Answer
Verified
417.6k+ views
Hint: The formula that we can use to find the area of a triangle with vertices A \[\left( {{x}_{1}},{{y}_{1}} \right)\] , B \[\left( {{x}_{2}},{{y}_{2}} \right)\] and C \[\left( {{x}_{3}},{{y}_{3}} \right)\] is as follows
\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\]
If area of triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\] = 0 and the points A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) are collinear.
Complete Step-by-Step solution:
As mentioned in the question, we have to find the area of the triangle whose coordinates are given in the question.
Now, we can find the area of this triangle by using the formula that is mentioned in the hint as follows
As the coordinates are A \[\left( 2,0 \right)\] , B \[\left( 1,2 \right)\] and C \[\left( -1,6 \right)\] which is given in the question, we can get the area of the triangle as
\[\begin{align}
& =~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right] \\
& =\dfrac{1}{2}\left[ 2(2-6)+1(6-0)+-1(0-2) \right] \\
& =\dfrac{1}{2}\left[ -8+6+2 \right] \\
& =0 \\
\end{align}\]
As we get the area of this triangle as zero, we can use the information that is given in the hint which is
if the area of the triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then the points are collinear.
Hence, the points are collinear.
Note: The students can make an error if they don’t know about the formula for calculation of the area of a triangle whose vertices are given that is mentioned in the hint as follows
The formula that we can use to find the area of a triangle with vertices A \[\left( {{x}_{1}},{{y}_{1}} \right)\] , B \[\left( {{x}_{2}},{{y}_{2}} \right)\] and C \[\left( {{x}_{3}},{{y}_{3}} \right)\] is as follows
\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\]
\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\]
If area of triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\] = 0 and the points A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) are collinear.
Complete Step-by-Step solution:
As mentioned in the question, we have to find the area of the triangle whose coordinates are given in the question.
Now, we can find the area of this triangle by using the formula that is mentioned in the hint as follows
As the coordinates are A \[\left( 2,0 \right)\] , B \[\left( 1,2 \right)\] and C \[\left( -1,6 \right)\] which is given in the question, we can get the area of the triangle as
\[\begin{align}
& =~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right] \\
& =\dfrac{1}{2}\left[ 2(2-6)+1(6-0)+-1(0-2) \right] \\
& =\dfrac{1}{2}\left[ -8+6+2 \right] \\
& =0 \\
\end{align}\]
As we get the area of this triangle as zero, we can use the information that is given in the hint which is
if the area of the triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$) is zero, then the points are collinear.
Hence, the points are collinear.
Note: The students can make an error if they don’t know about the formula for calculation of the area of a triangle whose vertices are given that is mentioned in the hint as follows
The formula that we can use to find the area of a triangle with vertices A \[\left( {{x}_{1}},{{y}_{1}} \right)\] , B \[\left( {{x}_{2}},{{y}_{2}} \right)\] and C \[\left( {{x}_{3}},{{y}_{3}} \right)\] is as follows
\[Area=~\dfrac{1}{2}~\left[ {{x}_{1}}({{y}_{2}}-~{{y}_{3}}~)+{{x}_{2}}({{y}_{3}}-{{y}_{1}}~)+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right]\]
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE