
Consider the set of all lines $px+qy+r=0$ such that $3p+2q+4r=0$. Which one of the following statements is true?
(a) The lines are all parallel.
(b) Each line passes through the origin.
(c) The lines are not concurrent.
(d) The lines are concurrent at the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
Answer
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Hint: We start solving the problem by converting the given condition $3p+2q+4r=0$ into the form that coincides with $px+qy+r=0$. After converting, we can see that the given set of lines $px+qy+r=0$ passes through a fixed point which makes us recall the definition of concurrent lines. Using this definition, we check the options to get the correct one.
Complete step-by-step answer:
Given that we have a set of lines $px+qy+r=0$ which satisfies $3p+2q+4r=0$. We need to find the properties of the set of these lines.
We can see that $px+qy+r=0$ represents a set of lines. So, we need to find the properties of lines so that for any values of p, q and r the equation $px+qy+r=0$ represents a set of lines.
So, we have given $3p+2q+4r=0$. We divide the total equation with 4.
$\Rightarrow \dfrac{3p+2q+4r}{4}=\dfrac{0}{4}$.
\[\Rightarrow \dfrac{3p}{4}+\dfrac{2q}{4}+\dfrac{4r}{4}=0\].
\[\Rightarrow p\left( \dfrac{3}{4} \right)+q\left( \dfrac{2}{4} \right)+r=0\].
\[\Rightarrow p\left( \dfrac{3}{4} \right)+q\left( \dfrac{1}{2} \right)+r=0\] ---(1).
If we compare equation (1) with the set of lines $px+qy+r=0$, we get the values of x and y as $\dfrac{3}{4}$ and $\dfrac{1}{2}$.
We get to see that the given set of lines $px+qy+r=0$ passes through the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
We know that if the set of lines in a plane passes through or intersects at a fixed point, then the given set of lines are said to be concurrent. Using this concept, we can say that the given set of lines $px+qy+r=0$ are said to be concurrent at the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
∴ The given set of lines $px+qy+r=0$ are concurrent at the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
So, the correct answer is “Option d”.
Note: We should not think that the set of lines depends on variables x and y. The set of lines are actually depending on the variables p, q and r. If the given set of lines are said to be parallel, then the slopes of all the sets of lines should be equal. We should remember that the equation of the lines passing through origin is of the form $y=mx$.
Complete step-by-step answer:
Given that we have a set of lines $px+qy+r=0$ which satisfies $3p+2q+4r=0$. We need to find the properties of the set of these lines.
We can see that $px+qy+r=0$ represents a set of lines. So, we need to find the properties of lines so that for any values of p, q and r the equation $px+qy+r=0$ represents a set of lines.
So, we have given $3p+2q+4r=0$. We divide the total equation with 4.
$\Rightarrow \dfrac{3p+2q+4r}{4}=\dfrac{0}{4}$.
\[\Rightarrow \dfrac{3p}{4}+\dfrac{2q}{4}+\dfrac{4r}{4}=0\].
\[\Rightarrow p\left( \dfrac{3}{4} \right)+q\left( \dfrac{2}{4} \right)+r=0\].
\[\Rightarrow p\left( \dfrac{3}{4} \right)+q\left( \dfrac{1}{2} \right)+r=0\] ---(1).
If we compare equation (1) with the set of lines $px+qy+r=0$, we get the values of x and y as $\dfrac{3}{4}$ and $\dfrac{1}{2}$.
We get to see that the given set of lines $px+qy+r=0$ passes through the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
We know that if the set of lines in a plane passes through or intersects at a fixed point, then the given set of lines are said to be concurrent. Using this concept, we can say that the given set of lines $px+qy+r=0$ are said to be concurrent at the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
∴ The given set of lines $px+qy+r=0$ are concurrent at the point $\left( \dfrac{3}{4},\dfrac{1}{2} \right)$.
So, the correct answer is “Option d”.
Note: We should not think that the set of lines depends on variables x and y. The set of lines are actually depending on the variables p, q and r. If the given set of lines are said to be parallel, then the slopes of all the sets of lines should be equal. We should remember that the equation of the lines passing through origin is of the form $y=mx$.
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