Question

Draw the graph of the system of equations x+ y=5 and 2x -y +2 =0. Shade the region bounded by these lines and the x- axis. Find the area of the shaded region.
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Answer 1

EXPLANATION.

Graph of the equation.

⇒ x + y = 5. - - - - - (1).

⇒ 2x - y + 2 = 0. - - - - - (2).

As we know that,

From equation (1), we get.

⇒ x + y = 5. - - - - - (1).

Put the values of x = 0 in the equation, we get.

⇒ (0) + y = 5.

⇒ y = 5.

Their Co-ordinates = (0,5).

Put the values of y = 0 in the equation, we get.

⇒ x + (0) = 5.

⇒ x = 5.

Their Co-ordinates = (5,0).

From equation (2), we get.

⇒ 2x - y + 2 = 0. - - - - - (2).

Put the values of x = 0 in the equation, we get.

⇒ 2(0) - y + 2 = 0.

⇒ - y + 2 = 0.

⇒ - y = - 2.

⇒ y = 2.

Their Co-ordinates = (0,2).

Put the values of y = 0 in the equation, we get.

⇒ 2x - (0) + 2 = 0.

⇒ 2x + 2 = 0.

⇒ x = - 1.

Their Co-ordinates = (-1,0).

Both curves intersects at a point = (1,4).

As we know that,

Area of triangle = 1/2 x base x height.

Height = 4 cm.

Base = 5 - (-1) = 6 cm.

Area of triangle = 1/2 x 6 x 4.

Area of triangle = 12 sq. units.

Answer 2

Given, two equations of line.

$\rm \: x + y = 5 \: \: \: - - - - (1) \\ \\ \rm \: 2x - y + 2 = 0 \: \: \: - - - - (2)$

We have to draw the graph of these equations

and shade the region bounded by these lines and the x- axis. Find the area of the shaded region.

Solution :

We know that,

To draw a graph of line we have to need at least two point . If we get that, we can easily draw the graph by adding those two point.

Drawing graph for eqn.(1)

Let, x = 0 and put this in eqn(1)

$0 + y = 5 \\ = > y = 5$

So, co-ordinates is A ( 0,5)

again,

Let, y = 0 and put this in eqn(1)

$x + 0 = 5 \\ \\ = > x = 0$

So co-ordinates is (5,0)

Add this two point. [See red line from the attachment ]

Drawing graph for eqn(2)

Let, x=0, and put this value in eqn (2)

$2 \times 0 - y + 2 = 0 \\ = > - y = - 2 = > y = 2$

So, co-ordinates is (0,2)

again,

let , y=0 and put this value in eqn (2)

$2x - 0 + 2 = 0 \\ = > 2x = - 2 \\ = > x = - 1$

So, co-ordinates is C(-1,0)

Add this two point. [See blue line from the attachment ]

Finally, from the graph we can see that,

The intersection point of eqn(1) and eqn(2) is

B(1,4)

Finding area of green shaded region :

From the graph,

A(5,0)

B(1,4)

C(-1,0)

Area of triangle ABC = 12 unit^2

[ See the attachment please ]