Question

1. q(-11,5), r(-8,4), s(8,-7), t(-1, -4)

parallel perpendicular neither

1. q(-3,8), r(1,15), s(10,-9), t(3,-5)

parallel perpendicular neither

Explanation:

Step1: Calculate slope of QR

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
For $Q(-11,5), R(-8,4)$:
$m_{QR}=\frac{4-5}{-8-(-11)}=\frac{-1}{3}=-\frac{1}{3}$

Step2: Calculate slope of ST

For $S(8,-7), T(-1,-4)$:
$m_{ST}=\frac{-4-(-7)}{-1-8}=\frac{3}{-9}=-\frac{1}{3}$

Step3: Compare slopes (Q13)

$m_{QR}=m_{ST}$, so lines are parallel.

Step4: Calculate slope of QR (Q14)

For $Q(-3,8), R(1,15)$:
$m_{QR}=\frac{15-8}{1-(-3)}=\frac{7}{4}$

Step5: Calculate slope of ST (Q14)

For $S(10,-9), T(3,-5)$:
$m_{ST}=\frac{-5-(-9)}{3-10}=\frac{4}{-7}=-\frac{4}{7}$

Step6: Compare slopes (Q14)

Slopes are not equal, and $m_{QR} \times m_{ST}=\frac{7}{4} \times -\frac{4}{7}=-1$ is not true? Correction: $\frac{7}{4} \times -\frac{4}{7}=-1$? Wait no, $\frac{7}{4} \times -\frac{4}{7}=-1$, but wait no, $\frac{7}{4} \times -\frac{4}{7}=-1$, but that would be perpendicular? Wait no, recalculate:
$m_{QR}=\frac{15-8}{1-(-3)}=\frac{7}{4}$
$m_{ST}=\frac{-5 - (-9)}{3-10}=\frac{4}{-7}=-\frac{4}{7}$
$\frac{7}{4} \times -\frac{4}{7}=-1$, but wait that is perpendicular? Wait no, I made a mistake. Wait no:
Wait $15-8=7$, $1 - (-3)=4$, correct. $-5 - (-9)=4$, $3-10=-7$, correct. $\frac{7}{4} \times -\frac{4}{7}=-1$, so they are perpendicular? Wait no, wait the problem says Q14: Q(-3,8), R(1,15), S(10,-9), T(3,-5). Wait no, slope of ST is $\frac{-9 - (-5)}{10-3}=\frac{-4}{7}=-\frac{4}{7}$? Oh! I mixed up the points. Slope formula is $\frac{y_T - y_S}{x_T - x_S}$, which is $\frac{-5 - (-9)}{3-10}=\frac{4}{-7}=-\frac{4}{7}$, or $\frac{y_S - y_T}{x_S - x_T}=\frac{-9 - (-5)}{10-3}=\frac{-4}{7}=-\frac{4}{7}$. That is correct. Then $\frac{7}{4} \times -\frac{4}{7}=-1$, which means they are perpendicular? Wait no, wait 7/4 times -4/7 is -1, which is the condition for perpendicular lines. Wait I made a mistake earlier. Wait no, let's recheck:
Wait Q(-3,8) to R(1,15): rise is 15-8=7, run is 1 - (-3)=4, slope 7/4.
S(10,-9) to T(3,-5): rise is -5 - (-9)=4, run is 3-10=-7, slope 4/-7=-4/7.
7/4 * (-4/7) = -1, so they are perpendicular? Wait no, that's the definition of perpendicular lines (product of slopes is -1). Wait but I thought it was neither, that was wrong. Wait no, let's recalculate:
Wait 7/4 multiplied by -4/7: the 7 cancels, 4 cancels, gives -1. Yes, that's correct. So Q14 is Perpendicular? Wait no, wait maybe I mixed up the points. Wait the problem says Q14: Q(-3,8), R(1,15), S(10,-9), T(3,-5). So lines QR and ST.
Slope QR: (15-8)/(1 - (-3))=7/4.
Slope ST: (-5 - (-9))/(3-10)=4/-7=-4/7.
7/4 * (-4/7) = -1, so they are perpendicular. Oh I made a mistake earlier.
Wait no, wait the user's problem says "Parallel Perpendicular Neither" for each. So correcting:

Step1: Slope of QR (Q13)

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
$m_{QR}=\frac{4-5}{-8-(-11)}=-\frac{1}{3}$

Step2: Slope of ST (Q13)

$m_{ST}=\frac{-4-(-7)}{-1-8}=-\frac{1}{3}$

Step3: Classify Q13

Equal slopes → Parallel.

Step4: Slope of QR (Q14)

$m_{QR}=\frac{15-8}{1-(-3)}=\frac{7}{4}$

Step5: Slope of ST (Q14)

$m_{ST}=\frac{-5-(-9)}{3-10}=-\frac{4}{7}$

Step6: Classify Q14

Product of slopes: $\frac{7}{4} \times -\frac{4}{7}=-1$ → Perpendicular.

Answer:

1. Parallel
2. Neither