Geometry and Trigonometry Practice Questions — JAMB UTME — Page 8

Question 141Question →

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 142Question →

A coastal monitoring station at point $O$ tracks two vessels on horizontal water. Vessel $A$ is located $15\text{ km}$ from $O$ on a bearing of $070^\circ$ , while Vessel $B$ is located $20\text{ km}$ from $O$ on a bearing of $160^\circ$ . What is the direct distance between Vessel $A$ and Vessel $B$ in kilometers?

Show answer & explanation

Answer: 25

Answer

The direct distance between Vessel A and Vessel B is 25 km.

The difference between the two bearings (

160^\circ - 070^\circ = 90^\circ

) establishes that triangle

AOB

is a right-angled triangle at station

O

. Applying Pythagoras' theorem yields

AB = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ km}

.

Step-by-Step Solution

1

Find the angle between the lines of sight to the two vessels

\angle AOB = 160^\circ - 070^\circ = 90^\circ

Subtracting the smaller bearing angle from the larger bearing angle from the same origin point gives the included angle.

2

Set up the equation for distance AB using Pythagoras' theorem

AB^2 = 15^2 + 20^2 = 225 + 400 = 625

Since the included angle is 90 degrees, the three points form a right-angled triangle where AB is the hypotenuse.

3

Calculate the principal square root of 625

AB = \sqrt{625} = 25\text{ km}

Taking the square root converts the squared distance into the direct linear distance between the vessels.

Key Concept

Calculating the distance between two points using bearings and right-angled triangle properties (Pythagoras' theorem).

Estimated Time:1m 30s

Question 143Question →

A line segment joins the points $A(1, 4)$ and $B(7, 10)$ . Point $P$ divides the line segment $AB$ internally in the ratio $1:2$ . A second line $L_2$ passes through $P$ and is perpendicular to $AB$ . If line $L_2$ intersects the y-axis at the point $(0, k)$ , find the value of $k$ .

Show answer & explanation

Answer: 9

Answer

The value of k is 9.

Using the section formula for internal division in a 1:2 ratio, the coordinates of point P are found to be (3, 6). The gradient of the segment AB is 1, which means the perpendicular line L_2 has a gradient of -1. Writing the equation of line L_2 passing through (3, 6) yields y = -x + 9. Evaluating at x = 0 gives the y-intercept k = 9.

Step-by-Step Solution

1

Calculate the coordinates of point P dividing segment AB internally in the ratio 1:2.

P = (3, 6)

Using the section formula x = (m x_2 + n x_1) / (m + n) and y = (m y_2 + n y_1) / (m + n) with ratio m:n = 1:2.

2

Calculate the gradient m_1 of the line segment AB.

m_1 = 1

Applying the gradient formula m = (y_2 - y_1) / (x_2 - x_1) gives (10 - 4) / (7 - 1) = 1.

3

Determine the gradient m_2 of the perpendicular line L_2.

m_2 = -1

Perpendicular lines satisfy m_1 * m_2 = -1, hence m_2 = -1 / 1 = -1.

4

Find the equation of line L_2 passing through P(3, 6) with gradient -1.

y = -x + 9

Using point-slope form y - y_1 = m(x - x_1) yields y - 6 = -1(x - 3).

5

Determine the y-intercept value k by setting x = 0.

k = 9

Substituting x = 0 into y = -x + 9 gives y = 9.

Key Concept

Section formula, perpendicular line gradients, and y-intercept determination

Estimated Time:3m 0s

Question 144Question →

A straight line $L_1$ passes through the points $(k, 2)$ and $(4, 8)$ . If $L_1$ is perpendicular to the line $2x - 3y + 6 = 0$ , what is the value of $k$ ?

-5

0

8

13

Show answer & explanation

Answer: 8

Answer

The value of

k

is

8

.

The line

2x - 3y + 6 = 0

has a gradient of

\frac{2}{3}

. A line perpendicular to it must have a gradient of

-\frac{3}{2}

. Calculating the gradient of

L_1

using the points

(k, 2)

and

(4, 8)

gives

\frac{8-2}{4-k} = \frac{6}{4-k}

. Equating

\frac{6}{4-k} = -\frac{3}{2}

and solving for

k

gives

k = 8

.

Step-by-Step Solution

1

Find the gradient of the given line

2x - 3y + 6 = 0

Rearranging into

y = mx + c

form gives

3y = 2x + 6 \implies y = \frac{2}{3}x + 2

. Therefore, the gradient

m_2 = \frac{2}{3}

.

To find the perpendicular gradient, we first need the gradient of the given line.

2

Determine the gradient of line

L_1

Since

L_1

is perpendicular to the given line, its gradient

m_1

satisfies

m_1 \cdot m_2 = -1

. Thus,

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

.

Perpendicular lines have gradients that are negative reciprocals of each other.

3

Express the gradient of

L_1

using the coordinates

(k, 2)

and

(4, 8)

and solve for

k

Gradient formula:

m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{4 - k} = \frac{6}{4 - k}

. Setting this equal to

-\frac{3}{2}

:

\frac{6}{4 - k} = -\frac{3}{2}

6 \cdot 2 = -3(4 - k)

12 = -12 + 3k

3k = 24 \implies k = 8

.

Equating the slope calculated from points to the perpendicular slope allows solving for the unknown coordinate

k

.

Key Concept

Perpendicular gradients and line slope formula

Question 145Question →

Five of the interior angles of a convex polygon are each equal to $140^\circ$ , while the remaining interior angles are each equal to $160^\circ$ . Calculate the number of sides of the polygon.

Show answer & explanation

Answer: 13

Answer

The number of sides of the polygon is 13.

Each

140^\circ

interior angle has an exterior angle of

40^\circ

, contributing

5 \times 40^\circ = 200^\circ

to the exterior angle sum. Each

160^\circ

interior angle has an exterior angle of

20^\circ

, contributing

(n - 5) \times 20^\circ

. Since the sum of exterior angles of any convex polygon is

360^\circ

, setting

200 + 20(n - 5) = 360

yields

20n = 260

, giving

n = 13

.

Step-by-Step Solution

1

Calculate the exterior angle measures

The exterior angles are

180^\circ - 140^\circ = 40^\circ

(for 5 vertices) and

180^\circ - 160^\circ = 20^\circ

(for the remaining

n - 5

vertices).

Interior and exterior angles at each vertex of a polygon form a linear pair and sum to

180^\circ

.

2

Apply the sum of exterior angles property

5(40^\circ) + (n - 5)(20^\circ) = 360^\circ

.

The sum of exterior angles of any convex polygon is always constant and equal to

360^\circ

.

3

Solve the linear equation for

n

200 + 20n - 100 = 360 \implies 20n = 260 \implies n = 13

.

Expanding terms and isolating

n

gives the exact number of sides.

Key Concept

Exterior angle sum property of convex polygons

Question 146Question →

A circle of radius $7\text{ cm}$ is inscribed inside a square. What is the area of the region inside the square but outside the circle? (Take $\pi = \frac{22}{7}$ )

42\text{ cm}^2

154\text{ cm}^2

196\text{ cm}^2

12\text{ cm}^2

Show answer & explanation

Answer: $42\text{ cm}^2$

Answer

The area of the region inside the square but outside the circle is

42\text{ cm}^2

.

Because the circle is inscribed inside the square, the diameter of the circle is equal to the side length of the square:

s = 2r = 2(7) = 14\text{ cm}

. The area of the square is

14^2 = 196\text{ cm}^2

. The area of the circle is

\pi r^2 = \frac{22}{7} \times 7^2 = 154\text{ cm}^2

. Subtracting the area of the circle from the area of the square gives the region inside the square but outside the circle:

196 - 154 = 42\text{ cm}^2

.

Step-by-Step Solution

1

Determine the side length of the square.

Side length

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

.

An inscribed circle touches all four sides of the square, so its diameter equals the side length of the square.

2

Calculate the area of the square.

Area of square

= s^2 = 14^2 = 196\text{ cm}^2

.

The area of a square with side length

s

is given by

s^2

.

3

Calculate the area of the inscribed circle.

Area of circle

= \pi r^2 = \frac{22}{7} \times 7^2 = 22 \times 7 = 154\text{ cm}^2

.

The area of a circle with radius

r

is given by

\pi r^2

.

4

Subtract the circle's area from the square's area to find the area of the shaded outer region.

Area

= 196\text{ cm}^2 - 154\text{ cm}^2 = 42\text{ cm}^2

.

The remaining area consists of the four corner regions between the circle and the bounding square.

Key Concept

Area of composite plane figures (shaded area between inscribed shape and container)

Question 147Question →

A trigonometric function is defined as $f(x) = a \sin(b x) + c$ , where $a > 0$ and $b > 0$ . The graph of $y = f(x)$ has a maximum value of $8$ , a minimum value of $-2$ , and a period of $120^\circ$ . What is the value of $a + b + c$ ?

Show answer & explanation

Answer: 11

Answer

The value of

a + b + c

is 11.

Using the maximum (

a + c = 8

) and minimum (

-a + c = -2

), solving simultaneously gives

a = 5

and

c = 3

. Using the period formula

T = \frac{360^\circ}{b} = 120^\circ

, we obtain

b = 3

. Adding these values together yields

5 + 3 + 3 = 11

.

Step-by-Step Solution

1

Set up a system of linear equations for the amplitude and vertical shift from the maximum and minimum bounds

a = 5

and

c = 3

Since

\sin(bx)

ranges from

-1

to

1

, the maximum is

a(1) + c = 8

and minimum is

a(-1) + c = -2

. Solving these simultaneous equations gives

c = 3

and

a = 5

.

2

Calculate the frequency coefficient

b

using the period formula

b = 3

For a sine curve specified in degrees, the period is

T = \frac{360^\circ}{b}

. Substituting

T = 120^\circ

yields

b = 3

.

3

Compute the requested sum

a + b + c

a + b + c = 11

Summing the calculated parameters

5 + 3 + 3 = 11

.

Key Concept

Determining parameters of a trigonometric graph from amplitude, vertical shift, and period.

Question 148Question →

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 149Question →

Determine the number of distinct solutions to the trigonometric equation $2\cos^2 \theta + \sin \theta - 1 = 0$ within the interval $0^\circ \le \theta \le 360^\circ$ .

Show answer & explanation

Answer: 3

Answer

The total number of distinct solutions in the given interval is 3.

Substituting

\cos^2 \theta = 1 - \sin^2 \theta

yields the quadratic equation

2\sin^2 \theta - \sin \theta - 1 = 0

. Factoring gives

\sin \theta = 1

and

\sin \theta = -\frac{1}{2}

. Within

0^\circ \le \theta \le 360^\circ

,

\sin \theta = 1

gives one solution (

90^\circ

), while

\sin \theta = -\frac{1}{2}

gives two solutions (

210^\circ

and

330^\circ

). In total, there are 3 distinct solutions.

Step-by-Step Solution

1

Substitute the identity

\cos^2 \theta = 1 - \sin^2 \theta

into the original equation

2(1 - \sin^2 \theta) + \sin \theta - 1 = 0

Converting the equation into a single trigonometric ratio allows for algebraic solving.

2

Simplify and arrange into quadratic form

2\sin^2 \theta - \sin \theta - 1 = 0

This puts the expression into standard quadratic form

au^2 + bu + c = 0

where

u = \sin \theta

.

3

Factor the quadratic equation

(2\sin \theta + 1)(\sin \theta - 1) = 0

Factoring determines the roots for

\sin \theta

.

4

Solve for possible values of

\sin \theta

\sin \theta = 1

or

\sin \theta = -\frac{1}{2}

By the zero-product property, at least one factor must equal zero.

5

Find the angles for each ratio in the interval

0^\circ \le \theta \le 360^\circ

\theta = 90^\circ, 210^\circ, 330^\circ

\sin \theta = 1

has one solution (

90^\circ

) and

\sin \theta = -0.5

has two solutions in the 3rd and 4th quadrants (

210^\circ

and

330^\circ

).

6

Count the solutions

3

There are 3 distinct values of

\theta

satisfying the condition.

Key Concept

Solving quadratic trigonometric equations using basic identities and quadrant analysis

Estimated Time:1m 30s

Question 150Question →

Find the acute angle $\theta$ , in degrees, that satisfies the trigonometric equation $\sqrt{3}\tan \theta - 3 = 0$ .

Show answer & explanation

Answer: 60

Answer

The acute angle

\theta

is

60^\circ

.

Rearranging the equation

\sqrt{3}\tan \theta - 3 = 0

gives

\sqrt{3}\tan \theta = 3

, so

\tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3}

. For an acute angle (

0^\circ < \theta < 90^\circ

), the angle with a tangent equal to

\sqrt{3}

is

60^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric ratio

\tan \theta

\tan \theta = \sqrt{3}

Add

3

to both sides and divide by

\sqrt{3}

, giving

\frac{3}{\sqrt{3}} = \sqrt{3}

.

2

Determine the value of the acute angle

\theta

\theta = 60^\circ

From special angle exact values,

\tan(60^\circ) = \sqrt{3}

.

Key Concept

Solving Simple Trigonometric Equations

Question 151Question →

A straight line $L$ has a $y$ -intercept of $-4$ and is perpendicular to the line segment connecting the points $P(-3, 1)$ and $Q(5, 5)$ . If the line $L$ intersects the $x$ -axis at $(a, 0)$ , what is the value of $a$ ?

-8

-2

2

8

Show answer & explanation

Answer: $-2$

Answer

The value of

a

is

-2

.

The gradient of

PQ

is calculated as

m_1 = \frac{5 - 1}{5 - (-3)} = \frac{1}{2}

. Because line

L

is perpendicular to

PQ

, its slope is

m_2 = -2

. Combining this with the given

y

-intercept of

-4

yields the equation

y = -2x - 4

. Setting

y = 0

to determine the

x

-intercept gives

0 = -2a - 4

, which solves to

a = -2

.

Step-by-Step Solution

1

Find the gradient

m_1

of the line segment

PQ

m_1 = \frac{5 - 1}{5 - (-3)} = \frac{4}{8} = \frac{1}{2}

The formula for the gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Calculate the gradient

m_2

of line

L

m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2

Perpendicular lines have gradients satisfying

m_1 \cdot m_2 = -1

.

3

Determine the equation of line

L

y = -2x - 4

Using the slope-intercept form

y = mx + c

, where

m = -2

and

c = -4

.

4

Find the

x

-intercept by substituting

y = 0

and

x = a

0 = -2a - 4 \implies 2a = -4 \implies a = -2

At the

x

-axis, the

y

-coordinate is always equal to

0

.

Key Concept

Perpendicular Line Slopes and Coordinate Intercepts

Question 152Question →

What is the gradient of a straight line that is perpendicular to the line passing through the points $(1, -2)$ and $(5, 6)$ ?

-\frac{1}{2}

2

-2

\frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

-\frac{1}{2}

The slope of the given line segment is computed as

\frac{6 - (-2)}{5 - 1} = 2

. The perpendicular gradient is the negative reciprocal of

2

, which gives

-\frac{1}{2}

.

Step-by-Step Solution

1

Calculate the gradient of the line passing through the two points

(1, -2)

and

(5, 6)

.

Using

m_1 = \frac{y_2 - y_1}{x_2 - x_1}

, we obtain

m_1 = \frac{6 - (-2)}{5 - 1} = \frac{8}{4} = 2

.

The slope of a straight line through two points is defined as the change in

y

divided by the change in

x

.

2

Apply the perpendicularity condition to find the perpendicular slope

m_2

.

Since

m_1 \cdot m_2 = -1

,

m_2 = -\frac{1}{m_1} = -\frac{1}{2}

.

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

.

Key Concept

Perpendicular Line Gradients

Estimated Time:45s

Question 153Question →

A frustum of a right circular cone has an upper base radius of $3\text{ cm}$ , a lower base radius of $6\text{ cm}$ , and a vertical height of $4\text{ cm}$ . What is the volume of the frustum?

84\pi\text{ cm}^3

252\pi\text{ cm}^3

60\pi\text{ cm}^3

81\pi\text{ cm}^3

Show answer & explanation

Answer: $84\pi\text{ cm}^3$

Answer

The volume of the frustum is

84\pi\text{ cm}^3

.

The correct answer is obtained by applying the frustum volume formula

V = \frac{1}{3}\pi h (R^2 + r^2 + R r)

. Substituting

R = 6

,

r = 3

, and

h = 4

gives

V = \frac{1}{3}\pi(4)(36 + 9 + 18) = 84\pi\text{ cm}^3

.

Step-by-Step Solution

1

Identify the given dimensions and formula for the volume of a frustum of a right circular cone.

Upper radius

r = 3\text{ cm}

, lower radius

R = 6\text{ cm}

, height

h = 4\text{ cm}

. Formula:

V = \frac{1}{3}\pi h (R^2 + r^2 + R r)

.

The volume of a frustum of a cone is derived by subtracting the top small cone from the total original cone.

2

Evaluate the terms inside the parentheses.

R^2 + r^2 + R r = 6^2 + 3^2 + (6 \times 3) = 36 + 9 + 18 = 63

.

Computing the effective area scaling factor of the frustum bases.

3

Multiply by

\frac{1}{3}\pi h

to find the total volume.

V = \frac{1}{3} \times \pi \times 4 \times 63 = 4 \times 21 \times \pi = 84\pi\text{ cm}^3

.

Completing the frustum volume calculation.

Key Concept

Volume of a Frustum of a Cone

Question 154Question →

What is the maximum value of the trigonometric function $y = 3\sin x + 2$ ?

Show answer & explanation

Answer: 5

Answer

The maximum value of the function is 5.

The basic sine function

\sin x

reaches a maximum value of

1

. Substituting

\sin x = 1

into

y = 3\sin x + 2

yields

y = 3(1) + 2 = 5

.

Step-by-Step Solution

1

Identify the maximum value of the sine term

The range of

\sin x

is

[-1, 1]

, so its maximum value is

1

.

The sine function oscillates between a minimum of

-1

and a maximum of

1

for all real numbers

x

.

2

Calculate the maximum value of the transformed function

y_{\text{max}} = 3(1) + 2 = 5

.

Multiplying by the positive amplitude coefficient

3

scales the peak to

3

, and adding the vertical shift of

2

raises the peak to

5

.

Key Concept

Maximum and Minimum Values of Trigonometric Functions

Question 155Question →

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 156Question →

In $\Delta KLM$ , side $k = 5\text{ cm}$ , side $l = 5\sqrt{3}\text{ cm}$ , and $\angle K = 30^\circ$ . If $\angle L$ is an obtuse angle, what is the measure of $\angle L$ ?

60^\circ

120^\circ

135^\circ

150^\circ

Show answer & explanation

Answer: $120^\circ$

Answer

The measure of angle

L

is

120^\circ

.

Applying the Sine Rule

\frac{k}{\sin K} = \frac{l}{\sin L}

gives

\sin L = \frac{5\sqrt{3} \cdot \sin 30^\circ}{5} = \frac{\sqrt{3}}{2}

. The inverse sine gives an acute angle of

60^\circ

. Because the problem specifies that angle

L

is obtuse, we find its supplementary angle in the second quadrant:

180^\circ - 60^\circ = 120^\circ

.

Step-by-Step Solution

1

Apply the Sine Rule relating sides

k, l

and angles

K, L

.

\frac{k}{\sin K} = \frac{l}{\sin L}

The Sine Rule connects the ratio of side lengths to the sines of their opposite angles.

2

Substitute the given values into the Sine Rule equation.

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{3}}{\sin L} \implies \frac{5}{0.5} = \frac{5\sqrt{3}}{\sin L} \implies 10 = \frac{5\sqrt{3}}{\sin L}

Known values are

k = 5\text{ cm}

,

l = 5\sqrt{3}\text{ cm}

, and

\sin 30^\circ = \frac{1}{2}

.

3

Solve for

\sin L

.

\sin L = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}

Isolating

\sin L

yields the principal ratio.

4

Determine the obtuse angle solution for

L

.

L = 180^\circ - 60^\circ = 120^\circ

Since

\sin L = \frac{\sqrt{3}}{2}

, the acute solution is

60^\circ

, so the supplementary obtuse solution is

180^\circ - 60^\circ = 120^\circ

.

Key Concept

Ambiguous Case of the Sine Rule

Estimated Time:1m 15s

Question 157Question →

The line $(k+1)x + 3y - 5 = 0$ is perpendicular to the line passing through the points $(2, 1)$ and $(4, 5)$ . What is the value of $k$ ?

-7

\frac{1}{2}

-\frac{5}{2}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The value of

k

is

\frac{1}{2}

.

The line passing through

(2, 1)

and

(4, 5)

has a gradient

m_1 = \frac{5-1}{4-2} = 2

. Rearranging

(k+1)x + 3y - 5 = 0

into

y = -\frac{k+1}{3}x + \frac{5}{3}

gives its gradient

m_2 = -\frac{k+1}{3}

. Using the perpendicular condition

m_1 \cdot m_2 = -1

, we get

2 \cdot \left(-\frac{k+1}{3}\right) = -1

, which simplifies to

2k + 2 = 3

and yields

k = \frac{1}{2}

.

Step-by-Step Solution

1

Calculate the gradient of the line passing through the points

(2, 1)

and

(4, 5)

.

m_1 = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Express the line

(k+1)x + 3y - 5 = 0

in slope-intercept form

y = mx + c

to find its gradient

m_2

.

3y = -(k+1)x + 5 \implies y = -\left(\frac{k+1}{3}\right)x + \frac{5}{3}

, so

m_2 = -\frac{k+1}{3}

The coefficient of

x

when solved for

y

represents the gradient of the line.

3

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

k

.

2 \cdot \left(-\frac{k+1}{3}\right) = -1 \implies -\frac{2(k+1)}{3} = -1 \implies 2(k+1) = 3 \implies 2k + 2 = 3 \implies k = \frac{1}{2}

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

.

Key Concept

Perpendicular Lines and Gradients

Estimated Time:1m 30s

Question 158Question →

A straight line $L$ passes through the point $(2, -3)$ and is perpendicular to the line $4x + 5y - 20 = 0$ . Calculate the $x$ -intercept of line $L$ .

Show answer & explanation

Answer: 4.4

Answer

The

x

-intercept of line

L

is

4.4

(or

\frac{22}{5}

).

The given line

4x + 5y - 20 = 0

has a slope of

-\frac{4}{5}

. The line perpendicular to it must have a slope equal to the negative reciprocal, which is

\frac{5}{4}

. Using the point-slope formula with point

(2, -3)

, the equation of line

L

is

y + 3 = \frac{5}{4}(x - 2)

, simplifying to

y = \frac{5}{4}x - \frac{11}{2}

. Setting

y = 0

yields

\frac{5}{4}x = \frac{11}{2}

, giving an

x

-intercept of

x = \frac{22}{5} = 4.4

.

Step-by-Step Solution

1

Determine the gradient of the given line

4x + 5y - 20 = 0

Gradient

m_1 = -\frac{4}{5}

Converting to slope-intercept form

y = -\frac{4}{5}x + 4

reveals the slope.

2

Calculate the perpendicular gradient for line

L

Gradient

m = \frac{5}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Derive the equation of line

L

using point

(2, -3)

y = \frac{5}{4}x - \frac{11}{2}

Substitute the point

(2, -3)

and gradient

m = \frac{5}{4}

into point-slope form.

4

Solve for the

x

-intercept by setting

y = 0

x = 4.4

The

x

-intercept is defined as the point where the line crosses the

x

-axis (

y = 0

).

Key Concept

Perpendicular line slope relationships and x-intercept calculations

Estimated Time:1m 30s

Question 159Question →

At which values of $x$ within the domain $0^\circ \le x \le 360^\circ$ do the graphs of $f(x) = 2\sin x$ and $g(x) = \tan x$ intersect?

0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

60^\circ, 300^\circ

0^\circ, 30^\circ, 180^\circ, 330^\circ, 360^\circ

0^\circ, 60^\circ, 120^\circ, 180^\circ, 360^\circ

Show answer & explanation

Answer: $0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ$

Answer

The graphs intersect at

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

.

Equating

2\sin x = \frac{\sin x}{\cos x}

yields

\sin x (2\cos x - 1) = 0

. Setting

\sin x = 0

gives

x = 0^\circ, 180^\circ, 360^\circ

, and setting

\cos x = \frac{1}{2}

gives

x = 60^\circ, 300^\circ

. Combining these gives the full solution set.

Step-by-Step Solution

1

Set the two trigonometric expressions equal to each other to find points of intersection.

2\sin x = \tan x

Intersection points occur where

f(x) = g(x)

.

2

Rewrite

\tan x

in terms of sine and cosine.

2\sin x = \frac{\sin x}{\cos x}

, for

x \neq 90^\circ, 270^\circ

Using the quotient identity

\tan x = \frac{\sin x}{\cos x}

allows simplification.

3

Rearrange the equation and factor out

\sin x

.

2\sin x \cos x - \sin x = 0 \implies \sin x(2\cos x - 1) = 0

Factoring prevents losing solutions that occur when

\sin x = 0

.

4

Solve each factor separately within

0^\circ \le x \le 360^\circ

.

Factor 1:

\sin x = 0 \implies x = 0^\circ, 180^\circ, 360^\circ

.
Factor 2:

2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = 60^\circ, 300^\circ

.

Cosine is positive in the first and fourth quadrants.

5

Combine all valid solutions.

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

All five values satisfy the original equation and lie within the given domain.

Key Concept

Solving trigonometric equations by factoring and finding all roots in a given domain

Estimated Time:2m 0s

Question 160Question →

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

Geometry and Trigonometry

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