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15. q(9, -3), r(9, 4), s(-2, 10), t(-2, 6)

Question

  1. q(9, -3), r(9, 4), s(-2, 10), t(-2, 6)

Explanation:

Step1: Calculate slope of QR

Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
$m_{QR} = \frac{4 - (-3)}{9 - 9} = \frac{7}{0}$ (undefined, vertical line)

Step2: Calculate slope of ST

$m_{ST} = \frac{6 - 10}{-2 - (-2)} = \frac{-4}{0}$ (undefined, vertical line)
Wait, correction: Wait, no—wait, vertical lines are parallel. Wait, no, let's recheck:
Wait, Q(9,-3), R(9,4): x-coordinates are same, so vertical line (undefined slope). S(-2,10), T(-2,6): x-coordinates are same, vertical line. But wait, the handwritten calculation was wrong. Wait, no, the question is to determine if QR and ST are parallel, perpendicular, or neither.
Wait, correct steps:

Step1: Find slope of $\overleftrightarrow{QR}$

$m_{QR} = \frac{y_R - y_Q}{x_R - x_Q} = \frac{4 - (-3)}{9 - 9} = \frac{7}{0}$ (undefined, vertical line)

Step2: Find slope of $\overleftrightarrow{ST}$

$m_{ST} = \frac{y_T - y_S}{x_T - x_S} = \frac{6 - 10}{-2 - (-2)} = \frac{-4}{0}$ (undefined, vertical line)

Step3: Classify the lines

Vertical lines are parallel to each other. Wait, but the handwritten work was incorrect. Wait, no—wait, the original problem's handwritten work had $\frac{4--3}{-2+9}$ which is wrong. Correctly, for QR, x's are 9 and 9, so denominator 0. For ST, x's are -2 and -2, denominator 0. So both are vertical, so parallel.

Wait, let's correct:

Step1: Calculate slope of $\overleftrightarrow{QR}$

Use slope formula: $m = \frac{y_2-y_1}{x_2-x_1}$
$m_{QR} = \frac{4 - (-3)}{9 - 9} = \frac{7}{0}$ (undefined, vertical line)

Step2: Calculate slope of $\overleftrightarrow{ST}$

$m_{ST} = \frac{6 - 10}{-2 - (-2)} = \frac{-4}{0}$ (undefined, vertical line)

Step3: Compare line types

All vertical lines are parallel to one another.

Answer:

Neither