A straight line L_1 passes through the points (k, 2) and (3, 8). If L_1 is perpendicular t… — JAMB UTME

A straight line $L_1$ passes through the points $(k, 2)$ and $(3, 8)$ . If $L_1$ is perpendicular to the line $L_2$ defined by the equation $2x - 3y + 5 = 0$ , what is the value of $k$ ?

$7$ Cevap
B
$-6$
C
$-1$
D
$12$

Cevap

The value of

k

7

Rearranging the line equation

2x - 3y + 5 = 0

gives

y = \frac{2}{3}x + \frac{5}{3}

, so its gradient is

\frac{2}{3}

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

-\frac{3}{2}

. Equating this to the slope formula

\frac{8 - 2}{3 - k}

gives

\frac{6}{3 - k} = -\frac{3}{2}

. Cross-multiplying yields

12 = -9 + 3k

, which simplifies to

3k = 21

, giving the correct answer

7

Adım Adım Çözüm

Find the gradient (

m_2

) of the line

L_2

Rearrange

2x - 3y + 5 = 0

into

y = mx + c

form:

3y = 2x + 5 \implies y = \frac{2}{3}x + \frac{5}{3}

. Thus,

m_2 = \frac{2}{3}

The coefficient of

x

in slope-intercept form gives the gradient of

L_2

Determine the gradient (

m_1

) of the perpendicular line

L_1

Using the perpendicular condition

m_1 \cdot m_2 = -1

m_1 = -\frac{1}{m_2} = -\frac{3}{2}

Perpendicular lines have gradients that are negative reciprocals of each other.

Express the gradient of

L_1

using the two given points

(k, 2)

and

(3, 8)

m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{3 - k} = \frac{6}{3 - k}

The gradient of a line through

(x_1, y_1)

and

(x_2, y_2)

is given by

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

Equate the two gradient expressions and solve for

k

\frac{6}{3 - k} = -\frac{3}{2} \implies 6 \cdot 2 = -3(3 - k) \implies 12 = -9 + 3k \implies 3k = 21 \implies k = 7

Solving the linear algebraic equation yields the required value of the unknown coordinate

k

Anahtar Kavram

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

(

m_1 \cdot m_2 = -1

Tahmini Süre:2m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru