In the xy-plane, line k has a slope of -\frac{3}{4} and intersects the positive x-axis at … — GRE General Test

In the $xy$ -plane, line $k$ has a slope of $-\frac{3}{4}$ and intersects the positive $x$ -axis at $(a, 0)$ and the positive $y$ -axis at $(0, b)$ . If the distance between the two intercept points $(a, 0)$ and $(0, b)$ is $15$ , what is the value of $a$ ?

A
$9$
$12$ Answer
C
$15$
D
$16$
E
$20$

Answer

The value of

a

12

Using the slope formula between

(a,0)

and

(0,b)

, we find

m = -\frac{b}{a} = -\frac{3}{4}

, giving

b = \frac{3}{4}a

. Applying the distance formula yields

\sqrt{a^2 + b^2} = 15

, or

a^2 + b^2 = 225

. Substituting

b = \frac{3}{4}a

produces

a^2 + \frac{9}{16}a^2 = 225

, which simplifies to

\frac{25}{16}a^2 = 225

. Multiplying by 16 and dividing by 25 yields

a^2 = 144

, giving

a = 12

since

a > 0

Step-by-Step Solution

Express

b

in terms of

a

using the slope formula.

The slope of line

k

passing through

(a, 0)

and

(0, b)

m = \frac{b - 0}{0 - a} = -\frac{b}{a} = -\frac{3}{4}

, which simplifies to

b = \frac{3}{4}a

The slope of a line through two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

Set up the distance equation between

(a, 0)

and

(0, b)

\sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} = 15

, so

a^2 + b^2 = 225

The distance formula between two points in the coordinate plane is derived from the Pythagorean theorem.

Substitute

b = \frac{3}{4}a

into the distance equation and solve for

a

a^2 + \left(\frac{3}{4}a\right)^2 = 225 \implies a^2 + \frac{9}{16}a^2 = 225 \implies \frac{25}{16}a^2 = 225 \implies a^2 = 144 \implies a = 12

Since

(a, 0)

is on the positive

x

-axis,

a

must be positive.

Key Concept

Slope and Distance in Coordinate Geometry

Estimated Time:1m 30s

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