
The equation of lines passing through origin and parallel to the lines $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$ is given as-
A. ${{m}_{1}}{{m}_{2}}{{x}^{2}}-\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{y}^{2}}=0$
B. ${{m}_{1}}{{m}_{2}}{{x}^{2}}+\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{y}^{2}}=0$
C. ${{m}_{1}}{{m}_{2}}{{y}^{2}}-\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{x}^{2}}=0$
D. ${{m}_{1}}{{m}_{2}}{{y}^{2}}+\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{x}^{2}}=0$
Answer
161.7k+ views
Hint: If two lines are passing through origin then they have slope multiplied by $x$ axis and the value of constant is zero.
Formula Used:$y=mx+c$
Complete step by step solution:The given equation of the lines are $y={{m}_{1}}x+{{c}_{1}}$ and$y={{m}_{2}}x+{{c}_{2}}$ .
Here ${{m}_{1}},{{m}_{2}}$ are the slopes of the lines respectively.
As the line are passing through origin so, the value of the constant term is zero. Thus the lines passing through the origin can be written as follows-
$y={{m}_{1}}x$
$y={{m}_{2}}x$
Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now the equation of the line parallel to the above given $y={{m}_{1}}x+{{c}_{1}}$ and$y={{m}_{2}}x+{{c}_{2}}$lines can be given as follows-
$\left( y-{{m}_{1}}x \right)\left( y-{{m}_{2}}x \right)=0$
Multiplying the two equations of the lines passing through the origin we will get the equation of the line which is parallel to the lines as follows-
$\left( y-{{m}_{1}}x \right)\left( y-{{m}_{2}}x \right)=0$
${{y}^{2}}-{{m}_{1}}xy-{{m}_{2}}xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$
${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$
Thus the equation of the line is ${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$.
Thus we can write that the equation of lines passing through origin and parallel to the lines $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$ is given as ${{m}_{1}}{{m}_{2}}{{x}^{2}}-\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{y}^{2}}=0$.
Option ‘A’ is correct
Note: The condition for two parallel line is that the multiplication of the slopes of the two lines is $-1$ . If two lines of equation $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$are parallel to each other then we can say that ${{m}_{1}}{{m}_{2}}=-1$ .
Formula Used:$y=mx+c$
Complete step by step solution:The given equation of the lines are $y={{m}_{1}}x+{{c}_{1}}$ and$y={{m}_{2}}x+{{c}_{2}}$ .
Here ${{m}_{1}},{{m}_{2}}$ are the slopes of the lines respectively.
As the line are passing through origin so, the value of the constant term is zero. Thus the lines passing through the origin can be written as follows-
$y={{m}_{1}}x$
$y={{m}_{2}}x$
Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now the equation of the line parallel to the above given $y={{m}_{1}}x+{{c}_{1}}$ and$y={{m}_{2}}x+{{c}_{2}}$lines can be given as follows-
$\left( y-{{m}_{1}}x \right)\left( y-{{m}_{2}}x \right)=0$
Multiplying the two equations of the lines passing through the origin we will get the equation of the line which is parallel to the lines as follows-
$\left( y-{{m}_{1}}x \right)\left( y-{{m}_{2}}x \right)=0$
${{y}^{2}}-{{m}_{1}}xy-{{m}_{2}}xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$
${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$
Thus the equation of the line is ${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}=0$.
Thus we can write that the equation of lines passing through origin and parallel to the lines $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$ is given as ${{m}_{1}}{{m}_{2}}{{x}^{2}}-\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{y}^{2}}=0$.
Option ‘A’ is correct
Note: The condition for two parallel line is that the multiplication of the slopes of the two lines is $-1$ . If two lines of equation $y={{m}_{1}}x+{{c}_{1}}$ and $y={{m}_{2}}x+{{c}_{2}}$are parallel to each other then we can say that ${{m}_{1}}{{m}_{2}}=-1$ .
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