A point P(x, y) moves in the Cartesian plane such that it is always equidistant from the t… — JAMB UTME

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from the two fixed points $A(2, 1)$ and $B(6, 5)$ . If the locus of $P$ intersects the line $2x + y = 14$ at the point $(x_0, y_0)$ , what is the value of $x_0$ ?

Answer: 7

Answer

The value of

x_0

7

The locus of points equidistant from

A(2, 1)

and

B(6, 5)

is the perpendicular bisector of line segment

AB

. The midpoint of

AB

(4, 3)

and its slope is

1

, giving the perpendicular bisector a slope of

-1

. The equation of this locus is

y - 3 = -1(x - 4)

, or

x + y = 7

. Subtracting

x + y = 7

from

2x + y = 14

directly gives

x_0 = 7

Step-by-Step Solution

Determine the equation of the locus of point P

The locus of P is the perpendicular bisector of segment AB, represented by

x + y = 7

The set of all points equidistant from two fixed points forms the perpendicular bisector of the line segment connecting those points.

Find the point of intersection with the line

2x + y = 14

Solving

x + y = 7

and

2x + y = 14

simultaneously gives

x_0 = 7

The intersection point of two geometric lines must satisfy both equations simultaneously.

Key Concept

Perpendicular Bisector Locus and Line Intersections

Question

Answer

Step-by-Step Solution

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