Question

If $$(0,\beta )$$lies on or inside the triangle with the sides $$y+3x+2=0,3y-2x-5=0$$ and $$4y+x-14=0$$, then
(a) $$0\le \beta \le {\displaystyle \frac{7}{2}}$$
(b) $$0\le \beta \le {\displaystyle \frac{5}{2}}$$
(c) $${\displaystyle \frac{5}{3}}\le \beta \le {\displaystyle \frac{7}{2}}$$
(d) None of these

Answer 1

Hint: Plot the given 3 line equations to form a triangle and find the point of intersection.

The figure for the given problem is as follows:

Now the given point $$(0,\beta )$$lies on the y-axis as its x-coordinate is zero.
From the above figure we see that the y-axis passes through the sides AC and BC.
Now we will substitute $$(0,\beta )$$ in the equation of side AC, i.e.,
$$4y+x-14=0$$
We get,
$$4(\beta )+0-14=0$$
$$4\beta =14$$
$$\beta ={\displaystyle \frac{14}{4}}$$
$$\beta ={\displaystyle \frac{7}{2}}$$
So the point of intersection of the y-axis and side AC is $$(0,{\displaystyle \frac{7}{2}})$$.
Similarly, we will substitute $$(0,\beta )$$ in the equation of side BC, i.e.,
$$3y-2x-5=0$$
We get,
$$3\beta -2(0)-5=0$$
$$3\beta =5$$
$$\beta ={\displaystyle \frac{5}{3}}$$
So, the point of intersection of the y-axis and side BC is $$(0,{\displaystyle \frac{5}{3}})$$.
Now as the given point lies on y-axis as well as on or inside of the triangle, so all the points between $$(0,{\displaystyle \frac{5}{3}})$$and $$(0,{\displaystyle \frac{7}{2}})$$, will satisfy the condition.
So, the value of $$\beta$$ will be,
$${\displaystyle \frac{5}{3}}\le \beta \le {\displaystyle \frac{7}{2}}$$
Hence, the correct answer is option (c).
Note: Here we can solve for the vertices of the triangle from the given equations of the sides. Then find the value of $$\beta$$. But it will be a lengthy process.