In the xy-plane, line k passes through the points (-1, 2) and (3, 10). Line m is perpendic… — GRE General Test

In the $xy$ -plane, line $k$ passes through the points $(-1, 2)$ and $(3, 10)$ . Line $m$ is perpendicular to line $k$ and passes through the point $(2, 4)$ . Line $m$ intersects the $x$ -axis at point $P$ and the $y$ -axis at point $Q$ . What is the distance between point $P$ and point $Q$ ?

A
$3\sqrt{5}$
B
$4\sqrt{5}$
C
$5\sqrt{3}$
$5\sqrt{5}$ Cevap
E
$15$

Cevap

5\sqrt{5}

The answer

5\sqrt{5}

is obtained by taking the negative reciprocal of the slope of line

k

(

m = 2

) to find line

m

's slope (

m = -1/2

), establishing line

m

's equation as

y = -1/2 x + 5

, finding the intercepts

P(10, 0)

and

Q(0, 5)

, and applying the distance formula.

Adım Adım Çözüm

Calculate the slope of line

k

The slope of line

k

m_k = \frac{10 - 2}{3 - (-1)} = \frac{8}{4} = 2

The slope formula is

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

Find the slope of line

m

, which is perpendicular to line

k

The slope of line

m

m_m = -\frac{1}{2}

Perpendicular lines have slopes that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

Determine the equation of line

m

using point-slope form with point

(2, 4)

y - 4 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 5

Substituting slope

m_m = -\frac{1}{2}

and point

(x_1, y_1) = (2, 4)

yields line

m

's slope-intercept equation.

Find the coordinates of the

x

-intercept

P

and

y

-intercept

Q

P = (10, 0)

and

Q = (0, 5)

Setting

y = 0

gives

0 = -\frac{1}{2}x + 5 \implies x = 10

. Setting

x = 0

gives

y = 5

Calculate the distance between

P(10, 0)

and

Q(0, 5)

d = \sqrt{(10 - 0)^2 + (0 - 5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}

Apply the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Anahtar Kavram

Perpendicular Line Slopes and Coordinate Distance

Tahmini Süre:2m 0s

Soru

Cevap

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