Question 343·Hard·Algebra

For a line $Ax+By=C$, the slope is $-A/B$ when $B\ne0$. If $4k+3=0$, the given line is vertical, so $\ell$ would have to be horizontal. Its points have different $y$-coordinates, so that cannot work. If $k-2=0$, the given line is horizontal, so $\ell$ would have to be vertical. Its points would have $x$-coordinates $3$ and $2$, so that cannot work either. For all remaining values, the given line has slope $-\frac{k-2}{4k+3}$, so flip the fraction and change its sign to get the slope $\ell$ needs: $m_{\ell}=\frac{4k+3}{k-2}$.

From $(3,-2)$ to $(k,6)$, the rise is $6-(-2)=8$, and the run is $k-3$. If $k=3$, $\ell$ is vertical, but the given line is not horizontal because $k-2=1$, so that cannot work. So the slope of $\ell$ is $m_{\ell}=\frac{8}{k-3}$. Keep the coordinate subtractions in the same order.

The slope from the points must equal the perpendicular slope from the equation. Both expressions describe $\ell$, so $\frac{8}{k-3}=\frac{4k+3}{k-2}$.

Cross-multiply to remove the denominators: $8(k-2)=(k-3)(4k+3)$. This gives an equation in one unknown, $k$.

Distribute each product, keeping the negative signs on the right: $8k-16=4k^2-9k-9$.

Move the left side to the right and combine like terms: $0=4k^2-17k+7$. The $k^2$ term makes this a quadratic, which can have two solutions.

Type $y=4x^2-17x+7$ and click its two $x$-intercepts, where $y=0$. Desmos shows about $(0.462,0)$ and $(3.788,0)$. The farther-right intercept gives the larger $k$, but its decimal does not yet identify the exact choice.

The choices with a minus sign are smaller than $\frac{17}{8}$, so compare the plus-sign choices. Type both expressions below. Desmos shows about $3.788$ for $\frac{17+\sqrt{177}}{8}$ and $3.454$ for $\frac{17+\sqrt{113}}{8}$. The first matches the larger intercept, so the larger possible value of $k$ is $\frac{17+\sqrt{177}}{8}$. Choice A.

Starting from $4k^2-17k+7=0$, the quadratic formula uses $a=4$, $b=-17$, and $c=7$. Keep the negative coefficient in parentheses: $k=\frac{-(-17)\pm\sqrt{(-17)^2-4(4)(7)}}{2(4)}$.

Evaluate the expression under the square root, called the discriminant: type $(-17)^2-4(4)(7)$ in Desmos. It shows $177$: $(-17)^2-4(4)(7)=289-112=177$.

So $k=\frac{17\pm\sqrt{177}}{8}$. Because $\sqrt{177}$ is positive, adding it gives the larger value, $\frac{17+\sqrt{177}}{8}$. Choice A.