Loci and Geometric Constructions Practice Questions — JAMB UTME

Question 1Question →

What is the equation of the locus of a point $P(x, y)$ that moves in a plane such that its distance from the origin $(0,0)$ is always 5 units?

Show answer & explanation

Answer: x^2 + y^2 = 25; x^2+y^2=25; x²+y²=25

Answer

x^2 + y^2 = 25

By definition, the locus of a point moving at a fixed distance of 5 units from the origin

(0,0)

is a circle centered at

(0,0)

with radius 5. Substituting into the standard circle equation

x^2 + y^2 = r^2

yields

x^2 + y^2 = 5^2 = 25

.

Step-by-Step Solution

1

Apply the distance formula between a general point

P(x, y)

and the origin

(0,0)

.

d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}

The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

.

2

Equate the distance formula to the given fixed distance of 5 units and square both sides.

\sqrt{x^2 + y^2} = 5 \implies x^2 + y^2 = 25

Squaring both sides eliminates the square root to give the algebraic equation of the locus.

Key Concept

The locus of points at a constant distance

r

from a fixed point

(h, k)

forms a circle with equation

(x-h)^2 + (y-k)^2 = r^2

.

Estimated Time:45s

Question 2Question →

Find the equation of the locus of a point $P(x, y)$ that moves such that it is equidistant from the fixed points $A(-1, 3)$ and $B(5, -1)$ .

Show answer & explanation

Answer: 3x - 2y - 4 = 0; 3x-2y-4=0; 3x - 2y = 4; 3x-2y=4; 12x - 8y - 16 = 0; y = (3/2)x - 2; y = 1.5x - 2

Answer

The equation of the locus is

3x - 2y - 4 = 0

(or

3x - 2y = 4

).

The locus of a point equidistant from two fixed points

A(-1, 3)

and

B(5, -1)

is the perpendicular bisector of the line segment joining them. Equating the squared distances

(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2

and simplifying yields the linear equation

3x - 2y - 4 = 0

.

Step-by-Step Solution

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}

Since point

P(x, y)

is equidistant from

A

and

B

,

PA = PB

.

2

Square both sides to remove the radical signs.

(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2

Squaring both sides eliminates square roots and simplifies polynomial expansion.

3

Expand all squared terms on both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1

Expanding allows gathering like terms.

4

Cancel

x^2

and

y^2

from both sides and collect all terms on one side.

(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0

Combining like terms simplifies the locus equation into standard linear form.

5

Divide the entire equation by the common factor of 4.

3x - 2y - 4 = 0

Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment

AB

.

Key Concept

Locus equidistant from two fixed points (Perpendicular Bisector)

Estimated Time:2m 0s

Question 3Question →

Determine the equation of the locus of a point $P(x, y)$ that is equidistant from the two parallel lines $2x - 3y + 6 = 0$ and $2x - 3y - 4 = 0$ .

Show answer & explanation

Answer: 2x - 3y + 1 = 0; 2x-3y+1=0; 2x - 3y = -1; 2x-3y=-1

Answer

2x - 3y + 1 = 0

The locus of points equidistant from two parallel lines

ax + by + c_1 = 0

and

ax + by + c_2 = 0

is a parallel line midway between them, defined by

ax + by + \frac{c_1 + c_2}{2} = 0

. Substituting

c_1 = 6

and

c_2 = -4

yields

\frac{6 + (-4)}{2} = 1

, giving the equation

2x - 3y + 1 = 0

.

Step-by-Step Solution

1

Identify the geometric principle for the locus between two parallel lines.

The locus of points equidistant from two parallel lines

ax + by + c_1 = 0

and

ax + by + c_2 = 0

is a third parallel line given by

ax + by + \frac{c_1 + c_2}{2} = 0

.

Points equidistant from two parallel lines lie on a parallel line midway between them.

2

Calculate the average of the constant terms

c_1 = 6

and

c_2 = -4

.

c_{mid} = \frac{6 + (-4)}{2} = \frac{2}{2} = 1

.

The midpoint constant term is the arithmetic mean of the two original constants.

3

Construct the equation of the locus line.

2x - 3y + 1 = 0

.

Combining the common linear coefficients

2x - 3y

with the calculated midpoint constant

1

gives the required equation.

Key Concept

Locus equidistant from two parallel lines

Estimated Time:1m 30s

Question 4Question →

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

x - y + 1 = 0

x^2 + y^2 - 6x - 8y - 7 = 0

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

Show answer & explanation

Answer: $x^2 + y^2 - 6x - 8y + 17 = 0$

Answer

The equation of the locus of

P

is

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

.

Step-by-Step Solution

1

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

.

2

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.

3

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.

Key Concept

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

Estimated Time:1m 30s

Question 5Question →

Find the equation of the locus of a point $P(x, y)$ that moves such that its distance from the fixed point $(3, 0)$ is always equal to its perpendicular distance from the vertical line $x = -3$ .

Show answer & explanation

Answer: y^2 = 12x; y^2 - 12x = 0; y^2=12x; y^2 - 12x = 0

Answer

The equation of the locus is

y^2 = 12x

(or

y^2 - 12x = 0

).

Equating the distance from

P(x, y)

to

(3, 0)

, which is

\sqrt{(x - 3)^2 + y^2}

, and the distance from

P(x, y)

to

x = -3

, which is

|x + 3|

, squaring both sides gives

x^2 - 6x + 9 + y^2 = x^2 + 6x + 9

. Subtracting

x^2 + 9

from both sides yields

y^2 = 12x

.

Step-by-Step Solution

1

Express the distance from

P(x, y)

to the point

(3, 0)

using the distance formula.

d_1 = \sqrt{(x - 3)^2 + (y - 0)^2} = \sqrt{(x - 3)^2 + y^2}

The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

.

2

Express the perpendicular distance from

P(x, y)

to the line

x = -3

.

d_2 = |x - (-3)| = |x + 3|

The perpendicular distance from a point

(x, y)

to a vertical line

x = k

is given by

|x - k|

.

3

Set the two distance expressions equal according to the locus condition.

\sqrt{(x - 3)^2 + y^2} = |x + 3|

The locus condition states that the distance to

(3, 0)

is equal to the distance to line

x = -3

.

4

Square both sides and simplify to obtain the Cartesian equation.

(x - 3)^2 + y^2 = (x + 3)^2 \implies x^2 - 6x + 9 + y^2 = x^2 + 6x + 9 \implies y^2 = 12x

Squaring eliminates the square root and absolute value signs, leading to the algebraic representation of the locus.

Key Concept

Definition and equation of a parabola as the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix)

Alternative Method

Recognize that the definition of a parabola is the locus of points equidistant from a focus

(a, 0)

and a directrix

x = -a

. Here

a = 3

, so the standard equation

y^2 = 4ax

directly gives

y^2 = 4(3)x = 12x

.

Estimated Time:1m 30s

Question 6Question →

A point $P(x, y)$ moves in a Cartesian plane such that the sum of the squares of its distances from two fixed points $A(0, 0)$ and $B(8, 0)$ is equal to $82$ , defining a locus $L_1$ . A second locus $L_2$ is the set of all points equidistant from the parallel lines $y = -1$ and $y = 7$ . Given that $L_1$ and $L_2$ intersect at two distinct points $M$ and $N$ , what is the length of the line segment $MN$ ?

Show answer & explanation

Answer: 8

Answer

The length of the line segment

MN

is 8 units.

The locus

L_1

simplifies to the circle

(x - 4)^2 + y^2 = 25

with center

(4,0)

and radius

5

. The locus

L_2

is the horizontal line

y = 3

. Substituting

y = 3

into the circle equation yields

(x - 4)^2 = 16

, giving intersection points at

(0,3)

and

(8,3)

. The distance between these points is

8

units.

Step-by-Step Solution

1

Formulate the algebraic expression for locus

L_1

The distance squared

PA^2 = x^2 + y^2

and

PB^2 = (x - 8)^2 + y^2

. Setting

PA^2 + PB^2 = 82

gives

x^2 + y^2 + (x - 8)^2 + y^2 = 82

.

By the distance formula, the sum of squared distances from

P(x, y)

to

A(0,0)

and

B(8,0)

equals 82.

2

Simplify and complete the square for locus

L_1

Expanding yields

2x^2 - 16x + 64 + 2y^2 = 82 \implies 2x^2 - 16x + 2y^2 = 18 \implies x^2 - 8x + y^2 = 9

. Completing the square gives

(x - 4)^2 + y^2 = 25

.

This shows

L_1

is a circle with center

(4, 0)

and radius

R = 5

.

3

Find the equation for locus

L_2

The locus of points equidistant from two parallel lines

y = -1

and

y = 7

is the line

y = \frac{-1 + 7}{2} = 3

.

The locus equidistant from two parallel lines is a parallel line running midway between them.

4

Find the intersection points

M

and

N

Substitute

y = 3

into

(x - 4)^2 + y^2 = 25

:

(x - 4)^2 + 3^2 = 25 \implies (x - 4)^2 = 16 \implies x - 4 = \pm 4

, so

x = 8

or

x = 0

. Thus,

M = (0, 3)

and

N = (8, 3)

.

Solving the circle and line equations simultaneously gives the

x

-coordinates of the intersection points.

5

Compute the distance between

M

and

N

Since both points share the ordinate

y = 3

, distance

MN = 8 - 0 = 8

.

The length of a horizontal segment is the difference between its endpoints'

x

-coordinates.

Key Concept

Intersection of loci involving circles and parallel lines

Question 7Question →

A point $P(x, y)$ moves such that it is equidistant from two parallel lines $3x - 4y + 11 = 0$ and $3x - 4y - 1 = 0$ , defining locus $L_1$ . A second locus $L_2$ consists of all points that are at a constant distance of $5$ units from the fixed point $(1, 7)$ . Calculate the distance between the two points of intersection of locus $L_1$ and locus $L_2$ .

Show answer & explanation

Answer: 6

Answer

The distance between the two points of intersection of locus L1 and locus L2 is 6 units.

Locus L1 is the line midway between the two given parallel lines, having the equation 3x - 4y + 5 = 0. Locus L2 is a circle centered at (1, 7) with radius 5. The perpendicular distance from the center (1, 7) to line L1 is 4 units. Applying Pythagoras' theorem to the right triangle formed by the radius, perpendicular distance, and half-chord gives a half-chord length of sqrt(5^2 - 4^2) = 3. Therefore, the distance between the two intersection points (the full chord length) is 2 * 3 = 6 units.

Step-by-Step Solution

1

Determine the equation of locus L1

3x - 4y + 5 = 0

The locus of points equidistant from two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is a line parallel to both and midway between them, given by ax + by + (c1 + c2)/2 = 0.

2

Identify the geometric representation of locus L2

A circle centered at (1, 7) with radius r = 5

The locus of points at a constant distance r from a fixed point (x0, y0) is a circle.

3

Calculate the perpendicular distance d from the circle's center (1, 7) to line L1

d = 4 units

Using the perpendicular distance formula d = |ax0 + by0 + c| / sqrt(a^2 + b^2), we get d = |3(1) - 4(7) + 5| / sqrt(3^2 + (-4)^2) = |-20| / 5 = 4.

4

Compute the total distance between the two intersection points

6 units

The line intersects the circle along a chord. Using Pythagoras' theorem, half the chord length is sqrt(r^2 - d^2) = sqrt(5^2 - 4^2) = 3. Multiplying by 2 gives the full distance of 6 units.

Key Concept

Intersection of Loci (Parallel Line Bisector and Circle)

Estimated Time:2m 30s

Question 8Question →

A point $P(x, y)$ moves in the Cartesian plane such that its distance from the origin $O(0, 0)$ is always half of its distance from the fixed point $Q(6, 0)$ . Which of the following equations represents the locus of $P$ ?

x^2 + y^2 + 4x - 12 = 0

x^2 + y^2 - 4x - 12 = 0

x^2 + y^2 + 12x - 36 = 0

x^2 + y^2 - 16x + 48 = 0

Show answer & explanation

Answer: $x^2 + y^2 + 4x - 12 = 0$

Answer

The equation of the locus of

P

is

x^2 + y^2 + 4x - 12 = 0

.

The correct equation

x^2 + y^2 + 4x - 12 = 0

is derived by expressing the condition

PO = \frac{1}{2} PQ

as

2 \cdot PO = PQ

, squaring both sides to get

4(x^2 + y^2) = (x - 6)^2 + y^2

, and simplifying to standard circle form.

Step-by-Step Solution

1

Express the distance condition algebraically.

The distance from

P(x,y)

to

O(0,0)

is

PO = \sqrt{x^2 + y^2}

, and the distance from

P(x,y)

to

Q(6,0)

is

PQ = \sqrt{(x-6)^2 + y^2}

. Given

PO = \frac{1}{2} PQ

, we have

2 \cdot PO = PQ

.

Translate the geometric distance description into algebraic expressions.

2

Square both sides of the equation to eliminate square roots.

4(x^2 + y^2) = (x - 6)^2 + y^2

.

Squaring removes the radicals; note that

(2 \cdot PO)^2 = 4 \cdot PO^2

.

3

Expand and group like terms.

4x^2 + 4y^2 = x^2 - 12x + 36 + y^2 \implies 3x^2 + 12x + 3y^2 - 36 = 0

.

Expand

(x - 6)^2 = x^2 - 12x + 36

and collect terms on one side.

4

Divide the entire equation by 3 to standard form.

x^2 + y^2 + 4x - 12 = 0

.

Simplifying by the common factor of 3 gives the equation of a circle representing the Circle of Apollonius.

Key Concept

Locus of a point with a constant ratio of distances from two fixed points (Circle of Apollonius)

Estimated Time:2m 0s

Question 9Question →

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

7x - y - 5 = 0

x + 7y + 15 = 0

7x + y + 5 = 0

Show answer & explanation

Answer: $x - 7y + 15 = 0$

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

1

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.

2

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

.

3

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

or

A = -B

.

4

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.

5

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 10Question →

A point $P$ moves in a plane such that its distance from a fixed point $O$ is always $7\text{ cm}$ . What is the diameter, in $\text{cm}$ , of the geometric locus traced out by point $P$ ?

Show answer & explanation

Answer: 14

Answer

The diameter of the locus traced out by point

P

is

14\text{ cm}

.

The locus of a point that maintains a constant distance from a fixed point is a circle. The fixed point

O

is the center of the circle, and the constant distance of

7\text{ cm}

is its radius (

r

). Since the diameter (

D

) of a circle is twice its radius (

D = 2r

), the diameter is

2 \times 7 = 14\text{ cm}

.

Step-by-Step Solution

1

Identify the shape of the geometric locus defined by the condition

A circle centered at point

O

with radius

r = 7\text{ cm}

By definition, the set of all points at a fixed distance from a single point forms a circle.

2

Calculate the diameter using the radius

D = 2 \times 7\text{ cm} = 14\text{ cm}

The diameter of a circle is equal to twice its radius.

Key Concept

Locus of a point at a constant distance from a fixed point

Estimated Time:45s

Question 11Question →

A point $P(x, y)$ moves in a Cartesian plane such that the square of its distance from $A(3, 0)$ exceeds the square of its distance from $B(-1, 2)$ by $4$ units. Which of the following equations represents the locus of $P$ ?

2x - y = 0

2x + y - 4 = 0

x - 2y + 1 = 0

x^2 + y^2 - 4x + 2y - 4 = 0

Show answer & explanation

Answer: $2x - y = 0$

Answer

The equation representing the locus of

P

is

2x - y = 0

.

Using the distance formula,

PA^2 = (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2

and

PB^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4

. Subtracting

PB^2

from

PA^2

yields

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4

, which simplifies to

-8x + 4y = 0

, or

2x - y = 0

.

Step-by-Step Solution

1

Express the square of the distance from

P(x, y)

to

A(3, 0)

using the distance formula

PA^2 = (x - 3)^2 + (y - 0)^2 = x^2 - 6x + 9 + y^2

The distance formula gives

PA^2 = (x - x_A)^2 + (y - y_A)^2

.

2

Express the square of the distance from

P(x, y)

to

B(-1, 2)

PB^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

Expand both algebraic squares for

PB^2

.

3

Set up the locus condition

PA^2 - PB^2 = 4

and simplify

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4 \implies -8x + 4y + 4 = 4

Subtracting

PB^2

from

PA^2

cancels the quadratic

x^2

and

y^2

terms.

4

Rearrange into standard linear equation form

-8x + 4y = 0 \implies 2x - y = 0

Divide the entire equation by

-4

to simplify.

Key Concept

Locus defined by difference of squares of distances to two fixed points

Estimated Time:1m 30s

Question 12Question →

A point $P(x, y)$ moves in the Cartesian plane such that its distance from the fixed point $A(2, 3)$ is equal to its distance from the fixed point $B(4, 1)$ . What is the equation of the locus of $P$ ?

Show answer & explanation

Answer: x - y - 1 = 0; x - y = 1; y = x - 1; x-y-1=0; x-y=1; y=x-1

Answer

The equation of the locus of

P

is

x - y - 1 = 0

(or

y = x - 1

).

The set of all points equidistant from two fixed points

A(2, 3)

and

B(4, 1)

forms the perpendicular bisector of segment

AB

. Equating the squared distances

(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2

and simplifying yields the linear equation

x - y - 1 = 0

(or

y = x - 1

).

Step-by-Step Solution

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - 2)^2 + (y - 3)^2} = \\sqrt{(x - 4)^2 + (y - 1)^2}

The locus of a point

P(x,y)

equidistant from two points

A

and

B

satisfies

PA = PB

.

2

Square both sides to eliminate the square roots and expand the terms.

(x - 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 1)^2 \\implies x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1

Squaring removes the radical sign, allowing algebraic simplification.

3

Subtract

x^2 + y^2

from both sides and collect linear terms.

-4x - 6y + 13 = -8x - 2y + 17

The quadratic terms cancel out since the locus equidistant from two points is a linear equation (perpendicular bisector).

4

Rearrange all terms to one side and simplify.

(-4x + 8x) + (-6y + 2y) + (13 - 17) = 0 \\implies 4x - 4y - 4 = 0 \\implies x - y - 1 = 0

Dividing the entire linear equation by

4

gives the equation in simplest form.

Key Concept

Locus equidistant from two fixed points (Perpendicular Bisector of a line segment)

Alternative Method

Find the midpoint

M

of

AB

,

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

. Calculate the gradient of

AB

,

m_1 = \\frac{1 - 3}{4 - 2} = \\frac{-2}{2} = -1

. The perpendicular gradient is

m_2 = -\\frac{1}{-1} = 1

. Use point-slope form:

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

or

x - y - 1 = 0

.

Estimated Time:1m 30s

Question 13Question →

A fixed line segment $AB$ has a length of $8\text{ cm}$ . A point $P$ moves in the plane such that the area of $\triangle PAB$ is always $20\text{ cm}^2$ . Which of the following best describes the locus of $P$ ?

A pair of parallel lines at a perpendicular distance of

5\text{ cm}

on opposite sides of

AB

A circle of radius

5\text{ cm}

centered at the midpoint of

AB

A pair of parallel lines at a perpendicular distance of

2.5\text{ cm}

on opposite sides of

AB

The perpendicular bisector of the line segment

AB

Show answer & explanation

Answer: A pair of parallel lines at a perpendicular distance of $5\text{ cm}$ on opposite sides of $AB$

Answer

A pair of parallel lines at a perpendicular distance of

5\text{ cm}

on opposite sides of

AB

The area of

\triangle PAB

is given by

\frac{1}{2} \times \text{base} \times \text{height}

. With a base

AB = 8\text{ cm}

, an area of

20\text{ cm}^2

requires a constant height

h = 5\text{ cm}

. The geometric locus of all points at a constant distance from a given straight line consists of two parallel lines situated at that distance on either side of the line.

Step-by-Step Solution

1

Express the area formula of the triangle in terms of base and height.

\text{Area} = \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 8 \times h = 4h

The base of

\triangle PAB

is fixed as the length of segment

AB

, which is

8\text{ cm}

.

2

Calculate the constant perpendicular height

h

.

4h = 20 \implies h = 5\text{ cm}

Setting the calculated area equal to the given constant area of

20\text{ cm}^2

gives the required height.

3

Determine the geometric locus corresponding to a constant perpendicular height.

The locus of points at a fixed distance

h = 5\text{ cm}

from line

AB

is a pair of parallel lines running on either side of

AB

at a distance of

5\text{ cm}

.

Any point

P

lying on either of these two parallel lines maintains a perpendicular distance of

5\text{ cm}

from

AB

, ensuring

\text{Area}(\triangle PAB) = 20\text{ cm}^2

.

Key Concept

Locus at a constant distance from a straight line

Question 14Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is at all times equidistant from two parallel lines given by the equations $3x + 4y - 12 = 0$ and $3x + 4y + 4 = 0$ . The locus of $P$ intersects the straight line $x - 2y - 8 = 0$ at the point $(a, b)$ . What is the value of $a - b$ ?

Show answer & explanation

Answer: 6

Answer

The value of

a - b

is 6.

The locus of a point moving equidistant between two parallel lines is the parallel line lying midway between them. Combining the parallel line equations

3x + 4y - 12 = 0

and

3x + 4y + 4 = 0

yields the locus line

3x + 4y - 4 = 0

. Solving the system formed by this locus line and

x - 2y - 8 = 0

gives

x = 4

and

y = -2

. Therefore,

a = 4

and

b = -2

, so

a - b = 4 - (-2) = 6

.

Step-by-Step Solution

1

Find the equation of the locus of point P

Locus equation:

3x + 4y - 4 = 0

The locus of points equidistant from two parallel lines

ax + by + c_1 = 0

and

ax + by + c_2 = 0

is a parallel line midway between them, given by

ax + by + \frac{c_1 + c_2}{2} = 0

.

2

Solve the simultaneous equations to find the intersection point

(a, b)

a = 4

and

b = -2

Substitute

x = 2y + 8

into

3x + 4y - 4 = 0

to get

3(2y + 8) + 4y - 4 = 0

, which yields

10y = -20

, so

y = -2

and

x = 4

.

3

Calculate the difference

a - b

6

Subtract

b = -2

from

a = 4

to obtain

4 - (-2) = 6

.

Key Concept

Locus of points equidistant from two parallel lines and intersection of straight lines

Question 15Question →

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 $k$ ?

Show answer & explanation

Answer: 2

Answer

The value of

k

is

2

.

The locus of points equidistant from

A(-2, 3)

and

B(4, 1)

is the perpendicular bisector of

AB

. The midpoint of

AB

is

(1, 2)

and the slope of

AB

is

-\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

1

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.

2

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.

3

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

.

4

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

.

5

Determine the value of

k

at

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

Question 16Question →

A point $P(x, y)$ moves in the Cartesian plane such that it maintains a constant distance of $10\text{ units}$ from a fixed point $C(2, -3)$ . If the locus of $P$ intersects the vertical line $x = 8$ at two points $A$ and $B$ , what is the distance between $A$ and $B$ ?

Show answer & explanation

Answer: 16

Answer

The distance between the intersection points A and B is 16 units.

The locus of point P moving at a constant distance of 10 units from C(2, -3) forms a circle

(x - 2)^2 + (y + 3)^2 = 100

. Substituting

x = 8

yields

(y + 3)^2 = 64

, giving

y = 5

and

y = -11

. The distance between the two points (8, 5) and (8, -11) along the vertical line

x = 8

is

5 - (-11) = 16

units.

Step-by-Step Solution

1

Determine the equation representing the locus of point P

(x - 2)^2 + (y + 3)^2 = 100

The locus of a point moving at a fixed distance from a fixed point is a circle with equation

(x - h)^2 + (y - k)^2 = r^2

.

2

Substitute

x = 8

into the locus equation to find the

y

-coordinates of the intersection points

(8 - 2)^2 + (y + 3)^2 = 100 \implies 36 + (y + 3)^2 = 100 \implies (y + 3)^2 = 64

The intersection points lie on both the locus circle and the vertical line

x = 8

.

3

Solve for the two possible values of

y

y + 3 = \pm 8 \implies y_1 = 5

and

y_2 = -11

Taking the square root gives both positive and negative solutions for the vertical coordinate.

4

Calculate the vertical distance between points

A(8, 5)

and

B(8, -11)

\text{Distance} = 5 - (-11) = 16\text{ units}

Since both points have the same

x

-coordinate (

x = 8

), the distance is simply the absolute difference between their

y

-coordinates.

Key Concept

Locus of a point at a constant distance from a fixed point (Circle)

Question 17Question →

What is the equation of the locus of a point $P(x, y)$ that is always equidistant from the point $(0, 4)$ and the line $y = -4$ ?

Show answer & explanation

Answer: x^2 = 16y; x^2 - 16y = 0; x^2-16y=0; y = x^2/16; y = \frac{x^2}{16}; x^{2}=16y; x^{2}-16y=0

Answer

The equation of the locus is

x^2 = 16y

(or

x^2 - 16y = 0

).

Equating the distance from

P(x, y)

to

(0, 4)

, which is

\sqrt{x^2 + (y-4)^2}

, to the perpendicular distance to the line

y = -4

, which is

|y + 4|

, and squaring both sides gives

x^2 + (y-4)^2 = (y+4)^2

. Expanding yields

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16

, which simplifies directly to

x^2 = 16y

or

x^2 - 16y = 0

.

Step-by-Step Solution

1

Formulate the distance expressions from point

P(x, y)

to the given point

(0, 4)

and line

y = -4

.

Distance to

(0, 4) = \sqrt{(x - 0)^2 + (y - 4)^2}

. Perpendicular distance to

y = -4

is

|y - (-4)| = |y + 4|

.

By definition of geometric locus, the distance from

P(x, y)

to the fixed point must equal its distance to the fixed line.

2

Equate the two distance expressions.

\sqrt{x^2 + (y - 4)^2} = |y + 4|

The point

P(x, y)

is equidistant from both geometric entities.

3

Square both sides of the equation to clear the square root and absolute value.

x^2 + (y - 4)^2 = (y + 4)^2

Squaring eliminates radical and absolute value signs while preserving algebraic equality.

4

Expand both squared binomial expressions and simplify.

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16 \implies x^2 = 16y

Subtracting

y^2 + 16

from both sides leaves

x^2 - 8y = 8y

, which simplifies to

x^2 = 16y

.

Key Concept

The locus of points equidistant from a fixed point (focus) and a fixed straight line (directrix) forms a parabola.

Estimated Time:1m 30s

Question 18Question →

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 $x_0$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x_0

is

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

is

(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

1

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.

2

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 19Question →

Two fixed points in a Cartesian plane are given as $A(-1, 2)$ and $B(3, 4)$ . A point $P(x, y)$ moves in the plane such that $PA^2 + PB^2 = 26$ . Which of the following equations represents the locus of $P$ ?

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

x^2 + y^2 + 2x + 6y + 2 = 0

x^2 + y^2 - 4x - 12y + 4 = 0

x^2 + y^2 - 2x - 6y + 30 = 0

Show answer & explanation

Answer: $x^2 + y^2 - 2x - 6y + 2 = 0$

Answer

The equation representing the locus of

P

is

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

.

Using the Cartesian coordinate distance formula,

PA^2 = (x+1)^2 + (y-2)^2

and

PB^2 = (x-3)^2 + (y-4)^2

. Adding these together yields

2x^2 + 2y^2 - 4x - 12y + 30

. Setting this equal to 26 gives

2x^2 + 2y^2 - 4x - 12y + 4 = 0

. Dividing the entire equation by 2 yields

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

, which is the correct locus equation.

Step-by-Step Solution

1

Express

PA^2

using the distance formula between

P(x, y)

and

A(-1, 2)

.

PA^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

The square of the distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

(x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Express

PB^2

using the distance formula between

P(x, y)

and

B(3, 4)

.

PB^2 = (x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 = x^2 + y^2 - 6x - 8y + 25

Expanding the distance squared formula for point

B

.

3

Sum

PA^2

and

PB^2

and equate to the given constant 26.

(x^2 + y^2 + 2x - 4y + 5) + (x^2 + y^2 - 6x - 8y + 25) = 26 \implies 2x^2 + 2y^2 - 4x - 12y + 30 = 26

Substitute the algebraic expressions into the locus condition

PA^2 + PB^2 = 26

.

4

Rearrange into general circle form and divide through by 2.

2x^2 + 2y^2 - 4x - 12y + 4 = 0 \implies x^2 + y^2 - 2x - 6y + 2 = 0

Subtract 26 from both sides and divide all terms by 2 to obtain the standard Cartesian equation.

Key Concept

Locus of a point with sum of squared distances to two fixed points equal to a constant

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