# 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) \[0\le \beta \le \dfrac{7}{2}\]

(b) \[0\le \beta \le \dfrac{5}{2}\]

(c) \[\dfrac{5}{3}\le \beta \le \dfrac{7}{2}\]

(d) None of these

Answer

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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,

\[4(\beta )+0-14=0\]

\[4\beta =14\]

\[\beta =\dfrac{14}{4}\]

\[\beta =\dfrac{7}{2}\]

So the point of intersection of the y-axis and side AC is \[\left( 0,\dfrac{7}{2} \right)\].

Similarly, we will substitute \[(0,\beta )\] in the equation of side BC, i.e.,

\[3y-2x-5=0\]

We get,

\[3\beta -2(0)-5=0\]

\[3\beta =5\]

\[\beta =\dfrac{5}{3}\]

So, the point of intersection of the y-axis and side BC is \[\left( 0,\dfrac{5}{3} \right)\].

Now as the given point lies on y-axis as well as on or inside of the triangle, so all the points between \[\left( 0,\dfrac{5}{3} \right)\]and \[\left( 0,\dfrac{7}{2} \right)\], will satisfy the condition.

So, the value of \[\beta \] will be,

\[\dfrac{5}{3}\le \beta \le \dfrac{7}{2}\]

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.

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,

\[4(\beta )+0-14=0\]

\[4\beta =14\]

\[\beta =\dfrac{14}{4}\]

\[\beta =\dfrac{7}{2}\]

So the point of intersection of the y-axis and side AC is \[\left( 0,\dfrac{7}{2} \right)\].

Similarly, we will substitute \[(0,\beta )\] in the equation of side BC, i.e.,

\[3y-2x-5=0\]

We get,

\[3\beta -2(0)-5=0\]

\[3\beta =5\]

\[\beta =\dfrac{5}{3}\]

So, the point of intersection of the y-axis and side BC is \[\left( 0,\dfrac{5}{3} \right)\].

Now as the given point lies on y-axis as well as on or inside of the triangle, so all the points between \[\left( 0,\dfrac{5}{3} \right)\]and \[\left( 0,\dfrac{7}{2} \right)\], will satisfy the condition.

So, the value of \[\beta \] will be,

\[\dfrac{5}{3}\le \beta \le \dfrac{7}{2}\]

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.

Last updated date: 17th Sep 2023

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