Question 104·Hard·Algebra

Compute the slope of $\overline{RS}$ using $R$ and $S$. A line perpendicular to it has slope equal to the negative reciprocal.

Once you know the intersection point $(x,y)$ and the slope $m$, use $b=y-mx$.

Enter the two lines determined by the graph points:

$y=-x+5$

$y=\frac{2}{3}x-\frac{7}{3}$

Click the intersection point of the two lines and record its coordinates $(x,y)$.

Compute the slope of $\overline{RS}$ from $R(-4,-5)$ and $S(2,-1)$ as $\frac{2}{3}$, then take the negative reciprocal to get $m=-\frac{3}{2}$.

Use the recorded intersection point and evaluate $b=y-mx$ in Desmos. The resulting value matches one answer choice.

From the graph, line $r$ passes through $P(-1,6)$ and $Q(5,0)$, so its slope is

and using $Q(5,0)$ gives $0=-5+b_r$, so $r$ is

Line $s$ passes through $R(-4,-5)$ and $S(2,-1)$, so its slope is

and using $S(2,-1)$ gives $-1=\frac{4}{3}+b_s$, so $s$ is

Set the expressions for $y$ equal:

Multiply by $3$:

So $22=5x$, hence

Substitute into $y=-x+5$:

The intersection point is $\left(\frac{22}{5},\frac{3}{5}\right)$.

The slope of $\overline{RS}$ is the slope of the line through $R(-4,-5)$ and $S(2,-1)$:

A line perpendicular to this has slope

Line $t$ passes through $\left(\frac{22}{5},\frac{3}{5}\right)$ and has slope $m=-\frac{3}{2}$. Substitute into $y=mx+b$:

Compute $\left(-\frac{3}{2}\right)\left(\frac{22}{5}\right)=-\frac{33}{5}$, so

Therefore, the value of $b$ is $\frac{36}{5}$.