Question

Two sides of a triangle have the joint equation $(x-3y + 2) (x+y-2) = 0$. The third side which is variable always passes through the point (-5,-1), then not possible values of slope of third side such that origin is an interior point of triangle is/are:

• A. (-oo, -1] U [1/5, oo)

Answer 1

Answer: (A)

(-oo, -1] U [1/5, oo)

Answer 2

Let the two given sides of the triangle be $L_1$ and $L_2$. $L_1: x - 3y + 2 = 0$ $L_2: x + y - 2 = 0$

Let the third side be $L_3$. It passes through the point $P(-5, -1)$. Let its slope be $m$. The equation of $L_3$ is $y - (-1) = m(x - (-5))$, which simplifies to $y + 1 = m(x + 5)$, or $mx - y + 5m - 1 = 0$.

First, find the vertices of the triangle.

1. Vertex A (Intersection of $L_1$ and $L_2$): Solving $x - 3y + 2 = 0$ and $x + y - 2 = 0$: Subtracting the second equation from the first: $(x - 3y + 2) - (x + y - 2) = 0 \Rightarrow -4y + 4 = 0 \Rightarrow y = 1$. Substituting $y=1$ into $x+y-2=0$: $x+1-2=0 \Rightarrow x=1$. So, vertex $A$ is $(1, 1)$.

2. Vertex B (Intersection of $L_1$ and $L_3$): $x - 3y + 2 = 0$ and $mx - y + 5m - 1 = 0$. From $L_3$, $y = mx + 5m - 1$. Substitute into $L_1$: $x - 3(mx + 5m - 1) + 2 = 0$ $x - 3mx - 15m + 3 + 2 = 0$ $x(1 - 3m) = 15m - 5 \Rightarrow x_B = \frac{15m - 5}{1 - 3m}$. $y_B = m\left(\frac{15m - 5}{1 - 3m}\right) + 5m - 1 = \frac{15m^2 - 5m + 5m(1-3m) - (1-3m)}{1 - 3m} = \frac{15m^2 - 5m + 5m - 15m^2 - 1 + 3m}{1 - 3m} = \frac{3m - 1}{1 - 3m} = -1$. So, vertex $B$ is $\left(\frac{15m - 5}{1 - 3m}, -1\right)$. (This is valid if $m \neq 1/3$)

3. Vertex C (Intersection of $L_2$ and $L_3$): $x + y - 2 = 0$ and $mx - y + 5m - 1 = 0$. From $L_3$, $y = mx + 5m - 1$. Substitute into $L_2$: $x + (mx + 5m - 1) - 2 = 0$ $x(1 + m) = 3 - 5m \Rightarrow x_C = \frac{3 - 5m}{1 + m}$. $y_C = m\left(\frac{3 - 5m}{1 + m}\right) + 5m - 1 = \frac{3m - 5m^2 + 5m(1+m) - (1+m)}{1 + m} = \frac{3m - 5m^2 + 5m + 5m^2 - 1 - m}{1 + m} = \frac{7m - 1}{1 + m}$. So, vertex $C$ is $\left(\frac{3 - 5m}{1 + m}, \frac{7m - 1}{1 + m}\right)$. (This is valid if $m \neq -1$)

For the origin $O(0,0)$ to be an interior point of the triangle $ABC$, it must lie on the same side of each line as the opposite vertex.

1. Origin and Vertex A must be on the same side of $L_3$: $L_3(x,y) = mx - y + 5m - 1$. $L_3(0,0) = 5m - 1$. $L_3(A) = L_3(1,1) = m(1) - 1 + 5m - 1 = 6m - 2$. For them to be on the same side, $L_3(0,0) \cdot L_3(A) > 0$: $(5m - 1)(6m - 2) > 0$ $2(5m - 1)(3m - 1) > 0$ $(5m - 1)(3m - 1) > 0$. This inequality holds when $m < 1/5$ or $m > 1/3$.

2. Origin and Vertex B must be on the same side of $L_2$: $L_2(x,y) = x + y - 2$. $L_2(0,0) = 0 + 0 - 2 = -2$. $L_2(B) = x_B + y_B - 2 = \frac{15m - 5}{1 - 3m} + (-1) - 2 = \frac{15m - 5}{1 - 3m} - 3 = \frac{15m - 5 - 3(1 - 3m)}{1 - 3m} = \frac{15m - 5 - 3 + 9m}{1 - 3m} = \frac{24m - 8}{1 - 3m}$. For them to be on the same side, $L_2(0,0) \cdot L_2(B) > 0$: $(-2) \cdot \frac{24m - 8}{1 - 3m} > 0$ $\frac{24m - 8}{1 - 3m} < 0$ $\frac{8(3m - 1)}{1 - 3m} < 0$. This simplifies to $\frac{-8(1 - 3m)}{1 - 3m} < 0$, which means $-8 < 0$. This is always true, provided $1 - 3m \neq 0$, i.e., $m \neq 1/3$.

3. Origin and Vertex C must be on the same side of $L_1$: $L_1(x,y) = x - 3y + 2$. $L_1(0,0) = 0 - 3(0) + 2 = 2$. $L_1(C) = x_C - 3y_C + 2 = \frac{3 - 5m}{1 + m} - 3\left(\frac{7m - 1}{1 + m}\right) + 2 = \frac{3 - 5m - 21m + 3 + 2(1 + m)}{1 + m} = \frac{6 - 26m + 2 + 2m}{1 + m} = \frac{8 - 24m}{1 + m}$. For them to be on the same side, $L_1(0,0) \cdot L_1(C) > 0$: $2 \cdot \frac{8 - 24m}{1 + m} > 0$ $\frac{8 - 24m}{1 + m} > 0$ $\frac{8(1 - 3m)}{1 + m} > 0$. This inequality holds when $(1 - 3m)(1 + m) > 0$. The roots are $m = 1/3$ and $m = -1$. This inequality holds when $-1 < m < 1/3$.

Combining all conditions for $m$:

1. $m < 1/5$ or $m > 1/3$
2. $m \neq 1/3$
3. $-1 < m < 1/3$

Let's find the intersection of these conditions: The interval $(-1, 1/3)$ from condition 3. Intersect this with $(-\infty, 1/5) \cup (1/3, \infty)$ from condition 1. The intersection is $(-1, 1/5)$. Condition 2 ($m \neq 1/3$) is already satisfied by this interval. Also, the conditions for vertices B and C to be well-defined ($m \neq 1/3$ and $m \neq -1$) are satisfied by this interval.

So, the possible values of the slope $m$ for which the origin is an interior point of the triangle are $m \in (-1, 1/5)$.

The question asks for the "not possible values of slope of third side such that origin is an interior point of triangle". This means we need to find the complement of the interval $(-1, 1/5)$ on the real number line. The complement is $(-\infty, -1] \cup [1/5, \infty)$.

The final answer is $\boxed{(-\infty, -1] \cup [1/5, \infty)}$.

Explanation of the solution:

1. Identify the equations of the two given sides ($L_1, L_2$) and the general equation of the third side ($L_3$) with slope $m$.
2. Calculate the coordinates of the three vertices of the triangle ($A, B, C$) by finding the intersection points of these lines.
3. For the origin $(0,0)$ to be an interior point, it must lie on the same side of each line as the vertex opposite to that line. This is checked using the sign of the line equation evaluated at the origin and the opposite vertex.
   • Origin and vertex $A$ must be on the same side of $L_3$.
   • Origin and vertex $B$ must be on the same side of $L_2$.
   • Origin and vertex $C$ must be on the same side of $L_1$.
4. Set up inequalities based on these conditions.
5. Solve the inequalities for $m$.
6. Find the intersection of all valid ranges of $m$. This gives the possible values of $m$.
7. The question asks for "not possible values", so find the complement of the derived range.

Answer:

The not possible values of slope of third side such that origin is an interior point of triangle is $(-\infty, -1] \cup [1/5, \infty)$.

The final answer is $\boxed{(-\infty, -1] \cup [1/5, \infty)}$.