A straight line L passes through the point (4, -2) and is perpendicular to the line define… — JAMB UTME

A straight line $L$ passes through the point $(4, -2)$ and is perpendicular to the line defined by the equation $3x - 2y + 8 = 0$ . What is the $y$ -intercept of line $L$ ?

$\frac{2}{3}$ Cevap
B
$-8$
C
$4$
D
$-\frac{14}{3}$

Cevap

The

y

-intercept of line

L

\frac{2}{3}

The given line

3x - 2y + 8 = 0

has a gradient of

\frac{3}{2}

. A perpendicular line must have a gradient equal to the negative reciprocal, which is

-\frac{2}{3}

. Substituting the point

(4, -2)

into

y - y_1 = m(x - x_1)

gives

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

, which simplifies to

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

. Setting

x = 0

yields the

y

-intercept of

\frac{2}{3}

Adım Adım Çözüm

Find the gradient of the given line.

Expressing

3x - 2y + 8 = 0

in slope-intercept form

y = mx + c

gives

2y = 3x + 8 \implies y = \frac{3}{2}x + 4

. The gradient

m_1 = \frac{3}{2}

The gradient of the given line is required to determine the slope of line

L

Determine the gradient of line

L

Since line

L

is perpendicular to the given line,

m_L = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_L = -1

Find the equation of line

L

using point-slope form.

Using

(x_1, y_1) = (4, -2)

and

m_L = -\frac{2}{3}

y - y_1 = m_L(x - x_1)

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

y + 2 = -\frac{2}{3}x + \frac{8}{3}

y = -\frac{2}{3}x + \frac{8}{3} - 2

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

To find the

y

-intercept, we need the complete equation of line

L

Identify the

y

-intercept.

Comparing

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

y = mx + c

, the

y

-intercept

c = \frac{2}{3}

The constant term

c

y = mx + c

represents the

y

-intercept.

Anahtar Kavram

Perpendicular lines in coordinate geometry have gradients that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

Soru

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