Question

Determine the range of values of θ ∈ [0, 2 π] for which the point (cos θ, sin θ) lies inside the triangle formed by the lines x + y = 2 ; x − y = 1 & 6x + 2y − 10 = 0.

Answer 1

Answer on Question #56877, Math, Geometry

**Problem.**

Determine the range of values of $\theta \in [0, 2\pi]$ for which the point (cos $\theta$, sin $\theta$) lies inside the triangle formed by the lines $x + y = 2$; $x - y = 1$ & $6x + 2y - 10 = 0$.

**Answer.**

These lines don't form a triangle.

**Solution.**

Let's determine the coordinates of the points of intersection of the lines given.

$$\left\{ \begin{array}{l} x + y = 2, \\ x - y = 1; \end{array} \right. \left\{ \begin{array}{l} 2x = 3, \\ 2y = 1; \end{array} \right. \left\{ \begin{array}{l} x = 1.5, \\ y = 0.5. \end{array} \right. \quad (1.5, 0.5).$$ $$\left\{ \begin{array}{c} x + y = 2, \\ 6x + 2y - 10 = 0; \end{array} \right. \left\{ \begin{array}{l} x + y = 2, \\ 3x + y = 5; \end{array} \right. \left\{ \begin{array}{l} 2x = 3, \\ x + y = 2; \end{array} \right. \left\{ \begin{array}{l} x = 1.5, \\ y = 0.5. \end{array} \right. \quad (1.5, 0.5).$$ $$\left\{ \begin{array}{c} x - y = 1, \\ 6x + 2y - 10 = 0; \end{array} \right. \left\{ \begin{array}{l} x - y = 1, \\ 3x + y = 5; \end{array} \right. \left\{ \begin{array}{l} 4x = 6, \\ x - y = 1; \end{array} \right. \left\{ \begin{array}{l} x = 1.5, \\ y = 0.5. \end{array} \right. \quad (1.5, 0.5).$$

So, the three lines are concurrent, thus don't form the triangle.

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