Question

Find the integral value of a for which the point (-2, a) lies strictly in the interior of the triangle formed by the lines $y=x$, $y=-x$ and $2x+3y=6$

• A. -1
• B. 0
• C. 1
• D. -1, 0, 1

Answer 1

Answer: (D)

-1, 0, 1

Answer 2

The triangle is formed by the lines: $y = x$, $y = -x$, $2x+3y = 6$.

Step 1: Find the vertices of the triangle.

1. Intersection of $y = x$ and $2x+3y=6$: $2x+3x=6 \implies 5x=6 \implies x=\frac{6}{5}$, $y=\frac{6}{5}$.

2. Intersection of $y = -x$ and $2x+3y=6$: $2x+3(-x)=6 \implies -x=6 \implies x=-6$, $y=6$.

3. Intersection of $y = x$ and $y = -x$: $x=-x \implies x=0$, $y=0$.

Thus, the vertices are $(\frac{6}{5},\frac{6}{5})$, $(-6,6)$, and $(0,0)$.

Step 2: Determine the conditions for a point $(-2,a)$ to lie inside the triangle.

(a) Condition from lines $y=x$ and $y=-x$:

At $x=-2$:

• On $y=x$: $y=-2$.
• On $y=-x$: $y=2$.

For the point $(-2,a)$ to lie between these two lines: $-2 < a < 2$.

(b) Condition from line $2x+3y=6$:

Since the triangle lies on the side where $2x+3y <6$ (this can be verified by testing the centroid of the triangle), substitute $x=-2$: $2(-2)+3a<6 \implies -4+3a<6 \implies 3a<10 \implies a<\frac{10}{3}$.

At $x=-2$, $\frac{10}{3} \approx 3.33$ which is less restrictive than $a<2$.

Step 3: Find the integral values satisfying the conditions.

From (a): $-2 < a < 2$.

The integers satisfying this inequality are: $a = -1, 0, 1$.

Thus, the integral values of $a$ for which the point $(-2,a)$ lies strictly in the interior of the triangle are $-1, 0, 1$.