In the standard (x, y) coordinate plane, line L_1 has a negative slope and passes through … — ACT

In the standard $(x, y)$ coordinate plane, line $L_1$ has a negative slope and passes through the point $(1, 4)$ . Line $L_2$ has a positive slope and passes through the point $(7, 3)$ . Both lines intersect the $x$ -axis at the same point $P$ . If the product of the slopes of $L_1$ and $L_2$ is $-\frac{4}{3}$ , and line $L_3$ is perpendicular to $L_1$ and passes through the point $(2, -2)$ , what is the $y$ -intercept of $L_3$ ?

Answer: -3.5

Answer

The

y

-intercept of line

L_3

-3.5

By writing the slope of the first line as

\frac{-4}{x_p - 1}

and the slope of the second line as

\frac{-3}{x_p - 7}

, their product is set to

-\frac{4}{3}

. Solving the resulting quadratic equation

(x_p - 4)^2 = 0

yields

x_p = 4

. Substituting this back gives a slope of

-\frac{4}{3}

for the first line. The perpendicular line must have a slope of

\frac{3}{4}

. Using the point-slope formula with point

(2, -2)

and slope

\frac{3}{4}

gives the line

y = \frac{3}{4}x - 3.5

, which crosses the

y

-axis at

-3.5

Step-by-Step Solution

Express the slopes of

L_1

and

L_2

in terms of the unknown

x

-coordinate of point

P

m_1 = \frac{0 - 4}{x_p - 1} = \frac{-4}{x_p - 1}

and

m_2 = \frac{0 - 3}{x_p - 7} = \frac{-3}{x_p - 7}

Since

P

lies on the

x

-axis, its coordinates are

(x_p, 0)

. We apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to the points on each line.

Set up the equation for the product of the slopes.

\left(\frac{-4}{x_p - 1}\right)\left(\frac{-3}{x_p - 7}\right) = -\frac{4}{3} \implies \frac{12}{(x_p - 1)(x_p - 7)} = -\frac{4}{3}

The problem states that the product of the slopes of

L_1

and

L_2

-\frac{4}{3}

Solve the equation for

x_p

x_p = 4

Cross-multiplying gives

36 = -4(x_p^2 - 8x_p + 7)

, which simplifies to

-9 = x_p^2 - 8x_p + 7 \implies x_p^2 - 8x_p + 16 = 0 \implies (x_p - 4)^2 = 0

Find the slope of

L_1

m_1 = \frac{-4}{4 - 1} = -\frac{4}{3}

Substituting

x_p = 4

back into the expression for

m_1

yields the slope of

L_1

Find the slope of

L_3

m_3 = \frac{3}{4}

Since

L_3

is perpendicular to

L_1

, its slope is the negative reciprocal of

m_1

Write the linear equation for

L_3

and determine its

y

-intercept.

y = \frac{3}{4}x - 3.5

, so the

y

-intercept is

-3.5

Using the point-slope form with point

(2, -2)

and slope

m_3 = \frac{3}{4}

, we get

y - (-2) = \frac{3}{4}(x - 2) \implies y + 2 = \frac{3}{4}x - 1.5 \implies y = \frac{3}{4}x - 3.5

Key Concept

The slope of a line is defined as

m = \frac{y_2 - y_1}{x_2 - x_1}

. Perpendicular lines have slopes that are negative reciprocals of each other, satisfying

m_1 \cdot m_2 = -1

. The

y

-intercept of a line is the value of

y

when

x = 0

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