The line (k+1)x + 3y - 5 = 0 is perpendicular to the line passing through the points (2, 1… — JAMB UTME

The line $(k+1)x + 3y - 5 = 0$ is perpendicular to the line passing through the points $(2, 1)$ and $(4, 5)$ . What is the value of $k$ ?

The value of

k

\frac{1}{2}

The line passing through

(2, 1)

and

(4, 5)

has a gradient

m_1 = \frac{5-1}{4-2} = 2

. Rearranging

(k+1)x + 3y - 5 = 0

into

y = -\frac{k+1}{3}x + \frac{5}{3}

gives its gradient

m_2 = -\frac{k+1}{3}

. Using the perpendicular condition

m_1 \cdot m_2 = -1

, we get

2 \cdot \left(-\frac{k+1}{3}\right) = -1

, which simplifies to

2k + 2 = 3

and yields

k = \frac{1}{2}

Calculate the gradient of the line passing through the points

(2, 1)

and

(4, 5)

m_1 = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

Express the line

(k+1)x + 3y - 5 = 0

in slope-intercept form

y = mx + c

to find its gradient

m_2

3y = -(k+1)x + 5 \implies y = -\left(\frac{k+1}{3}\right)x + \frac{5}{3}

, so

m_2 = -\frac{k+1}{3}

The coefficient of

x

when solved for

y

represents the gradient of the line.

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

k

2 \cdot \left(-\frac{k+1}{3}\right) = -1 \implies -\frac{2(k+1)}{3} = -1 \implies 2(k+1) = 3 \implies 2k + 2 = 3 \implies k = \frac{1}{2}

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

-1

Perpendicular Lines and Gradients

Estimated Time:1m 30s

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