Answer to Question #228765 in Analytic Geometry for Bless

Question #228765

A triangle is formed by the lines y=x, y=2x-1 and 3y-4x=1. Find the area of the triangle


1
Expert's answer
2021-09-02T16:38:46-0400

Given equation are y=x,y=2x1,3y4x=1y=x,y=2x-1,3y-4x=1

Let y=x...(1)y=x...(1)

y=2x1(2)y=2x-1…(2)

3y4x=1(3)3y-4x=1…(3)

Solving (1) and (2), we get:

(x,y)=(1,1)(x,y)=(1,1)

Solving (2) and (3), we get:

3(2x1)4x=13(2x-1)-4x=1

2x=4⇒2x=4 

x=2⇒x=2

y=3∴y=3

(x,y)=(2,3)(x,y)=(2,3)

Solving (1) and (3), we get:

(x,y)=(1,1)(x,y)=(-1,-1)

So, let coordinates of points A, B, C of ABC\triangle ABC be (1,1),(2,3),(1,1)(1,1),(2,3),(-1,-1) respectively.

We know that area of triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1,y_1),(x_2,y_2),(x_3,y_3) is given as:

=12x1y11x2y21x3y31\triangle =|\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}|

\therefore Area of ABC=12111231111\triangle ABC=|\frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ -1 & -1 & 1 \end{vmatrix}|

=12[1(3+1)1(2+1)+1(2+3)]=\frac{1}{2} [1(3+1)-1(2+1)+1(-2+3)]

=12[42+1]=\frac{1}{2} [4-2+1]

=32=\frac{3}{2} square units.


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