Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

4x - 3y + 4 = 0,

4x + 3y - 20 = 0.

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Solution

The given system of equation is 4x - 3y + 4 = 0 and 4x + 3y - 20 = 0

The vertices are (-1,0),(5,0),(2,4)

$A r e a = \frac{1}{2} b h$

b = 6 units

h = 4 units

$A r e a = \frac{1}{2} b h = \frac{1}{2} \times 6 \times 4 = 12 s q . u n i t s$

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Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis: 4x - 3y + 4 = 0, 4x + 3y - 20 = 0.

The given system of equation is 4x - 3y + 4 = 0 and 4x + 3y - 20 = 0 The vertices are (-1,0),(5,0),(2,4) Area=12bh b = 6 units h = 4 units Area=12bh=12×6×4=12 sq.units

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

4x - 3y + 4 = 0,

4x + 3y - 20 = 0.

The given system of equation is 4x - 3y + 4 = 0 and 4x + 3y - 20 = 0

The vertices are (-1,0),(5,0),(2,4)

$A r e a = \frac{1}{2} b h$

b = 6 units

h = 4 units

$A r e a = \frac{1}{2} b h = \frac{1}{2} \times 6 \times 4 = 12 s q . u n i t s$