Points P(k, 2) and Q(3, 8) lie on a straight line L_1. If L_1 is perpendicular to the line… — JAMB UTME

Points $P(k, 2)$ and $Q(3, 8)$ lie on a straight line $L_1$ . If $L_1$ is perpendicular to the line $L_2$ given by $4x + 3y - 12 = 0$ , what is the value of $k$ ?

Answer: -5

Answer

The value of

k

-5

To determine

k

, the gradient of

L_2

(

4x + 3y - 12 = 0

) is found to be

-\frac{4}{3}

. Using the perpendicularity rule

m_1 \cdot m_2 = -1

, the gradient of

L_1

\frac{3}{4}

. Equating this to the slope formula

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

gives

\frac{6}{3 - k} = \frac{3}{4}

, which simplifies to

k = -5

Step-by-Step Solution

Find the gradient

m_2

of line

L_2

m_2 = -\frac{4}{3}

Converting

4x + 3y - 12 = 0

y = mx + c

form gives

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

Apply the perpendicular line condition to find

m_1

m_1 = \frac{3}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

Express the gradient

m_1

using the coordinates of

P

and

Q

m_1 = \frac{6}{3 - k}

Using the slope formula

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

for points

P(k, 2)

and

Q(3, 8)

Solve for

k

k = -5

Equating

\frac{6}{3 - k} = \frac{3}{4}

yields

3(3 - k) = 24 \implies 9 - 3k = 24 \implies k = -5

Key Concept

Perpendicular Lines and Gradient Formula

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