How are the slopes of perpendicular lines related? If one line has slope $-\frac12$, what number must you use for the other slope so that their product is $-1$?

Once you know the slope of line $p$, substitute that slope and the coordinates $(4,-3)$ into $y=mx+b$ and solve the resulting equation for $b$.

In Desmos, type y = -1/2 x + 6 to see the given line and confirm its slope is $-\tfrac12$.

Type y = 2x + b. Desmos will create a slider for $b$. This line has slope $2$, the negative reciprocal of $-\tfrac12$, so it is perpendicular to the original line.

Add the point (4,-3) in Desmos. Adjust the slider for $b$ until the line y = 2x + b passes exactly through the point (4,-3). The value of $b$ at that moment is the $y$-intercept you need.

In a new Desmos line, type -3 = 2*4 + b and let Desmos solve for $b$. The solution it displays for $b$ is the correct $y$-intercept.

The given line is in slope-intercept form $y=mx+b$, where $m$ is the slope.

For the line $y=-\frac12 x+6$, the slope is $m=-\frac12$.

Slopes of perpendicular (non-vertical) lines are negative reciprocals. This means if one slope is $m$, the other is $-\dfrac{1}{m}$.

Here, the original slope is $-\frac12$, so the perpendicular slope $m_p$ is:

So line $p$ has slope $2$.

Line $p$ has slope $2$ and passes through $(4,-3)$. Use the slope-intercept form $y=mx+b$ with $m=2$ and the point $(x,y)=(4,-3)$:

Now solve this equation for $b$.

From the equation

subtract $8$ from both sides:

So the $y$-intercept of line $p$ is $b=-11$.