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A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? If yes, then write the coordinates of point of intersection of lines a and b. 

Also, find the area enclosed between these lines and Y-axis. 

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Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1.

 Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0) and Y - axis at (0,2) and so the given line b is correct.

Thus, the student is right .

Now, substituting y = 1 in equation of line b, we get 

2x + 3 (1) 6  ⇒ 2x = 6 - 3 = 3  ⇒ x = 3 / 2 

Here, the point of intersection is (3/2,1)

 ∴ Area enclosed between these lines and Y - axis .

= Area of ΔABC = 1/2 × Base × Height 

= 1/2 × BC × AC = 1/2 ×× 3 / 2 = 3 / 4 sq unit.

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