A point P(x, y) moves such that its perpendicular distance from the straight line L_1: 4x … — JAMB UTME

A point $P(x, y)$ moves such that its perpendicular distance from the straight line $L_1: 4x - 3y + 5 = 0$ is equal to its perpendicular distance from the straight line $L_2: 3x + 4y - 10 = 0$ . Which of the following equations represents one of the straight lines constituting the locus of $P$ ?

$x - 7y + 15 = 0$ Answer
B
$7x - y - 5 = 0$
C
$x + 7y + 15 = 0$
D
$7x + y + 5 = 0$

Answer

The equation

x - 7y + 15 = 0

represents one of the lines constituting the locus.

The locus of a point equidistant from two intersecting straight lines is the pair of angle bisectors between those lines. Setting the perpendicular distance formulas equal yields

4x - 3y + 5 = \pm(3x + 4y - 10)

. Solving the positive branch yields

x - 7y + 15 = 0

, which correctly represents one of the component lines of the locus.

Step-by-Step Solution

Write the perpendicular distance formulas from point

P(x, y)

to both given lines.

d_1 = \frac{|4x - 3y + 5|}{\sqrt{4^2 + (-3)^2}} = \frac{|4x - 3y + 5|}{5}

and

d_2 = \frac{|3x + 4y - 10|}{\sqrt{3^2 + 4^2}} = \frac{|3x + 4y - 10|}{5}

The locus of points equidistant from two intersecting lines consists of the angle bisectors of the angles between the lines.

Set the two perpendicular distances equal to each other.

\frac{|4x - 3y + 5|}{5} = \frac{|3x + 4y - 10|}{5} \implies |4x - 3y + 5| = |3x + 4y - 10|.

Since the point is equidistant from both lines,

d_1 = d_2

Remove absolute values by considering both positive and negative cases.

4x - 3y + 5 = \pm(3x + 4y - 10)

Absolute value equality

|A| = |B|

implies

A = B

A = -B

Evaluate Case 1 (positive sign) to find the first line equation.

4x - 3y + 5 = 3x + 4y - 10 \implies (4x - 3x) + (-3y - 4y) + (5 + 10) = 0 \implies x - 7y + 15 = 0

Grouping like terms yields the linear equation for the first angle bisector.

Evaluate Case 2 (negative sign) to find the second line equation.

4x - 3y + 5 = -(3x + 4y - 10) \implies 4x - 3y + 5 = -3x - 4y + 10 \implies 7x + y - 5 = 0

Grouping like terms yields the linear equation for the second angle bisector.

Key Concept

Locus equidistant from two intersecting lines (Angle Bisectors)

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic