Question

Statement 1: Area of \[\Delta ABC\] where, A (20,22), B (21,24), C (22,23) and area of \[\Delta PQR\] where P(0,0), Q(1,2), R(2,1) is equal.
Statement 2: The area of the triangle be constant with respect to parallel transformation of coordinate axis.
(a) Statement 1 is true, Statement 2 is the correct explanation for Statement 1.
(b) Statement 1 is true, Statement 2 is not a correct explanation for Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is false, Statement 2 is true.

Answer 1

Hint: Find the area of both the triangles with the formula,
Area =\[\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|\] and compare the areas of \[\Delta ABC\] and \[\Delta PQR\].

Complete step-by-step answer:
Considering Statement -1:
Let us find the area of \[\Delta ABC\] whose coordinates be:
A (20,22) = \[\left( {{x}_{1,}}{{y}_{1}} \right)\]
B (21,24) = \[\left( {{x}_{2,}}{{y}_{2}} \right)\]
C (22,23) = \[\left( {{x}_{3,}}{{y}_{3}} \right)\]
Area of \[\Delta ABC\] is given as:
Area = \[\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|\]
Substituting the respective coordinates, we have:
 =\[\dfrac{1}{2}\left| 20(1)+21(1)+22(-2) \right|\]
=\[\dfrac{1}{2}\left| 41-44 \right|\]
=\[\dfrac{3}{2}\] Square units
Now let us find the area of \[\Delta PQR\] in a similar fashion:
Where,
P = (0,0)
Q = (1,2)
R = (2,1)
So, area of \[\Delta PQR\] is:
Area = \[\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|\]
Substituting the coordinates in the formula, we get:
=\[\dfrac{1}{2}\left| 0+1(1)+2(-2) \right|\]
=\[\dfrac{1}{2}\left| -3 \right|\]
=\[\dfrac{3}{2}\] Square units
Hence, we can show that:
 Area of \[\Delta ABC\]= Area of \[\Delta PQR\]=\[\dfrac{3}{2}\] Square units.
Now, considering statement - 2:
we already know that , according to parallel axis of transformation, if the origin is shifted from (0,0) to the new location \[\left( {{x}^{1}},{{y}^{1}} \right)\], then the point ‘p’\[\left( x,y \right)\] on coordinate axis will become \[\left( {{x}_{1}},{{y}_{1}} \right)\]
\[{{x}^{1}}=x-{{x}_{1}}\]
\[{{y}^{1}}=y-{{y}_{1}}\]
The quantity of area is generally considered as a vector function. But, one of the most important aspects of operations of vectors is their invariance, which means independent of a particular coordinate system. Therefore, as the area vector is invariant, the area of the triangle does not change due to the transformation of the axis. So, the area of \[\Delta ABC\] and \[\Delta PQR\] remains the same.
So, statement 1 is true and statement 2 is the correct explanation of statement 1.
Hence, option (A) is the correct answer.
Note: We need to consider the modulus in the formula, as the area of a closed figure cannot take a negative value. Also remember that, the area enclosed by a polygon does not change with respect to the transformation of the axes, as the lengths of sides and angles between them still remain the same.