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

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from two fixed points $A(-2, 3)$ and $B(4, 1)$ . If the locus of $P$ intersects the horizontal line $y = 5$ at the point $(k, 5)$ , what is the value of $k$ ?

Answer: 2

Answer

The value of

k

2

The locus of points equidistant from

A(-2, 3)

and

B(4, 1)

is the perpendicular bisector of

AB

. The midpoint of

AB

(1, 2)

and the slope of

AB

-\frac{1}{3}

, giving a perpendicular slope of

3

. The equation of the locus is

3x - y = 1

. Substituting

y = 5

gives

3k - 5 = 1

, which yields

k = 2

Step-by-Step Solution

Identify the nature of the locus

The locus of points equidistant from two fixed points

A

and

B

is the perpendicular bisector of the line segment

AB

By definition, the set of points equidistant from two fixed points forms a straight line perpendicular to the segment joining the two points at its midpoint.

Find the midpoint of segment

AB

Midpoint

M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2)

The perpendicular bisector passes through the midpoint of the line segment.

Calculate the gradient of

AB

and the perpendicular gradient

Gradient of

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

. Thus, the perpendicular gradient is

3

Perpendicular lines have gradients whose product is

-1

Derive the equation of the locus

y - 2 = 3(x - 1) \implies y = 3x - 1 \implies 3x - y = 1

Use the point-slope form of a line equation with point

(1, 2)

and slope

3

Determine the value of

k

y = 5

3k - 5 = 1 \implies 3k = 6 \implies k = 2

Substitute the point

(k, 5)

into the locus equation.

Key Concept

Perpendicular Bisector as a Locus

Estimated Time:1m 30s

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