Question

The area of the triangle formed by the lines
x + y - 3 = 0, x - 3y + 9 = 0 and 3x – 2y + 1 = 0
(1) 16/7 sq. units
3) 10/7 sq. units
(2) 4 sq. units
(4) 9 sq. units​

Answer 1

Answer:

Option (3) 10/7 sq units

Step-by-step explanation:

Equation of the lines

$L_1 : x+y=3$

$L_2 : x-3y=-9$

$L_3 : 3x-2y=-1$

Intersection point of the lines $L_1, L_2, L_3$ are $(0, 3), (1,2), (15/7, 26/7)$

The area of triangle is given by

$\boxed{A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}$

Therefore, area is

$A=\frac{1}{2}[0(2-\frac{26}{7})+1(\frac{26}{7}-3)+\frac{15}{7}(3-2)]$

$A=\frac{1}{2}[\frac{5}{7}+\frac{15}{7}]$

$\implies A=\frac{10}{7}$

Hope this answer is helpful.

Answer 2

The area of the triangle is $\frac{10}{7}\ \text{sq.units}$.

Step-by-step explanation:

Let the equations of lines,

$A : x+y=3$

$B : x-3y=-9$

$C : 3x-2y=-1$

Plot these equations in the graph.

The intersection point of the lines are (0, 3), (1,2), $(\frac{15}{7},\frac{26}{7})$

Refer the attached figure below.

The area of triangle is given by

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here, $(x_1,y_1)=(0,3)$, $(x_2,y_2)=(1,2)$, $(x_3,y_3)=(\frac{15}{7},\frac{26}{7})$

Substitute the value in the formula,

$A=\frac{1}{2}[0(2-\frac{26}{7})+1(\frac{26}{7}-3)+\frac{15}{7}(3-2)]$

$A=\frac{1}{2}[\frac{5}{7}+\frac{15}{7}]$

$A=\frac{10}{7}$

Therefore, option 2 is correct.

The area of the triangle is $\frac{10}{7}\ \text{sq.units}$.

#Learn more

The area of triangle formed by the line x/a + y/b = 1 with the coordinate axes is-1) ab2)2ab3)1/2ab4)2/2ab

https://brainly.in/question/1218683