We have two lines $3x - y + 1 = 0$ and $x + 2y - 5 = 0$. The region $R$ is the interior region between these lines that contains the origin. We are given a point $(a^2, a+1)$ and need to find all values of $a$ for which this point lies in $R$. Given: - Line $L_1$: $3x - y + 1 = 0$ - Line $L_2$: $x + 2y - 5 = 0$ - Point: $(a^2, a+1)$ - Region $R$ contains the origin We need to find: the set of all values of $a$ such that $(a^2, a+1)$ lies in $R$. **Strategy:** Our plan is to use the same-side test. For the point to lie in the same region as the origin, it must be on the same side of both lines as the origin. We will substitute the origin into each line to determine the sign, then substitute $(a^2, a+1)$ and require the same sign, giving us two inequalities to solve and intersect. **Step 1: Same-side test for $L_1$: $3x - y + 1 = 0$** We first check which side the origin is on by substituting $(0, 0)$ into $3x - y + 1$. $3(0) - 0 + 1 = 1 > 0$ So the origin gives a positive value. For the point $(a^2, a+1)$ to be on the same side, we need: $3(a^2) - (a+1) + 1 > 0$ $3a^2 - a - 1 + 1 > 0$ $3a^2 - a > 0$ $a(3a - 1) > 0$ This product is positive when both factors are positive or both are negative: - $a < 0$ and $3a - 1 < 0$ (true when $a < 0$), or - $a > 0$ and $3a - 1 > 0$ (true when $a > 1/3$) So $a \in (-\infty, 0) \cup (1/3, \infty)$. This means the point is on the same side as the origin for $L_1$ when $a < 0$ or $a > 1/3$. **Step 2: Same-side test for $L_2$: $x + 2y - 5 = 0$** Substituting the origin $(0, 0)$ into $x + 2y - 5$: $0 + 0 - 5 = -5 < 0$ So the origin gives a negative value. For the point $(a^2, a+1)$ to be on the same side: $a^2 + 2(a+1) - 5 < 0$ $a^2 + 2a + 2 - 5 < 0$ $a^2 + 2a - 3 < 0$ Using factoring: $(a + 3)(a - 1) < 0$ This product is negative when exactly one factor is negative, which happens when $-3 < a < 1$. So $a \in (-3, 1)$. This means the point is on the same side as the origin for $L_2$ when $-3 < a < 1$. **Step 3: Find the intersection of both conditions** We need both conditions to hold simultaneously: - From $L_1$: $a \in (-\infty, 0) \cup (1/3, \infty)$ - From $L_2$: $a \in (-3, 1)$ Intersecting these: $(-\infty, 0) \cap (-3, 1) = (-3, 0)$ and $(1/3, \infty) \cap (-3, 1) = (1/3, 1)$. So $a \in (-3, 0) \cup (1/3, 1)$. **Verification:** We test $a = -1$ (should be in $R$). Point $= (1, 0)$. For $L_1$: $3(1) - 0 + 1 = 4 > 0$. Same sign as origin. For $L_2$: $1 + 0 - 5 = -4 < 0$. Same sign as origin. Checks out. **Summary:** We used the same-side test for each line by checking the sign the origin produces and requiring the given point to produce the same sign. This gave us two inequalities in $a$, and their intersection yielded the final answer $(-3, 0) \cup (1/3, 1)$.

Straight Lines PYQs