In the xy-plane, line L_1 is defined by the equation 2x - 3y = 6. Line L_2 is perpendicula… — GRE General Test

In the $xy$ -plane, line $L_1$ is defined by the equation $2x - 3y = 6$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $P(4, -1)$ . Point $Q(a, b)$ lies on line $L_2$ such that the distance between point $P$ and point $Q$ is $\sqrt{13}$ . Which of the following statements regarding point $Q$ or line $L_2$ could be true? Select all such statements.

Point $Q$ lies in Quadrant I.Answer
The sum of the coordinates of point $Q$ , $a + b$ , is equal to $4$ .Answer
The distance from point $Q$ to the origin is $2\sqrt{13}$ .Answer
D
The $y$ -intercept of line $L_2$ is $(0, -6)$ .
E
Point $Q$ could have coordinates $(4, 2)$ .

Answer

The correct statements are that point Q can lie in Quadrant I, the sum of the coordinates of point Q can equal 4, and the distance from point Q to the origin can be 2√13.

Solving the system formed by line

L_2

(

3x + 2y = 10

) and the distance constraint

(a-4)^2 + (b+1)^2 = 13

yields two possible points:

(2, 2)

and

(6, -4)

. The point

(2, 2)

lies in Quadrant I and has a coordinate sum of

2 + 2 = 4

. The point

(6, -4)

has a distance to the origin of

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

Step-by-Step Solution

Determine the slope and equation of line

L_2

Line

L_1

has slope

m_1 = \frac{2}{3}

. Therefore, perpendicular line

L_2

has slope

m_2 = -\frac{3}{2}

. Using point

P(4, -1)

, the equation of

L_2

y - (-1) = -\frac{3}{2}(x - 4)

, which simplifies to

y = -\frac{3}{2}x + 5

3x + 2y = 10

Perpendicular lines in the coordinate plane have negative reciprocal slopes.

Express the distance constraint between

P(4, -1)

and

Q(a, b)

algebraically.

Since

Q(a, b)

lies on

L_2

b = -\frac{3}{2}a + 5

. The distance squared is

(a - 4)^2 + (b + 1)^2 = 13

. Substituting

b + 1 = -\frac{3}{2}(a - 4)

yields

(a - 4)^2 + \left(-\frac{3}{2}(a - 4)\right)^2 = 13

, which simplifies to

\frac{13}{4}(a - 4)^2 = 13

, so

(a - 4)^2 = 4

Applying the distance formula and substituting the line equation reduces the problem to a quadratic equation in one variable.

Solve for the possible coordinates of point

Q

Taking square roots gives

a - 4 = 2

a - 4 = -2

, resulting in

a = 6

a = 2

. Correspondingly,

b = -4

b = 2

. Thus,

Q

can be

(6, -4)

(2, 2)

Quadratic equations of the form

(x-h)^2 = k

have two real solutions.

Evaluate each statement against the possible coordinates

Q(6, -4)

and

Q(2, 2)

For

Q(2, 2)

: it lies in Quadrant I (valid), its coordinate sum is

2 + 2 = 4

(valid). For

Q(6, -4)

: its distance to the origin is

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

(valid). The

y

-intercept of

L_2

(0, 5)

, and

(4, 2)

is not on

L_2

Direct verification confirms which properties hold for the two solved points.

Key Concept

Perpendicular Slopes and Distance Formula in Coordinate Geometry

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic