Question

Find a range of values of $c$ such that the lines $x-y=2$ and $cx+y=3$ intersect in the first quadrant.

Answer 1

Step 1: Define the problem

Given two lines $x-y=2$ and $cx+y=3$.

Since the intersection happens in the first quadrant, $x>0$ and $y>0$.

From the first equation,

$$\begin{gathered} x-y=2 \\ \Rightarrow x=y+2 \end{gathered}$$

Step 2: Compute range of $c$ for $y>0$

Substituting $x=y+2$ in $cx+y=3$,

$$\begin{gathered} c(y+2)+y=3 \\ \Rightarrow cy+2c+y=3 \\ \Rightarrow y(c+1)=3-2c \\ \Rightarrow y=\frac{3-2c}{c+1} \end{gathered}$$

Therefore, for $y>0$, $c\in \left(-1,\frac{3}{2}\right)$

Step 3: Compute range of $c$ for $x>0$

Substituting $y=\frac{3-2c}{c+1}$ in $x-y=2$

$$\begin{gathered} x-\frac{3-2c}{c+1}=2 \\ \Rightarrow x=\frac{5}{c+1} \end{gathered}$$

Therefore, for $x>0$, $c\in \left(-1,\infty \right)$

Step 4: Compute range of values for $c$

$$\begin{gathered} c\in \left(-1,\frac{3}{2}\right)\cap \left(-1,\infty \right) \\ \Rightarrow c\in \left(-1,\frac{3}{2}\right) \end{gathered}$$

Therefore, the range of values for $c$ such that the two lines intersect in first quadrant is, $-1<c<\frac{3}{2}$.