The parallel lines L_1: 3x - 4y + 25 = 0 and L_2: 3x - 4y = 0 are intersected by a straigh… — JAMB UTME

The parallel lines $L_1: 3x - 4y + 25 = 0$ and $L_2: 3x - 4y = 0$ are intersected by a straight line $L_3$ with gradient $m > 1$ . If the length of the line segment of $L_3$ intercepted between $L_1$ and $L_2$ is $5\sqrt{5}$ units, find the value of $m$ .

Answer: 2

Answer

The value of the gradient

m

is 2.

The perpendicular distance between the parallel lines

L_1: 3x - 4y + 25 = 0

and

L_2: 3x - 4y = 0

d = \frac{25}{\sqrt{3^2 + (-4)^2}} = 5

units. The acute angle

\phi

between

L_3

and the parallel lines satisfies

\sin(\phi) = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}

, which gives

\tan(\phi) = \frac{1}{2}

. The gradient of

L_1

and

L_2

m_1 = \frac{3}{4}

. Using the tangent formula for the angle between two lines,

\tan(\phi) = \left|\frac{m - m_1}{1 + m m_1}\right| \implies \frac{1}{2} = \left|\frac{4m - 3}{4 + 3m}\right|

, which yields

m = 2

m = \frac{2}{11}

. Under the constraint

m > 1

, the unique value of

m

2

Step-by-Step Solution

Calculate the perpendicular distance

d

between the parallel lines

L_1

and

L_2

d = \frac{|25 - 0|}{\sqrt{3^2 + (-4)^2}} = \frac{25}{5} = 5

units

The perpendicular distance between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Determine the trigonometric relationship between the perpendicular distance, intercepted segment, and intersection angle

\phi

\sin(\phi) = \frac{\text{perpendicular distance}}{\text{intercepted segment}} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}

The perpendicular distance forms the opposite side of a right triangle whose hypotenuse is the intercepted segment.

Calculate

\tan(\phi)

using right-triangle trigonometry

Since

\sin(\phi) = \frac{1}{\sqrt{5}}

\cos(\phi) = \sqrt{1 - \sin^2(\phi)} = \frac{2}{\sqrt{5}}

, so

\tan(\phi) = \frac{1}{2}

The angle between lines formula requires

\tan(\phi)

Apply the angle between two lines formula and solve for

m

\tan(\phi) = \left| \frac{m - m_1}{1 + m m_1} \right| \implies \frac{1}{2} = \left| \frac{4m - 3}{4 + 3m} \right|

where

m_1 = \frac{3}{4}

. Case 1:

\frac{4m - 3}{4 + 3m} = \frac{1}{2} \implies 8m - 6 = 4 + 3m \implies 5m = 10 \implies m = 2

. Case 2:

\frac{4m - 3}{4 + 3m} = -\frac{1}{2} \implies 8m - 6 = -4 - 3m \implies 11m = 2 \implies m = \frac{2}{11}

Evaluating the absolute value produces two potential solutions.

Apply the domain constraint

m > 1

m = 2

The problem restricts

m > 1

, which excludes

m = \frac{2}{11}

Key Concept

Distance between parallel lines and angle of intersection between straight lines

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