If one line has slope $m_1$ and another line is perpendicular to it with slope $m_2$, what equation relates $m_1$ and $m_2$? Use this to express the slope of line $\ell$ in terms of $k$.

Compute the slope of $\ell$ using the points $(3,-2)$ and $(k,6)$. Then set that equal to your perpendicular slope expression and solve for $k$.

After you form a quadratic equation in $k$, use the quadratic formula and pay close attention to the discriminant $b^2 - 4ac$ so you don’t make a sign or arithmetic error.

In Desmos, type the equation y = 4x^2 - 17x + 7. Here, the variable x in Desmos plays the role of $k$ in the problem.

Look at where the parabola crosses the x-axis (its x-intercepts). Click on each intercept to see its x-coordinate; these two x-values are the two possible values of $k$ that satisfy the perpendicular and point conditions.

Compare the two x-intercepts you found in Desmos and identify the larger x-value. That larger x-coordinate is the value of $k$ the problem is asking for.

The given line is

This is in the form $Ax + By = C$, where $A = k-2$ and $B = 4k+3$.

For a line $Ax + By = C$, the slope is $m = -\dfrac{A}{B}$, so the slope of this line is

Line $\ell$ is perpendicular to the given line.

Perpendicular lines have slopes that are negative reciprocals, so if the first line has slope $m_1$, then the perpendicular slope $m_2$ satisfies $m_1 m_2 = -1$.

Since $m_1 = -\dfrac{k-2}{4k+3}$, the slope of line $\ell$ is

Line $\ell$ passes through the points $(3,-2)$ and $(k,6)$.

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

So for $(3,-2)$ and $(k,6)$,

Both expressions describe the same slope $m_2$, so set them equal:

Cross-multiply:

Expand both sides:

• Left side: $8(k - 2) = 8k - 16$.
• Right side:

Set them equal and bring all terms to one side:

So $k$ must satisfy the quadratic equation

Use the quadratic formula for $4k^2 - 17k + 7 = 0$, where $a = 4$, $b = -17$, $c = 7$:

Compute the discriminant:

So

There are two possible values of $k$, and the larger one is