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# The value of $m$ for which the lines $3x = y - 8$ and $6x + my + 16 = 0$ coincide isA.2B.$- 2$C.$\dfrac{1}{2}$D.$- \dfrac{1}{2}$

Last updated date: 22nd Jun 2024
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Hint: First we will find the slope of the tangent is the differentiation of the given curves with respect to $x$ and then we will use that for the given equations to coincide, the slope should be equal.

We are given that the lines $3x = y - 8$ and $6x + my + 16 = 0$ coincide.
Rewriting the given equations, we get
$y = 3x + 8{\text{ ......eq.(1)}}$
$y = - \dfrac{6}{m}x - \dfrac{{16}}{m}{\text{ ......eq(2)}}$
We know that the slope of the tangent is the differentiation of the given curve with respect to $x$.
Differentiating the equation (1) with respect to $x$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3x + 8} \right) \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}3x - \dfrac{d}{{dx}}8 \\ \Rightarrow \dfrac{{dy}}{{dx}} = 3{\text{ ......eq.(3)}} \\$
Differentiating the equation (2) with respect to $x$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{6}{m}x - \dfrac{{16}}{m}} \right) \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{6}{m}x} \right) - \dfrac{d}{{dx}}\left( { - \dfrac{{16}}{m}} \right) \\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{6}{m}{\text{ ......eq.(4)}} \\$
For the given equations to coincide, the slope should be equal.
So, taking equation (3) and equation (4) equal, we get
$\Rightarrow 3 = - \dfrac{6}{m}$
Cross-multiplying the above equation, we get
$\Rightarrow 3m = - 6$
Dividing the above equation by 3 on both sides, we get
$\Rightarrow m = - 2$
Hence, option B is correct.

Note: The slope equals the rise divided by the run. You can determine the slope of a line from its graph by looking at the rise and run. One characteristic of a line is that its slope is constant all the way along it. So, you can choose any 2 points along the graph of the line to figure out the slope. Avoid calculation mistakes.