Question

Let \((x_0,y_0)\in\mathbb Z^2\) be a point on straight line \(8x-3y=11\) which is equidistant from coordinate axes. Then point \((x_0,y_0)\) will lie only in:

• A. II quadrant
• B. III quadrant
• C. I quadrant
• D. IV quadrant

Hint:

A point equidistant from both coordinate axes always satisfies \( |x|=|y| \).

Answer 1

The Correct Option is D

Step 1: Translate equidistant from the axes.
The distance from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$. Equal distances mean $|x|=|y|$, so $y=x$ or $y=-x$.
Step 2: Bring in the line.
The point must also satisfy $8x-3y=11$, and we need integer coordinates. We test each case.
Step 3: Try $y=x$.
Then $8x-3x=11$ gives $5x=11$, so $x=\frac{11}{5}$, not an integer. This case fails.
Step 4: Try $y=-x$.
Then $8x+3x=11$ gives $11x=11$, so $x=1$ and $y=-1$. These are integers, so this works.
Step 5: Locate the point.
The point is $(1,-1)$: positive $x$ and negative $y$.
Step 6: Name the quadrant.
A point with $x>0$ and $y<0$ lies in the fourth quadrant.
\[ \boxed{\text{IV quadrant}} \]