Question

1. write an ordered pair for the point in quadrant iv that is 4 units away from point s(-1.5, -2.5) on the coordinate plane.

Explanation:

Step1: Recall Quadrant IV properties

In Quadrant IV, \(x>0\) and \(y<0\). Point \(S(-1.5, -2.5)\) is in Quadrant III (\(x<0,y<0\)). To move to Quadrant IV from \(S\), we need to increase the \(x\)-coordinate (since \(x\) in IV is positive) and can adjust \(y\) (but \(y\) remains negative). The distance is 4 units. Let's consider horizontal/vertical movement. Let the new point be \((x,y)\). The distance formula is \(d = \sqrt{(x - x_S)^2+(y - y_S)^2}\), but for simplicity, if we move horizontally (since moving right (positive \(x\)) from \(S\) to IV: \(x_S=-1.5\), so to get \(x>0\), we can move \(4\) units in \(x\)-direction? Wait, no, distance is 4. Let's think: from \(S(-1.5, -2.5)\), to get to Quadrant IV, \(x\) must be positive. Let's move horizontally (change \(x\), keep \(y\) same? Wait, \(y\) of \(S\) is \(-2.5\), so \(y\) in IV should be negative, so maybe keep \(y=-2.5\) and move \(x\) so that distance is 4. So \(|x - (-1.5)|=4\) (horizontal distance), since vertical distance is 0. So \(x + 1.5 = 4\) (since we want \(x>0\), so positive direction), so \(x = 4 - 1.5 = 2.5\). Then \(y=-2.5\) (same \(y\) to move horizontally). So the point is \((2.5, -2.5)\). Let's check distance: \(|2.5 - (-1.5)|=|4| = 4\), which matches. And \((2.5, -2.5)\) is in Quadrant IV (\(x>0,y<0\)).

Step2: Verify the point

Check quadrant: \(x = 2.5>0\), \(y=-2.5<0\) → Quadrant IV. Distance from \(S(-1.5, -2.5)\): \(\sqrt{(2.5 - (-1.5))^2+(-2.5 - (-2.5))^2}=\sqrt{(4)^2+0^2}=4\), which is correct.

Answer:

\((2.5, -2.5)\)