Line _1 has equation 3x - 2y = 1 and goes through A = (-1, -2). Line _2 has equation y = 1 and meets line _1 at point B. Line _3 has positive slope, goes through point A, and meets _2 at point C. The area of △ ABC is 3. What is the slope of _3?

2013 AMC 12B Problem 8

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Line

\ell_1

has equation

3x - 2y = 1

and goes through

A = (-1, -2).

Line

\ell_2

has equation

y = 1

and meets line

\ell_1

at point

B.

Line

\ell_3

has positive slope, goes through point

A,

and meets

\ell_2

at point

C.

The area of

\triangle ABC

3.

What is the slope of

\ell_3?

\dfrac{2}{3}

\dfrac{3}{4}

1

\dfrac{4}{3}

\dfrac{3}{2}

Find

B

by solving

3x - 2y = 1

together with

y = 1

The distance from

A

to line

\ell_2

3,

\tfrac12\cdot BC\cdot 3 = 3

gives

BC

Solving

3x - 2y = 1

with

y = 1

gives

B = (1, 1).

The distance from

A = (-1, -2)

to the line

y = 1

3,

\tfrac12\cdot BC\cdot 3 = 3

gives

BC = 2.

Then

C = (3, 1)

C = (-1, 1);

the latter makes

\ell_3

vertical, so

C = (3, 1)

and the slope is

\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34.

Thus, the correct answer is B.