Question

a triangle is formed by the straight lines x + 2y - 3=0, 3x - 2y + 7= 0 and y+1=0 find graphically
( 1) the coordinates of the vertices of the triangle
(2) the area of the triangle
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Answer 1

Step-by-step explanation:

bhaiya please determine that

what vertices are you want to be that formed on your axis or formed on x axis

and this is the correct way to solve this problem

Answer 2

Answer:

1) Coordinates of the vertices of the triangle are (-1,2) , (-3,-1) and (5,-1).

2)area of the triangle is 12$cm^{2}$

Step-by-step explanation:

Given,

straight lines are ,x +2y -3 =0 ,3x -2y +7 = 0 and y+1 = 0

Let x+2y =3 .........(i)

 3x -2y = -7 ........(ii)

  and y= -1 ......(iii)

For part 1:

Step 1:

For  x+2y = 3

put x = 0 we get y = 1.5,

put y = 0 we get x = 3

Step 2:

For  3x -2y = -7

put x = 0  we get y = 3.5

put y = 0 we get x = -2.33

Ploting the equation x+2y =3 , 3x -2y = -7 and y =-1

in the graph we get the vertices of the triangle are (-1,2) , (-3,-1) and (5,-1)

Part 2:

Name the vertices of the triangle  are A,Band C

so , ABC is a triangle in which A (-3,-1) , B (5,-1)and C(-1,2)

Step 1:

By distance formula

BC = $\sqrt{(-1-5)^{2} + (2+1)^{2} } = \sqrt{45}$

AC = $\sqrt{(-1+3)^{2}+ (2+1)^{2} } =\sqrt{13}$

where from the graph ABC is a triangle in which BC is the height and AC is the base

Therefore ,

Area of a triangle = $\frac{1}{2}$ ×base ×height

                              = $\frac{1}{2}$ × 3.6× 6.7

                              = 1.8 ×6.7

             Area of triangle ABC    = 12 .06  = 12$cm^{2}$

Final answer:

Hence , the coordinates of the vertices of the triangle is (-3,-1),(5,-1) and (-1,2) and the area of the triangle is  12 $cm^{2}$