Question

Determine the area of triangle formed by the lines 2x+y=6 and 2x-y+2=0 and x- axis​

Answer 1

Answer:

2x+y=6putx=0y=6putx=1y=4putx=2y=2and2x−y+2=0puty=0x=1putx=0y=2putx=2y=6andx−y=0x=0,y=0x=1,y=1x=2,y=2

Area of bounded region

21[1(−2−2)−2(2−4)+2(4+2)]21[−4+4+12]6sq.unit

Answer 2

Explanation:

Given:

$2x+y=6,2x-y+2 = 0$

To find: The area of a triangle formed by the given lines and the x-axis is ?

Solution:

Tip: Find the intersection of lines and intersection of lines with x-axis.

$\bf Ar(∆ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Step 1: Find intersection of lines.

$2x + y = 6 \\ 2x - y = - 2 \\ - - - - - - \\ 4x = 4 \\ x = 1$

put value of x in any equation

$2(1) + y = 6 \\ \\ y = 6 - 2 \\ \\ y = 4$

Let the intersection of lines is C(1,4)

Step 2: Find intersection of lines with x-axis.

as we know that y coordinate is zero everywhere on x-axis.

Put y=0

$2x + 0 = 6 \\ \\ x = \frac{6}{3} \\ \\ x = 3$

Let the point is B(3,0)

$2x - (0) = - 2 \\ \\ x = - 1$

let the point is A(-1,0).

Step 3: Find area of triangle.

Apply the formula to find Area of ∆ABC when three vertices are known.

$Ar(∆ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

A(-1,0) B(3,0) and C(1,4)

$= \frac{1}{2} | - 1(0 - 4) + 3(4 - 0) + 1(0 - 0)|$

$= \frac{1}{2} |4 + 12| \\ \\ = \frac{1}{2} |16|$

$Ar(∆ABC) = 8 \: sq - units$

Final answer:

Area of triangle is 8 sq-units.

To learn more on brainly:

Find the area of the triangle whose vertices are (1,0),(6,0)and(4,3)

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