If the line 3x + py - 7 = 0 is perpendicular to the line passing through the points (1, -2… — JAMB UTME

If the line $3x + py - 7 = 0$ is perpendicular to the line passing through the points $(1, -2)$ and $(4, 7)$ , what is the value of $p$ ?

The value of

p

9

The line passing through

(1, -2)

and

(4, 7)

has a gradient of

m_1 = \frac{7 - (-2)}{4 - 1} = 3

. The equation

3x + py - 7 = 0

can be rewritten as

y = -\frac{3}{p}x + \frac{7}{p}

, giving a gradient of

m_2 = -\frac{3}{p}

. For perpendicular lines, the product of their gradients must equal

-1

, so

3 \times \left(-\frac{3}{p}\right) = -1

, which simplifies to

p = 9

Calculate the gradient (

m_1

) of the line passing through

(1, -2)

and

(4, 7)

m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

Express the line

3x + py - 7 = 0

in slope-intercept form (

y = mx + c

) to find its gradient (

m_2

)

py = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}

, so

m_2 = -\frac{3}{p}

The coefficient of

x

when solved for

y

represents the gradient of the straight line.

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

p

3 \cdot \left(-\frac{3}{p}\right) = -1 \implies -\frac{9}{p} = -1 \implies p = 9

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

-1

Perpendicular Lines and Gradients

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