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? -Maths 9th

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? -Maths 9th
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Answer :

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 × 1 ×  3 / 2 = 3 / 4 sq unit.
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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? -Maths 9th

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Last Answer : 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 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

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