Question

Draw the graph of 2x +y =6 and 2x-y+2=0.Find point of intersection also find area of triangle formed by these two lines x-axis

Answer 1

SOLUTION:

We have,

We have,2x + y = 6 ⇒ y = 6 - 2x

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have y = 6 - 2 x 0 = 6

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have y = 6 - 2 x 0 = 6When x = 3, we have

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have y = 6 - 2 x 0 = 6When x = 3, we havey = 6 - 2 x 3 = 0

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have y = 6 - 2 x 0 = 6When x = 3, we havey = 6 - 2 x 3 = 0When x = 2, we have

We have,2x + y = 6 ⇒ y = 6 - 2xWhen x = 0, we have y = 6 - 2 x 0 = 6When x = 3, we havey = 6 - 2 x 3 = 0When x = 2, we havey = 6 - 2 x 2 = 2

• SEE IMAGE

Now, we plot the points A(0,6), B(3,0) and C(2,2) on the graph paper. We join A, B and C and extend it on both sides to obtain the graph of the equation 2x + y = 6.

We have,

We have,2x - y + 2 = 0

⇒ y = 2x + 2

⇒ y = 2x + 2When x = 0, we have

⇒ y = 2x + 2When x = 0, we havey = 2 x 0 + 2 = 2

⇒ y = 2x + 2When x = 0, we havey = 2 x 0 + 2 = 2When x = -1, we have

⇒ y = 2x + 2When x = 0, we havey = 2 x 0 + 2 = 2When x = -1, we havey = 2 x (-1) + 2 = 0

⇒ y = 2x + 2When x = 0, we havey = 2 x 0 + 2 = 2When x = -1, we havey = 2 x (-1) + 2 = 0When x = 1, we have

⇒ y = 2x + 2When x = 0, we havey = 2 x 0 + 2 = 2When x = -1, we havey = 2 x (-1) + 2 = 0When x = 1, we havey = 2 x 1 + 2 = 4

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. above

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. aboveThus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. aboveThus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.Clearly, we have

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. aboveThus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.Clearly, we haveFM = y-coordinate of point F(1,4) = 4 and BE = 4

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. aboveThus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.Clearly, we haveFM = y-coordinate of point F(1,4) = 4 and BE = 4∴ Area of the shaded region = Area of △FBE

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. aboveThus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.Clearly, we haveFM = y-coordinate of point F(1,4) = 4 and BE = 4∴ Area of the shaded region = Area of △FBE⇒ Area of the shaded region = 1/2(Base x Height) => 1/2(BE x FM)

2(BE x FM)= > (1/2×4×4) sq. units = 8 sq. units.