A point P(x, y) moves in a plane such that the line segment joining the fixed points A(1, … — JAMB UTME

A point $P(x, y)$ moves in a plane such that the line segment joining the fixed points $A(1, 2)$ and $B(5, 6)$ subtends a right angle at $P$ . Which of the following equations represents the locus of $P$ ?

$x^2 + y^2 - 6x - 8y + 17 = 0$ Cevap
B
$x - y + 1 = 0$
C
$x^2 + y^2 - 6x - 8y - 7 = 0$
D
$x^2 + y^2 + 6x + 8y + 17 = 0$

Cevap

The equation of the locus of

P

x^2 + y^2 - 6x - 8y + 17 = 0

The locus of a point

P

that subtends a

90^\circ

angle at two fixed points

A(1, 2)

and

B(5, 6)

forms a circle with

AB

as diameter. Using the gradient condition for perpendicular lines,

\frac{y-2}{x-1} \times \frac{y-6}{x-5} = -1

, which simplifies to

x^2 + y^2 - 6x - 8y + 17 = 0

Adım Adım Çözüm

Apply the perpendicularity condition for the line segments

AP

and

BP

Gradient of

AP = \frac{y - 2}{x - 1}

and gradient of

BP = \frac{y - 6}{x - 5}

. Since

\angle APB = 90^\circ

, their product must be

-1

\left(\frac{y - 2}{x - 1}\right) \cdot \left(\frac{y - 6}{x - 5}\right) = -1

Two perpendicular line segments have gradients whose product is

-1

Multiply out the denominators and numerators.

(y - 2)(y - 6) = -(x - 1)(x - 5) \implies y^2 - 8y + 12 = -(x^2 - 6x + 5).

Algebraic expansion of the equation obtained from the gradient product.

Rearrange all terms to one side to express in standard second-degree form.

x^2 + y^2 - 6x - 8y + 17 = 0.

Rearranging yields the Cartesian equation of the locus.

Anahtar Kavram

Locus of a point subtending a right angle at two fixed points

Tahmini Süre:1m 30s

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