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Question: The area of the triangle formed by the lines \[{x^2} - 4{y^2} = 0\] and \(x = a\), is \(A)2{a^2}\...

The area of the triangle formed by the lines x2−4y2=0{x^2} - 4{y^2} = 0 and x=ax = a, is
A)2a2A)2{a^2}
B)a22B)\dfrac{{{a^2}}}{2}
C)3a22C)\dfrac{{\sqrt 3 {a^2}}}{2}
D)2a23D)\dfrac{{2{a^2}}}{{\sqrt 3 }}

Explanation

Solution

First, we need to know about the concept of area and triangle.
Area means quantitative or else like a measurement. A triangle is three sides of closed vertices or like angles with 2D2D ( two-dimensional shapes).
The area of the triangle in which the total surface or else like a space enclosed by three lines of boundaries of the triangle is called as or known as area of the triangle.
Formula used:
Area of the triangle for a general method Area=12(base)(height)Area = \dfrac{1}{2}(base)(height)
Area of the triangle for the vertices using matrix method A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\\ {{x_2}}&{{y_2}}&1 \\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| and expressed as in the form of A=12[x1(y2−y3)−y1(x2−x3)+1(x2y3−x3y2)]A = \dfrac{1}{2}[{x_1}({y_2} - {y_3}) - {y_1}({x_2} - {x_3}) + 1({x_2}{y_3} - {x_3}{y_2})]

Complete step by step answer:
Since from the given that we have two equations which are x2−4y2=0{x^2} - 4{y^2} = 0 and x=ax = a
Since equation two itself a result as x=ax = a , so further proceed with the first equation x2−4y2=0{x^2} - 4{y^2} = 0
Separate this equation into factors using the separation of the equation method, which is (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2} and thus we get the result as x2−4y2=0⇒(x−2y)(x+2y)=0{x^2} - 4{y^2} = 0 \Rightarrow (x - 2y)(x + 2y) = 0
As per the multiplication operation, we have either (x−2y)=0(x - 2y) = 0 or (x+2y)=0(x + 2y) = 0
Also, we know that x=ax = a. Now substitute the value of x=ax = aat the above equation we get (x−2y)=0⇒y=a2(x - 2y) = 0 \Rightarrow y = \dfrac{a}{2} and (x+2y)=0⇒y=−a2(x + 2y) = 0 \Rightarrow y = \dfrac{{ - a}}{2}
Thus, we have two values as x=a,y=−a2x = a,y = \dfrac{{ - a}}{2} and x=a,y=a2x = a,y = \dfrac{a}{2}
If we simplify the equations (x−2y)=0(x - 2y) = 0 or (x+2y)=0(x + 2y) = 0 by elimination method or factorizing method, then we get 2x=0⇒x=02x = 0 \Rightarrow x = 0substituting this value in the above equation we get (x−2y)=0⇒y=0(x - 2y) = 0 \Rightarrow y = 0
Hence the third values are x=0,y=0x = 0,y = 0
Thus, we have the three vertices for the matrix, then we have from the area of the triangle matrix determinant formula; A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\\ {{x_2}}&{{y_2}}&1 \\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| and thus apply the values that we found in above we get A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\\ {{x_2}}&{{y_2}}&1 \\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| \Rightarrow \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1 \\\ a&{\dfrac{{ - {a}}}{2}}&1 \\\ a&{\dfrac{{{a}}}{2}}&1 \end{array}} \right|
Using the determinant formula for the matrix we have A=12[a×a2+a×a2]A = \dfrac{1}{2}[a \times \dfrac{a}{2} + a \times \dfrac{a}{2}]
Further solving we get A=12[a×a2+a×a2]⇒a22sq.unitA = \dfrac{1}{2}[a \times \dfrac{a}{2} + a \times \dfrac{a}{2}] \Rightarrow \dfrac{{{a^2}}}{2}sq.unit

So, the correct answer is “Option B”.

Note: We can also apply the A=12[x1(y2−y3)−y1(x2−x3)+1(x2y3−x3y2)]A = \dfrac{1}{2}[{x_1}({y_2} - {y_3}) - {y_1}({x_2} - {x_3}) + 1({x_2}{y_3} - {x_3}{y_2})] on assuming the three points which are found.
On finding the general area of triangle or length of the base or length of height we use the formulaArea=12(base)(height)Area = \dfrac{1}{2}(base)(height)
But the only difference is substituting the known values to get the unknown value. Suppose if the base is unknown, then substitute the area of the triangle and height values to find it.