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The complete set of real values of 'a' such that the point lies triangle p(a,sina) lies inside the triangle formed by the lines

a. x−2y+2=0;x+y=0 and x−y−π=0

1 Answer

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Equation x−2y+2=0 solution is a = − π − 2

Equation x+y=0 solution is y = − 1/2 π

Equation x−y−π=0 solution is x = 1/2 π

Let's solve the system of equations:

a−2y+2=0

x+y=0

x−y−π=0

We can solve the system of equations by first rearranging the terms in standard form, and then using elimination.

Rearrange terms in standard form:

a−2y =−2

y+x = 0

−y+x−π=0

Eliminate y:

a−2y = −2

y+x = 0

2x−π = 0

Solve for x:

x = 1/2 π

Substitute x back into the first equation to solve for y:

a − 2y = −2

y + 1/2 π =0

y = − 1/2 π

Substitute x and y back into the third equation to solve for a:

x−y−π=0

1/2 π - (1/2 π) − π=0

a = − π − 2

User Bernhard Koenig
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