Point $P$ lies on line $m$ and is 3 units from the origin. How can you use the distance formula $\sqrt{x^2 + y^2}$ along with the equation of line $m$ to find possible coordinates for $P$?

Once you have the two possible points for $P$, remember that the slope of a line perpendicular to slope $-2$ is $\tfrac12$. For each candidate point, consider what the $x$-intercept of the line with slope $\tfrac12$ through that point would be, and use the condition that it must be negative.

After you get the equation for line $n$ in the form $y = mx + b$, set $y=0$ to find the $x$-intercept and solve for $x$.

In Desmos, type y = -2x + 6 to graph line $m$. Then type x^2 + y^2 = 9 to graph the circle of radius 3 centered at the origin. Click on the intersection points of the line and the circle to see their coordinates; these are the possible positions of point $P$.

For each intersection point $(x_1,y_1)$, enter a line with slope $0.5$ through that point, for example y - y1 = 0.5(x - x1) (replacing x1 and y1 with the actual coordinates). This gives you the possible graphs for line $n$.

For each candidate line $n$, look where it crosses the $x$-axis (where $y=0$). Use Desmos’s trace or just read the $x$-coordinate of that intersection. Choose the line whose $x$-intercept is negative, and use that $x$-value as your answer.

Line $m$ passes through $(0,6)$ and $(4,-2)$.

Compute its slope:

Because it passes through $(0,6)$, its $y$-intercept is $6$, so the equation of line $m$ is

Point $P$ is on line $m$ and is 3 units from the origin, so its coordinates $(x,y)$ must satisfy both

• the line equation $y = -2x + 6$, and
• the circle equation $x^2 + y^2 = 9$ (distance 3 from the origin).

Substitute $y = -2x + 6$ into $x^2 + y^2 = 9$:

Solve this quadratic:

so $x = 3$ or $x = \tfrac{9}{5}$.

Find the corresponding $y$-values using $y = -2x + 6$:

• If $x=3$, then $y = -2(3) + 6 = 0$, giving $P_1 = (3,0)$.
• If $x=\tfrac{9}{5}$, then $y = -2\cdot \tfrac{9}{5} + 6 = -\tfrac{18}{5} + \tfrac{30}{5} = \tfrac{12}{5}$, giving $P_2 = \bigl(\tfrac{9}{5}, \tfrac{12}{5}\bigr)$.

Line $n$ is perpendicular to line $m$, so its slope is the negative reciprocal of $-2$:

If line $n$ passed through $P_1 = (3,0)$, then $(3,0)$ would already be its $x$-intercept, which is positive, contradicting the condition that the $x$-intercept of line $n$ is negative.

Therefore, $P$ must be the other intersection point:

Line $n$ has slope $\tfrac{1}{2}$ and passes through $P\bigl(\tfrac{9}{5}, \tfrac{12}{5}\bigr)$.

Use slope-intercept form $y = \tfrac12 x + b$ and substitute the coordinates of $P$ to solve for $b$:

Subtract $\tfrac{9}{10}$ from both sides:

So the equation of line $n$ is

The $x$-intercept occurs where $y = 0$.

Set $y = 0$ in the equation of line $n$ and solve for $x$:

So the $x$-intercept of line $n$ is $-3$, which corresponds to choice B.