Question 119·Medium·Linear Functions

Perpendicular lines have slopes that are negative reciprocals: if one slope is $m$, the other is $-\tfrac{1}{m}$. Apply this to the slope you just found.

Write the slope between $(-3,4)$ and $(5,k)$ using $\dfrac{y_2-y_1}{x_2-x_1}$, then set it equal to the perpendicular slope and solve for $k$.

Type y = 1/2 x + 7 to see the given line. From the equation, note the slope is $\tfrac{1}{2}$, so the perpendicular slope should be the negative reciprocal, $-2$.

Enter the equation y - 4 = -2(x + 3) or equivalently y = -2(x + 3) + 4. This is the line with slope $-2$ passing through $(-3,4)$, matching the conditions for line $\ell$.

Add a new expression: type y = -2(5 + 3) + 4 or substitute $x=5$ into the line’s equation in Desmos. The resulting value of $y$ is the value of $k$ for the point $(5,k)$ on line $\ell$.

The equation of the given line is $y=\tfrac{1}{2}x+7$.

In slope-intercept form $y=mx+b$, the coefficient of $x$ is the slope $m$.

So the slope of this line is $m=\tfrac{1}{2}$.

For two non-vertical lines to be perpendicular, their slopes must be negative reciprocals of each other. That means if one slope is $m$, the other must be $-\tfrac{1}{m}$.

Here the given slope is $\tfrac{1}{2}$, so the slope of line $\ell$ must be

So line $\ell$ has slope $-2$.

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is

For line $\ell$, the two points are $(-3,4)$ and $(5,k)$.

Let $(-3,4)$ be $(x_1,y_1)$ and $(5,k)$ be $(x_2,y_2)$:

• $x_1=-3$, $y_1=4$
• $x_2=5$, $y_2=k$

Then the slope of line $\ell$ is

Since this line is perpendicular to the given line, this slope must equal $-2$.

Set the slope from the two points equal to the perpendicular slope:

Multiply both sides by $8$:

Add $4$ to both sides:

So the value of $k$ is $-12$, which corresponds to choice B.