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:
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:
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A point equidistant from both coordinate axes always satisfies \( |x|=|y| \).
Updated On: Jun 11, 2026
- II quadrant
- III quadrant
- I quadrant
- IV quadrant
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The Correct Option is D
Solution and Explanation
Step 1: Use condition of equal distance from axes.
Distance from \(x\)-axis is \[ |y|. \] Distance from \(y\)-axis is \[ |x|. \] Hence \[ |x|=|y|. \] Therefore, \[ y=x \] or \[ y=-x. \]
Step 2: Substitute \(y=x\).
\[ 8x-3x=11 \] \[ 5x=11. \] No integer solution.
Step 3: Substitute \(y=-x\).
\[ 8x+3x=11 \] \[ 11x=11 \] \[ x=1. \] Thus \[ y=-1. \] Hence point is \[ (1,-1). \] This lies in \[ \boxed{\text{IV quadrant}}. \] Therefore answer is \[ \boxed{(D)}. \]
Distance from \(x\)-axis is \[ |y|. \] Distance from \(y\)-axis is \[ |x|. \] Hence \[ |x|=|y|. \] Therefore, \[ y=x \] or \[ y=-x. \]
Step 2: Substitute \(y=x\).
\[ 8x-3x=11 \] \[ 5x=11. \] No integer solution.
Step 3: Substitute \(y=-x\).
\[ 8x+3x=11 \] \[ 11x=11 \] \[ x=1. \] Thus \[ y=-1. \] Hence point is \[ (1,-1). \] This lies in \[ \boxed{\text{IV quadrant}}. \] Therefore answer is \[ \boxed{(D)}. \]
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