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Question

Draw the graphs of the following linear equations on the same graph paper:
2x + 3y = 12, x − y = 1

Find the coordinates of the vertices of the triangle formed by the two straight lines and the area bounded by these lines and x-axis.

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Solution

We are given,

We get,

Now, substituting in , we get

Substituting in , we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x

0

6

y

4

0

Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

y=x-1

Now, substituting in y=x-1,we get

y=-1

Substituting in y=x-1,we get

y=-2

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x

0

–1

y

-1

-2

Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on y axis.

Therefore,

AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis.

FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

A=12Base×Height =12AC×FE =125×3 =152 sq. units


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