}{{25}} + \dfrac{{4y}}{{25}} = \dfrac{{25}}{{25}}\\
 \Rightarrow \dfrac{x}{{3/25}} + \dfrac{y}{{4/25}} = 1\end{array}\]
We can see by comparing with the fourth formula that the 1st equation has negative intercepts while the 2nd equation has positive intercepts.
\[\therefore 3x + 4y = 25\] is the equation of the required line.

Option B is the correct option.

Note:
We can also find the solution by checking which equation from the option satisfies the conditions given in the question.
The equation given in option A does not have a distance of 5 units from the origin. So, option A is eliminated.
Out of options B, C and D; the equations in option C and D do not satisfy the pair of lines given in the question. So, they are also eliminated.
We can conclude that option B is the correct option.