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\[A(0,2),B(2,0)\]and \[C(4,4)\]

(a) \[\alpha \in \left( \dfrac{1}{3},\dfrac{2}{3} \right)\]

(b) \[\alpha \in \left( \dfrac{2}{3},1 \right)\]

(c) \[\alpha \in \left( \dfrac{2}{3},\dfrac{4}{3} \right)\]

(d) \[\alpha \in \left( \dfrac{1}{3},1 \right)\]

Answer
Verified

Hint: Find the equation of the lines which are forming the triangle.

The figure for the given problem is as follows:

Now we will find the equations of all the three sides of the triangle.

We know equation of line between the two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\]can be written as,

\[\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]

Applying the above formula, the equation of side AB is,

\[\dfrac{y-2}{0-2}=\dfrac{x-0}{2-0}\]

\[\dfrac{y-2}{-2}=\dfrac{x}{2}\]

\[\dfrac{y-2}{-1}=\dfrac{x}{1}\]

On cross multiplication, we get

\[y-2=-x\]

\[x+y-2=0........(i)\]

Similarly, the equation of side BC is,

\[\dfrac{y-0}{4-0}=\dfrac{x-2}{4-2}\]

\[\dfrac{y}{4}=\dfrac{x-2}{2}\]

\[\dfrac{y}{2}=\dfrac{x-2}{1}\]

On cross multiplication, we get

\[y=2x-4\]

\[2x-y-4=0........(ii)\]

And, the equation of side AC is,

\[\dfrac{y-2}{4-2}=\dfrac{x-0}{4-0}\]

\[\dfrac{y-2}{2}=\dfrac{x}{4}\]

\[\dfrac{y-2}{1}=\dfrac{x}{2}\]

On cross multiplication, we get

\[2y-4=x\]

\[x-2y+4=0........(iii)\]

Therefore the figure with equations is,

Two given points \[\left( {{x}_{1}},\text{ }{{y}_{1}} \right)\]and \[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)\]will lie on the same side of the line \[ax+by+c=0\] if \[a{{x}_{1}}+b{{y}_{1}}+c\] and \[a{{x}_{2}}+b{{y}_{2}}+c\] will have same signs.

From the above figure it is clear that the points B and D lie on the same side of the line AC. So, it should satisfy the above condition, i.e.,

\[2-2(0)+4=6>0\]

So, when we substitute the value of point D in line AC, it should be greater than zero, i.e.,

\[\alpha -2(2\alpha )+4>0\]

\[\alpha -4\alpha +4>0\]

\[-3\alpha +4>0\]

\[4>3\alpha \]

\[\Rightarrow \alpha <\dfrac{4}{3}.........(iv)\]

Now from the above figure it is clear that the points C and D lie on the same side of the line AB. Substitute value of point C in equation of line AB, we get

\[4+4-2=6>0\]

So, when we substitute the value of point D in line AB, it should be greater than zero, i.e.,

\[\alpha +2\alpha -2>0\]

\[3\alpha -2>0\]

\[3\alpha >2\]

\[\Rightarrow \alpha >\dfrac{2}{3}.........(v)\]

So, from equation (iv) and (v), we get

\[\alpha \in \left( \dfrac{2}{3},\dfrac{4}{3} \right)\]

Hence the correct answer is option (c).

Note: We can solve this by finding boundaries of x and y. But the options are given in fraction form, using this method it gives the exact answer.

The figure for the given problem is as follows:

Now we will find the equations of all the three sides of the triangle.

We know equation of line between the two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\]can be written as,

\[\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]

Applying the above formula, the equation of side AB is,

\[\dfrac{y-2}{0-2}=\dfrac{x-0}{2-0}\]

\[\dfrac{y-2}{-2}=\dfrac{x}{2}\]

\[\dfrac{y-2}{-1}=\dfrac{x}{1}\]

On cross multiplication, we get

\[y-2=-x\]

\[x+y-2=0........(i)\]

Similarly, the equation of side BC is,

\[\dfrac{y-0}{4-0}=\dfrac{x-2}{4-2}\]

\[\dfrac{y}{4}=\dfrac{x-2}{2}\]

\[\dfrac{y}{2}=\dfrac{x-2}{1}\]

On cross multiplication, we get

\[y=2x-4\]

\[2x-y-4=0........(ii)\]

And, the equation of side AC is,

\[\dfrac{y-2}{4-2}=\dfrac{x-0}{4-0}\]

\[\dfrac{y-2}{2}=\dfrac{x}{4}\]

\[\dfrac{y-2}{1}=\dfrac{x}{2}\]

On cross multiplication, we get

\[2y-4=x\]

\[x-2y+4=0........(iii)\]

Therefore the figure with equations is,

Two given points \[\left( {{x}_{1}},\text{ }{{y}_{1}} \right)\]and \[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)\]will lie on the same side of the line \[ax+by+c=0\] if \[a{{x}_{1}}+b{{y}_{1}}+c\] and \[a{{x}_{2}}+b{{y}_{2}}+c\] will have same signs.

From the above figure it is clear that the points B and D lie on the same side of the line AC. So, it should satisfy the above condition, i.e.,

\[2-2(0)+4=6>0\]

So, when we substitute the value of point D in line AC, it should be greater than zero, i.e.,

\[\alpha -2(2\alpha )+4>0\]

\[\alpha -4\alpha +4>0\]

\[-3\alpha +4>0\]

\[4>3\alpha \]

\[\Rightarrow \alpha <\dfrac{4}{3}.........(iv)\]

Now from the above figure it is clear that the points C and D lie on the same side of the line AB. Substitute value of point C in equation of line AB, we get

\[4+4-2=6>0\]

So, when we substitute the value of point D in line AB, it should be greater than zero, i.e.,

\[\alpha +2\alpha -2>0\]

\[3\alpha -2>0\]

\[3\alpha >2\]

\[\Rightarrow \alpha >\dfrac{2}{3}.........(v)\]

So, from equation (iv) and (v), we get

\[\alpha \in \left( \dfrac{2}{3},\dfrac{4}{3} \right)\]

Hence the correct answer is option (c).

Note: We can solve this by finding boundaries of x and y. But the options are given in fraction form, using this method it gives the exact answer.

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