The system of linear equations below represents two distinct lines in the -plane, where is a constant. For what value of are the two lines perpendicular?

Solution available with step-by-step explanation

SAT Linear Equations in Two Variables Question 21 | Free SAT Prep | Aniko

00:00

Question 21·Hard·Linear Equations in Two Variables

0 / 140

The system of linear equations below represents two distinct lines in the $xy$ -plane, where $k$ is a constant.

\begin{aligned} (k-3)x + 2y &= 7 \\ 4x - (k+1)y &= 5 \end{aligned}

For what value of $k$ are the two lines perpendicular?

For systems that depend on a parameter (like k) and ask when two lines are perpendicular or parallel, first isolate y in each equation to get the slopes in terms of the parameter. Then apply the correct slope condition: equal slopes for parallel lines, product of slopes equal to -1 for perpendicular lines. Solve the resulting simple equation for the parameter, and quickly check against the answer choices to avoid arithmetic mistakes.

Hints

Click me!

Focus on slopes

Perpendicular lines in the coordinate plane have a special relationship between their slopes. First, express both equations in the form $y=mx+b$ so you can clearly see each slope.

Write the slopes in terms of k

From each equation, solve for $y$ and identify the slope as the coefficient of $x$ . Keep the expressions in terms of $k$ rather than substituting answer choices right away.

Use the perpendicular condition

Once you have both slopes $m_1$ and $m_2$ , recall the condition that makes lines perpendicular and turn it into an equation involving $k$ . Then solve that equation.

Desmos Guide

Enter the general slope expressions

In Desmos, type (3-k)/2 on one line and 4/(k+1) on another line. Then add a slider for k so you can change its value.

Use the perpendicular condition numerically

On a new line, type ( (3-k)/2 ) * ( 4/(k+1) ) to represent the product of the slopes. Move the k slider or manually enter each answer choice for k and watch the value of this product.

Identify the correct k from the product

The value of k that makes the product of the slopes exactly -1 corresponds to the choice where the two lines are perpendicular.

Step-by-step Explanation

Rewrite each equation in slope-intercept form

Put each equation in the form $y=mx+b$ so you can see the slopes.

From $(k-3)x + 2y = 7$ :

Subtract $(k-3)x$ from both sides:
$2y = - (k-3)x + 7 = (3-k)x + 7$
Divide by $2$ :
$y = \dfrac{3-k}{2}x + \dfrac{7}{2}$

So the slope of the first line is
$m_1 = \dfrac{3-k}{2}$ .

From $4x - (k+1)y = 5$ :

Subtract $4x$ from both sides:
$-(k+1)y = 5 - 4x$
Multiply both sides by $-1$ :
$(k+1)y = 4x - 5$
Divide by $k+1$ (noting $k \neq -1$ so we can divide):
$y = \dfrac{4}{k+1}x - \dfrac{5}{k+1}$

So the slope of the second line is
$m_2 = \dfrac{4}{k+1}$ .

Use the perpendicular slopes condition

For two non-vertical lines to be perpendicular, the product of their slopes must be $-1$ .

So set

\frac{3-k}{2} \cdot \frac{4}{k+1} = -1.

Solve the equation for k

Simplify the left-hand side:

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

Set this equal to $-1$ and solve:

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

Multiply both sides by $k+1$ :

2(3-k) = -(k+1).

Distribute:

6 - 2k = -k - 1.

Add $2k$ to both sides and add $1$ to both sides:

6 + 1 = -k + 2k \\[0.7em] 7 = k.

So the value of $k$ that makes the two lines perpendicular is $\boxed{7}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation