Question

If the intercept made on the line y=mx by the line y=2 and y=6 is less than 5 then the range of values of m is.
A.$\left( -\infty ,-\dfrac{4}{3} \right)\cup \left( \dfrac{4}{3},\infty \right)$
B.$\left( -\dfrac{4}{3},\dfrac{4}{3} \right)$
C.$\left( -\dfrac{3}{4},\dfrac{3}{4} \right)$
D.None of these.

Answer 1

Hint: Start by finding the point of intersection of the lines y=2 and y=mx and also of the lines y=6 and y=mx in terms of m. Then using the distance formula, given by $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ , find the distance between the point of intersections and form an inequality such that the distance is less than 5.

Complete step-by-step answer:
Let us start by drawing a representative diagram of the situation given in the figure.

Now we can see that the point A is the point of intersection of the point y=2 and y=mx. So, putting the value y=2 in y=mx, we get
2=mx
$\Rightarrow x=\dfrac{2}{m}$
So, point A on the Cartesian plane is $A\left( \dfrac{2}{m},2 \right)$ .
Now we can also see that the point B is the point of intersection of the point y=6 and y=mx. So, putting the value y=6 in y=mx, we get
6=mx
$\Rightarrow x=\dfrac{6}{m}$
So, point B on the Cartesian plane is $B\left( \dfrac{6}{m},6 \right)$ .
Now, according to the distance formula, the distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
$\therefore AB=\sqrt{{{\left( \dfrac{6}{m}-\dfrac{2}{m} \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}=\sqrt{{{4}^{2}}\left( \dfrac{1}{{{m}^{2}}}+1 \right)}=4\sqrt{\dfrac{1}{{{m}^{2}}}+1}$
Now according to the question, the intercept AB must be less than 5.

$\therefore AB<5$
$\Rightarrow 4\sqrt{\dfrac{1}{{{m}^{2}}}+1}<5$
$\Rightarrow \sqrt{\dfrac{1}{{{m}^{2}}}+1}<\dfrac{5}{4}$
Now as we can see that both sides of the inequality are greater than 1, so we can square both sides of the inequality. On doing so, we get
$\dfrac{1}{{{m}^{2}}}+1<\dfrac{25}{16}$
$\Rightarrow \dfrac{1}{{{m}^{2}}}<\dfrac{9}{16}$
Again as the denominators of both the sides are positive, we can cross multiply without changing the signs.
$16<9{{m}^{2}}$
$\begin{aligned} & \Rightarrow 9{{m}^{2}}-16>0 \\\ & \Rightarrow \left( 3m-4 \right)\left( 3m+4 \right)>0 \\\ \end{aligned}$
So, looking at the result we can say that the inequality holds true when both 3m-4 and 3m+4 are either positive or both are negative.
So, for first case when both are positive m must be greater than $\dfrac{4}{3}$ , i.e. , $m\in \left( \dfrac{4}{3},\infty \right)$
For the next case, when both are negative, m must be less than $-\dfrac{4}{3}$ , i.e. , $m\in \left( -\infty ,-\dfrac{4}{3} \right)$
So, the overall answer is the union of the above two cases, which is equal to $m\in \left( -\infty ,-\dfrac{4}{3} \right)\cup \left( \dfrac{4}{3},\infty \right)$ . Therefore, the answer to the above question is option (a).

Note: Whenever dealing with an inequality be very careful while you multiply, square or perform other operations, as there are cases where the sign of inequality changes. For example: x>y implies $-y>-x$ , i.e. , when both sides of an inequality are multiplied by a negative number, the sign of inequality changes.