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
Verified
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.
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