
The values of \[\alpha \] if \[(\alpha ,2\alpha )\]lies inside the \[\Delta ABC\] if
\[A(0,2),B(2,0)\]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
557.4k+ views
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Whales are warmblooded animals which live in cold seas class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How much is 23 kg in pounds class 11 chemistry CBSE

Explain zero factorial class 11 maths CBSE
