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
388.8k+ 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
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is 1 divided by 0 class 8 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Convert compound sentence to simple sentence He is class 10 english CBSE

India lies between latitudes and longitudes class 12 social science CBSE

Why are rivers important for the countrys economy class 12 social science CBSE

Distinguish between Khadar and Bhangar class 9 social science CBSE
