Line & inequality

Geometry Level pending

If $x-y + 3 > 0$ and $-x-y+6 > 0$ , then find the area in the first quadrant described by these two inequalities.

\frac{29}{2}

\frac{63}{4}

\frac{17}{2}

\frac{45}{4}

\frac{9}{2}

1 solution

Tanveer Alam
Jan 12, 2016

x - y +3 = 0
- x - y + 6 = o

solving this two above equation we have A(3/2, 9/2).

we know that B(0, 3), C(0, 0), D(6, 0)

using this 4 coordinates we would be able to find the area & the area is 9/2.

you should mention why area bounded by A(3/2, 9/2) B(0, 3), C(0, 0), D(6, 0) is the only required area

Atul Shivam - 5 years, 5 months ago

In the problem, i told about 1st quadrant.isn't it anough for that?

Tanveer Alam - 5 years, 5 months ago

but why you have taken area below AD & AB only??? you should mention it in your solution

Atul Shivam - 5 years, 5 months ago

This solution is not correct. Just graphically you can see that the area has to be bigger than 8, given that two 2x2 squares are encompassed in the area.

Andrew Ellinor - 5 years, 4 months ago

Line & inequality

1 solution

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