Question

If the area of the triangle formed by point A (x, y), B (1, 2) and C (2, 1) is 6 square units, then show that x + y = 15.

Answer 1

Hint: Use the area formula which is,
Area = \[\dfrac{1}{2}\left| {{a}_{1}}\left( {{b}_{2}}-{{b}_{3}} \right)+{{a}_{2}}\left( {{b}_{3}}-{{b}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}-{{b}_{2}} \right) \right|\], where \[\left( {{a}_{1}},{{b}_{1}} \right)\], \[\left( {{a}_{2}},{{b}_{2}} \right)\], \[\left( {{a}_{3}},{{b}_{3}} \right)\] are points and then equate it with area equals to 6 to get the relation.

Complete step-by-step answer:
In the question we are given three vertices of triangle A, B, C which are (x, y), (1, 2) and (2, 1) respectively. The area of the ABC is given as 6 square units and we have to prove x + y = 15.
For finding the relation between x and y we will use the formula of area which is,
Area = \[\dfrac{1}{2}\left| {{a}_{1}}\left( {{b}_{2}}-{{b}_{3}} \right)+{{a}_{2}}\left( {{b}_{3}}-{{b}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}-{{b}_{2}} \right) \right|\], where points of vertices are \[\left( {{a}_{1}},{{b}_{1}} \right)\], \[\left( {{a}_{2}},{{b}_{2}} \right)\], \[\left( {{a}_{3}},{{b}_{3}} \right)\] whose area to be find out.
Here points are (x, y), (1, 2) and (2, 1) so area will be
\[\Rightarrow \]Area = \[\dfrac{1}{2}\left| x\left( 2-1 \right)+1\left( 1-y \right)+2\left( y-2 \right) \right|\]
\[\Rightarrow \]Area = \[\dfrac{1}{2}\left| x+1-y+2y-4 \right|\]
\[\Rightarrow \]Area = \[\dfrac{1}{2}\left| x+y-3 \right|\]
We know that area of triangle is 6 square units so we can write,
\[\Rightarrow \dfrac{1}{2}\left| x+y-3 \right|=6\]
On cross multiplication we get,
\[\Rightarrow \left| x+y-3 \right|=12\]
When \[\left| k \right|=l\], where l is constant then after the remaining mod sign we get, \[k=\pm l\].
So, we can write as,
\[\Rightarrow x+y-3=\pm 12\]
Or, \[x+y=3+12\] or 3 – 12
So, x + y = 15 or – 15
Hence the equations are x + y = 15 and x + y = - 15.
Hence proved x + y = 15.

Note: One can also use distance formula to find distance between all the points and then use Heron’s formula and equate it with the given area but this process will be very lengthy and tedious.