$6$ Consider the points $\mathrm{P}(4,7), \mathrm{Q}(8,4), \mathrm{R}(7,0)$ , and $\mathrm{S}(-1, t)$ . Find $t$ given that $\mathrm{PQ} \| \mathrm{SR}$ .

Full solution

6

Consider the points

\mathrm{P}(4,7), \mathrm{Q}(8,4), \mathrm{R}(7,0)

, and

\mathrm{S}(-1, t)

. Find

t

given that

\mathrm{PQ} \| \mathrm{SR}

Calculate Slope of PQ: Given points $P(4,7)$ , $Q(8,4)$ , $R(7,0)$ , and $S(-1,t)$ , we need to find the value of $t$ such that $PQ$ is parallel to $SR$ . For two lines to be parallel, their slopes must be equal. Let's first calculate the slope of $PQ$ . The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $Q(8,4)$ $0$ . Slope of $PQ$ = $Q(8,4)$ $2$ Slope of $PQ$ = $Q(8,4)$ $4$ Slope of $PQ$ = $Q(8,4)$ $6$
Calculate Slope of SR: Now, let's calculate the slope of SR using the coordinates of R $(7,0)$ and S $(-1,t)$ . $\text{Slope of SR} = \frac{t - 0}{-1 - 7}$ $\text{Slope of SR} = \frac{t}{-8}$ $\text{Slope of SR} = -\frac{t}{8}$
Set Equal Slopes: For $PQ$ to be parallel to $SR$ , their slopes must be equal. Therefore, we set the slope of $PQ$ equal to the slope of $SR$ . $\newline$ $-\frac{3}{4} = -\frac{t}{8}$
Solve for t: Now, we solve for t by cross-multiplying. $\newline$ $-3 \times 8 = -t \times 4$ $\newline$ $-24 = -4t$
Solve for t: Now, we solve for t by cross-multiplying. $\newline$ $-3 \times 8 = -t \times 4$ $\newline$ $-24 = -4t$ Divide both sides by $-4$ to isolate t. $\newline$ $t = (-24) / (-4)$ $\newline$ $t = 6$