Question

In a triangle \[ABC\], coordinates of \[A\] are \[\left( 1,2 \right)\] and the equation of the medians through \[B\] and \[C\] are \[x+y=5\] and \[x=4\] respectively, then the area of \[\Delta ABC\] (in sq. units) is
A. \[5\]
B. \[9\]
C. \[12\]
D. \[4\]

Answer 1

Hint: To solve the given question, we can find the intersection point of the two median by plotting the graphs of both the equations. Then we will need to find the coordinates of the point B and C by using the midpoint formula. After getting all the coordinates of the three vertices of the triangle, by using the formula of the area of the triangle with the three given vertices we will find the area of the triangle.

Formula used:
Mid-point of any given line with two point is,
 \[mid-po\operatorname{int}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] , where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.
The formula of the area of the triangle with the three given vertices is given by,
Area of the triangle = \[\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)\] ,
where \[\left( {{x}_{1,}}{{y}_{1}} \right),\ \left( {{x}_{2,}}{{y}_{2}} \right)\ and\ \left( {{x}_{3,}}{{y}_{3}} \right)\] are the coordinates of the three given vertices.

Complete step by step solution:
We have given that,
In a triangle \[ABC\] , coordinates of \[A\] are \[\left( 1,2 \right)\] and the equation of the medians through \[B\] and \[C\] are \[x+y=5\] and \[x=4\] respectively.
Thus,
The equation of the median through \[B\] is \[x+y=5\]. So the coordinates of the point \[B\] is \[\left( \alpha ,5-\alpha \right)\] .
And,
The equation of the median through \[C\] is \[x=4\] . So let the coordinates of the point \[C\] is \[\left( 4,\beta \right)\] .
Now,
The centroid of the triangle is given by the intersection of \[x+y=5\] and \[x=4\] .
Intersection of \[x+y=5\] and \[x=4\] shown in the graph below;

Therefore, the point of intersection is \[\left( 4,1 \right)\] .
Thus, let the centroid of the triangle at the point \[G=\left( 4,1 \right)\] .
Now,
The figure of the triangle \[\Delta ABC\] is as shown in the figure below;

Now,
The midpoint of the side of the triangle \[AC\] is the median of the \[\angle B\] i.e. \[x+y=5\] and it is given by,
 \[\Rightarrow N=\left( \dfrac{1+4}{2},\dfrac{2+\beta }{2} \right)\]
Thus,
 \[\Rightarrow \dfrac{1+4}{2}+\dfrac{2+\beta }{2}=5\]
 \[\Rightarrow \dfrac{5+2+\beta }{2}=5\]
 \[\Rightarrow 7+\beta =10\]
 \[\Rightarrow \beta =3\]
Hence, we got the coordinate of the point \[C\] is \[\left( 4,3 \right)\] .
Now,
We have the coordinates of the point \[A\ is\ \left( 1,2 \right)\] and \[B\ is\ \left( \alpha ,5-\alpha \right)\] .
The midpoint of the side of the triangle \[AB\] is the median of the \[\angle C\] i.e. \[x=4\] and it is given by,
 \[\Rightarrow M=\left( \dfrac{1+\alpha }{2},\dfrac{2+5-\alpha }{2} \right)=\left( \dfrac{1+\alpha }{2},\dfrac{7-\alpha }{2} \right)\]
We have given that \[x=4\] , then the x-coordinate of the mid-point ‘M’ is equal to 4.
Thus,
 \[\Rightarrow \dfrac{1+\alpha }{2}=4\]
 \[\Rightarrow 1+\alpha =8\]
 \[\Rightarrow \alpha =7\]
Now,
Substituting the value of \[\alpha =7\] in the co-ordinate of the point \[B\ is\ \left( \alpha ,5-\alpha \right)\] .
 \[\Rightarrow \left( \alpha ,5-\alpha \right)=\left( 7,5-7 \right)=\left( 7,-2 \right)\]
Hence, we got the co-ordinate of the point \[B\] is \[\left( 7,-2 \right)\] .
Therefore,
We have now a a triangle \[ABC\] , coordinates of \[A\] is \[\left( 1,2 \right)\] , coordinates of \[B\] is \[\left( 7,-2 \right)\] and the coordinates of \[C\] is \[\left( 4,3 \right)\] .
Using the formula of the area of the triangle with the three given vertices is given by,
Area of the triangle = \[\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)\] , where \[\left( {{x}_{1,}}{{y}_{1}} \right),\ \left( {{x}_{2,}}{{y}_{2}} \right)\ and\ \left( {{x}_{3,}}{{y}_{3}} \right)\] are the coordinates of the three given vertices.
Now,
Let the coordinate of the point \[A=\left( 1,2 \right)\] be \[\left( {{x}_{1,}}{{y}_{1}} \right)\] , the coordinate of the point \[B=\left( 7,-2 \right)\] be \[\left( {{x}_{2,}}{{y}_{2}} \right)\] and the coordinate of the point \[C=\left( 4,3 \right)\] be \[\left( {{x}_{3,}}{{y}_{3}} \right)\] .
Substituting the values in the formula of the area of triangle, we will obtain
Area of the triangle \[\Delta ABC\] = \[\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)\]
Substituting the values,
Area of the triangle \[\Delta ABC\] = \[\dfrac{1}{2}\left( 1\left( -2-3 \right)+7\left( 3-2 \right)+4\left( 2-\left( -2 \right) \right) \right)\]
Solving the numbers, we will get
Area of the triangle \[\Delta ABC\] = \[\dfrac{1}{2}\left( -5+7+16 \right)=\dfrac{1}{2}\times 18=9sq.units\]
Therefore,
The area of the triangle \[\Delta ABC\] is \[9sq.units\] .

Hence, the option (b) is the correct answer.

Note: One may note that the formula for the area of triangle in simplified form is given as:
Area of the triangle = \[\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)\] , where \[\left( {{x}_{1,}}{{y}_{1}} \right),\ \left( {{x}_{2,}}{{y}_{2}} \right)\ and\ \left( {{x}_{3,}}{{y}_{3}} \right)\] are the coordinates of the three given vertices.
Students always need to remember that if by substituting the values of given coordinates in the formula for area, the area turns out negative then we have to take the modulus to make it positive. And students can choose any of the vertices as \[\left( {{x}_{1,}}{{y}_{1}} \right),\ \left( {{x}_{2,}}{{y}_{2}} \right)\ and\ \left( {{x}_{3,}}{{y}_{3}} \right)\] because it will not change the answer.