What is the area (in unit squares) of the region enclosed by the graphs of the equations

2x – 3y + 6 = 0, 4x + y = 16 and y = 0?

Given:

2x – 3y + 6 = 0

4x + y = 16

y = 0

Formula used:

Area of triangle = \(\frac{1}{{2}} \) ∣x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)∣ 

Calculation:

2x – 3y + 6 = 0 

⇒ 2x – 3y = -6 ....(1)

4x + y = 16  ....(2)

Multiplying equation (2) by 3

⇒ 12x + 3y = 48 ...(3)

Now,

Adding (1) and (3), we get

⇒ (2x + 12x) = (48 – 6)

⇒ 14x = 42

⇒ x = 3

Now, Putting the value of x in equation (1)

⇒ 2 × 3 – 3y = -6

⇒ 6 – 3y = -6

⇒ -3y = -1

⇒ y = (-1 – 3) = 4

So,(x₁, y₁) = (3, 4)

Now,

We put y = 0 in equation (2)

⇒ 4x + 0 = 16

⇒ 4x = 16

⇒ x = 4

So,(x₂, y₂) = (4, 0)

Again,

We put y = 0 in equation (1)

⇒ 2x – 0 = -6

⇒ 2x = -6

⇒ x = -3

So,(x₃, y₃) = (-3, 0)

Now,

Area of triangle = \(\frac{1}{{2}} \) ∣x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)∣ 

⇒ \(\frac{1}{{2}} \)∣3(0 – 0) + 4(0 – 4) + (-3)(4 – 0)∣ 

⇒ \(\frac{1}{{2}} \)∣0 + (-16) + (-12)∣ 

⇒ \(\frac{1}{{2}} \)∣-16 – 12∣ 

⇒ \(\frac{1}{{2}} \)∣-28∣ 

⇒ 14  [Negative sign is always treated as positive in modulus]

∴ The required value is 14.

Alternate Method

 When we find the vertices from the given equations

Given:

2x – 3y + 6 = 0

4x + y = 16

y = 0

Area of triangle = \(\frac{1}{2}\) × Base × Height

Area of triangle = \(\frac{1}{2}\) × 7 × 4 = 14