Question

Find k, if one of the lines given by $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$ is perpendicular to the line $2x-y=5$.

Answer 1

Hint : First find the slope (m) of the line $2x-y=5$. Next, the pair of line $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$is passing from the origin, so the slope is ${{m}_{1}}=\dfrac{y}{x}$. Also ${{m}_{1}}=\dfrac{-1}{m}$and hence replacing the value of ${{m}_{1}}$in the equation$k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$, we will get the required value of k.

Complete step by step solution :
In the question, we have to find the value of k, when it is given that the line $2x-y=5$is perpendicular to one of the lines in the pair of lines equation $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$.
So at first we need to check if the pair of lines is passing from the origin (0,0) or not.
So at x=0 and y=0, we have the equation $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$ as
$\Rightarrow k\,{{(0)}^{2}}+10(0)(0)+8{{(0)}^{2}}=0$
So this shows that the pair of lines $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$passes from the origin.
So now the slope of this pair of line will be written as:
${{m}_{1}}=\dfrac{y}{x}$. Next, we divide the $k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$ by ${{x}^{2}}$both the sides, to get:

& \Rightarrow \dfrac{k\,{{x}^{2}}+10xy+8{{y}^{2}}}{{{x}^{2}}}=0 \\\ & \Rightarrow k+\dfrac{10y}{x}+\dfrac{8{{y}^{2}}}{{{x}^{2}}}=0 \\\ \end{aligned}$$ Next, replacing $${{m}_{1}}=\dfrac{y}{x}$$, we will get: $$\begin{aligned} & \Rightarrow k+\dfrac{10y}{x}+\dfrac{8{{y}^{2}}}{{{x}^{2}}}=0 \\\ & \Rightarrow k+10{{m}_{1}}+8{{({{m}_{1}})}^{2}}=0\,\,\,\,\,\,\,\,\,\,\,\,\,(eq-1) \\\ \end{aligned}$$ Next, we need to find this value of slope $${{m}_{1}}$$. That will be found using the fact that the line $$2x-y=5$$ is perpendicular to the line $$k\,{{x}^{2}}+10xy+8{{y}^{2}}=0$$ So the slope (m) of the line $$2x-y=5$$ is found by first converting it in the slope intercept form $$y=mx+c$$. Here m is the slope. So the $$2x-y=5$$ is rewritten as: $$\begin{aligned} & \Rightarrow 2x-y=5 \\\ & \Rightarrow y=2x-5 \\\ \end{aligned}$$ So here the slope of this line is $$m=2$$. Next, the line that is perpendicular to this line will have the slope $${{m}_{1}}=\dfrac{-1}{m}$$ S0 we have: $${{m}_{1}}=\dfrac{-1}{2}\,\,\,\,\,\,\,\because m=2$$ Next, we just need to put this value in the equation 1 to get the required value of k. So we have: $$\begin{aligned} & \Rightarrow k+10{{m}_{1}}+8{{({{m}_{1}})}^{2}}=0\, \\\ & \Rightarrow k+10\left( \dfrac{-1}{2}\, \right)+8{{\left( \dfrac{-1}{2}\, \right)}^{2}}=0\, \\\ & \Rightarrow k-5+2=0\, \\\ & \Rightarrow k=3 \\\ \end{aligned}$$ So this is the required value of k. **Note** : We should avoid making a calculation error while writing the equation of the line. When we have the pair of lines given to us, then it can pass through the origin, so we have to check that in order to write the slope as $${{m}_{1}}=\dfrac{y}{x}$$. So, if the line is not passing from the origin then we have to careful as in that case the slope $${{m}_{1}}\ne \dfrac{y}{x}$$