The lines with equations and are perpendicular in the -plane, where is a nonzero constant. What is the value of ?

Solution available with step-by-step explanation

SAT Linear Equations in Two Variables Question 99 | Free SAT Prep | Aniko

00:00

Question 99·Hard·Linear Equations in Two Variables

0 / 140

The lines with equations

(3k - 2)x + 4y = 6

and

2x + (k + 5)y = 9

are perpendicular in the $xy$ -plane, where $k$ is a nonzero constant.

What is the value of $k$ ?

For perpendicular-line questions with parameters, first recall that in standard form $Ax + By = C$ , the slope is $-A/B$ . Quickly write each slope in terms of the parameter (here, $k$ ), then use the fact that the product of the slopes of perpendicular non-vertical lines is $-1$ to set up a single equation. From there, focus on careful algebra: distribute correctly, combine like terms, and isolate the parameter without flipping fractions or dropping negative signs. This approach is faster and less error-prone than fully converting both equations to slope-intercept form.

Hints

Click me!

Think about slopes

The question is about perpendicular lines. What relationship do the slopes of two perpendicular non-vertical lines have?

Get slopes from standard form

Each equation is written as $Ax + By = C$ . How can you find the slope of a line directly from this form without fully solving for $y$ ?

Write and use the slope equation

Express the slope of each line in terms of $k$ , then set their product equal to $-1$ and solve that equation.

Algebra check

When you solve the equation in $k$ , be careful with distributing, combining like terms, and handling negative signs so you do not mix up the fraction or its sign.

Desmos Guide

Define the slope product as a function of k

In Desmos, treat $k$ as the variable $x$ . Enter the equation for the product of slopes:

Type: y = ((2 - 3x)/4)*(-2/(x + 5))

Graph the perpendicularity condition

Enter the horizontal line representing $-1$ (the required product for perpendicular lines):

Type: y = -1

Find the intersection to get k

Use the intersection tool (tap the point where the two graphs meet) and read the x-coordinate of the intersection. That x-value is the value of $k$ that makes the lines perpendicular.

Step-by-step Explanation

Find a formula for slope from standard form

Both equations are in standard form $Ax + By = C$ .

For any line in this form, the slope $m$ is

m = -\frac{A}{B}

where $A$ is the coefficient of $x$ and $B$ is the coefficient of $y$ .

Write the slope of each line in terms of k

First line: $(3k - 2)x + 4y = 6$

Here $A = 3k - 2$ and $B = 4$ , so the slope is

m_1 = -\frac{3k - 2}{4} = \frac{2 - 3k}{4}.

Second line: $2x + (k + 5)y = 9$

Here $A = 2$ and $B = k + 5$ , so the slope is

m_2 = -\frac{2}{k + 5}.

Use the perpendicular slope relationship

For two non-vertical perpendicular lines, the product of their slopes is $-1$ :

m_1 \cdot m_2 = -1.

Substitute the expressions you found:

\frac{2 - 3k}{4} \cdot \left(-\frac{2}{k + 5}\right) = -1.

Simplify the left side:

\frac{2 - 3k}{4} \cdot \left(-\frac{2}{k + 5}\right) = \frac{(2 - 3k)(-2)}{4(k+5)} = -\frac{2(2 - 3k)}{4(k+5)} = -\frac{2 - 3k}{2(k+5)}.

So the equation becomes

-\frac{2 - 3k}{2(k+5)} = -1.

Solve the equation for k

Remove the negative signs on both sides:

\frac{2 - 3k}{2(k+5)} = 1.

Now cross-multiply:

2 - 3k = 2(k + 5).

Distribute on the right:

2 - 3k = 2k + 10.

Move all terms to one side:

2 - 3k - 2k - 10 = 0 \\[0.7em] -5k - 8 = 0.

Solve for $k$ :

-5k = 8 \quad \Rightarrow \quad k = -\frac{8}{5}.

So the correct answer is $k = -\dfrac{8}{5}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation