OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Exercise 11 (B)

S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(b)

Question 1. Find the slope of a line whose inclination to the positive direction of x-axis in anticlockwise direction is given as :
(a) 30°
(b) 45°
(c) 60°
(e) 75°
Answer:
If the inclination of a line to the positive direction of the x-axis is \( \theta \), then the slope of that line will be \( \tan \theta \).
(a) Slope when angle is 30°: \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)
(b) Slope when angle is 45°: \( \tan 45^\circ = 1 \)
(c) Slope when angle is 60°: \( \tan 60^\circ = \sqrt{3} \)
(e) Slope when angle is 75°: \( \tan 75^\circ = 3.7321 \) (This value comes from a tangent table)
In simple words: The slope of a line tells you how steep it is. You can find it using the tangent of the angle the line makes with the x-axis. A larger angle generally means a steeper slope.

🎯 Exam Tip: Remember the standard tangent values for common angles (0°, 30°, 45°, 60°, 90°) as they are frequently used. For other angles, you might need to use a calculator or a trigonometric table.

Question 2. Find the slope and inclination of the line through each pair of the following points:
(a) (1, 2) and (5, 6)
(b) (0, 0) and \( (\sqrt{3}, 3) \)
Answer:
(a) For points (1, 2) and (5, 6):
The slope \( (m) = \frac{y_2-y_1}{x_2-x_1} = \frac{6-2}{5-1} = \frac{4}{4} = 1 \).
The inclination angle is \( \theta \). Since \( \tan \theta = 1 \), the angle of inclination \( \theta = 45^\circ \).
(b) For points (0, 0) and \( (\sqrt{3}, 3) \):
The slope \( (m) = \frac{y_2-y_1}{x_2-x_1} = \frac{3-0}{\sqrt{3}-0} = \frac{3}{\sqrt{3}} \).
To simplify, we multiply the numerator and denominator by \( \sqrt{3} \): \( \frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \).
The inclination angle is \( \theta \). Since \( \tan \theta = \sqrt{3} \), the angle of inclination \( \theta = 60^\circ \).
In simple words: To find the slope between two points, subtract the y-coordinates and divide by the difference in x-coordinates. Then, use the tangent function to find the angle that corresponds to that slope, which is the line's inclination.

🎯 Exam Tip: Always simplify radical expressions for slopes. If a slope is a recognized trigonometric value (like 1 or \( \sqrt{3} \)), you can directly state the angle of inclination; otherwise, you might need \( \arctan(m) \).

Question 3. The side BC of an equilateral ∆ABC is parallel to the x-axis. What are the slopes of its sides?
Answer:
In an equilateral triangle ABC, each angle is \( 60^\circ \). Side BC is parallel to the x-axis, so its inclination is \( 0^\circ \). Therefore, the slope of BC \( = \tan 0^\circ = 0 \). For side AB, its inclination with the positive x-axis is \( 60^\circ \) (since \( \angle B = 60^\circ \) and BC is horizontal). So, the slope of AB \( = \tan 60^\circ = \sqrt{3} \). For side AC, its inclination with the positive x-axis would be \( 180^\circ - 60^\circ = 120^\circ \) (because \( \angle C = 60^\circ \) and AC slopes downwards to the right from A). So, the slope of AC \( = \tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \).In simple words: Since it's an equilateral triangle, all its angles are 60 degrees. If one side lies flat on the x-axis, its slope is zero. The other sides will make angles of 60 and 120 degrees with the x-axis, giving slopes of \( \sqrt{3} \) and \( -\sqrt{3} \).

🎯 Exam Tip: When a side of a triangle is parallel to an axis, its slope is either 0 (parallel to x-axis) or undefined (parallel to y-axis), which simplifies finding the inclinations of the other sides.

Question 4. In a regular hexagon ABCDEF, AB || ED || the x-axis. What are the slopes of its sides?
Answer:
In a regular hexagon ABCDEF, all interior angles are \( 120^\circ \). Given that side AB is parallel to the x-axis, its inclination is \( 0^\circ \). So, the slope of AB \( = \tan 0^\circ = 0 \). Since ED is also parallel to the x-axis, its inclination is also \( 0^\circ \). So, the slope of ED \( = \tan 0^\circ = 0 \). The sides of a regular hexagon, in order, make angles with each other that are multiples of \( 60^\circ \) (relative to the previous side's direction). If AB is horizontal (\( 0^\circ \)), then: Slope of BC: The angle of BC with the x-axis would be \( 60^\circ \). So, \( \text{slope of BC} = \tan 60^\circ = \sqrt{3} \). Slope of CD: The angle of CD with the x-axis would be \( 120^\circ \). So, \( \text{slope of CD} = \tan 120^\circ = -\sqrt{3} \). Slope of DE: This is parallel to AB, so its angle is \( 180^\circ \) or \( 0^\circ \). So, \( \text{slope of DE} = \tan 0^\circ = 0 \). Slope of EF: The angle of EF with the x-axis would be \( 240^\circ \) or \( -120^\circ \), which is \( 60^\circ \) if measured from the negative x-axis. Or, it's parallel to BC. So, \( \text{slope of EF} = \tan 60^\circ = \sqrt{3} \). Slope of FA: The angle of FA with the x-axis would be \( 300^\circ \) or \( -60^\circ \), which is \( 120^\circ \) if measured from the negative x-axis. Or, it's parallel to CD. So, \( \text{slope of FA} = \tan 120^\circ = -\sqrt{3} \).In simple words: For a regular hexagon where two opposite sides are flat (parallel to the x-axis), their slopes are zero. The other sides will have slopes of \( \sqrt{3} \) or \( -\sqrt{3} \), depending on which way they are tilting.

🎯 Exam Tip: For regular polygons, remember the relationship between their interior angles and how sides are oriented. If one side's inclination is known, the inclinations of other sides can be found by adding or subtracting the polygon's exterior angle (360/n for an n-sided polygon).

Question 5. Find y if the slope of the line joining (- 8, 11), (2, y) is \( \frac{-4}{3} \).
Answer:
We know that the slope \( (m) = \frac{y_2-y_1}{x_2-x_1} \).
Given points are \( (x_1, y_1) = (-8, 11) \) and \( (x_2, y_2) = (2, y) \), and the slope \( m = \frac{-4}{3} \).
Substitute these values into the slope formula:
\( \frac{-4}{3} = \frac{y - 11}{2 - (-8)} \)
\( \implies \frac{-4}{3} = \frac{y - 11}{2 + 8} \)
\( \implies \frac{-4}{3} = \frac{y - 11}{10} \)
Now, cross-multiply:
\( -4 \times 10 = 3 \times (y - 11) \)
\( -40 = 3y - 33 \)
Add 33 to both sides to isolate the term with y:
\( -40 + 33 = 3y \)
\( -7 = 3y \)
Divide by 3 to find y:
\( y = \frac{-7}{3} \)
In simple words: We used the formula for finding the slope between two points. We put in the known numbers and the slope, and then solved the equation to find the missing 'y' value. This is a common way to find a missing coordinate.

🎯 Exam Tip: Be careful with signs, especially when subtracting negative numbers (e.g., \( 2 - (-8) \)). A small error in signs can lead to a completely different answer.

Question 6. Find the value of a, if the line passing through (-5, -8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a).
Answer:
If two lines are parallel, their slopes must be equal.
First, find the slope of the line passing through \( (-5, -8) \) and \( (3, 0) \). Let this be \( m_1 \).
\( m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{0 - (-8)}{3 - (-5)} = \frac{0 + 8}{3 + 5} = \frac{8}{8} = 1 \).
Next, find the slope of the line passing through \( (6, 3) \) and \( (4, a) \). Let this be \( m_2 \).
\( m_2 = \frac{a - 3}{4 - 6} = \frac{a - 3}{-2} \).
Since the two lines are parallel, their slopes are equal:
\( m_1 = m_2 \)
\( \implies 1 = \frac{a - 3}{-2} \)
Multiply both sides by -2:
\( -2 \times 1 = a - 3 \)
\( -2 = a - 3 \)
Add 3 to both sides to find 'a':
\( a = -2 + 3 \)
\( a = 1 \).
In simple words: When lines are parallel, they have the same slope. We calculated the slope for both lines and set them equal to each other to find the unknown value 'a'. This property helps us find missing coordinates or confirm if lines are parallel.

🎯 Exam Tip: Remember the key property of parallel lines (slopes are equal) and perpendicular lines (product of slopes is -1). This is fundamental for solving many coordinate geometry problems.

Question 7. Find the slope of a line perpendicular to the line whose slope is :
(a) 3
(b) \( \frac{-5}{6} \)
(c) 5
(d) \( \frac{-36}{7} \)
(e) 0
(f) Infinite
Answer:
If a line has a slope 'm', then the slope of a line perpendicular to it is \( \frac{-1}{m} \). This is because the product of slopes of two perpendicular lines is -1.
(a) Given slope \( m = 3 \).
Slope of perpendicular line \( = \frac{-1}{3} \).
(b) Given slope \( m = \frac{-5}{6} \).
Slope of perpendicular line \( = \frac{-1}{\frac{-5}{6}} = \frac{6}{5} \).
(c) Given slope \( m = 5 \).
Slope of perpendicular line \( = \frac{-1}{5} \).
(d) Given slope \( m = \frac{-36}{7} \).
Slope of perpendicular line \( = \frac{-1}{\frac{-36}{7}} = \frac{7}{36} \).
(e) Given slope \( m = 0 \). A line with slope 0 is a horizontal line.
A line perpendicular to a horizontal line is a vertical line. The slope of a vertical line is undefined, or "infinite".
(f) Given slope \( m = \text{infinite} \) (undefined). A line with infinite slope is a vertical line.
A line perpendicular to a vertical line is a horizontal line. The slope of a horizontal line is 0.
In simple words: To find the slope of a perpendicular line, you flip the original slope and change its sign. If the original slope is zero (a flat line), the perpendicular line is straight up and down, which has an infinite slope. If the original is straight up and down, the perpendicular is flat, with a slope of zero.

🎯 Exam Tip: Always remember that the slope of a horizontal line is 0 and the slope of a vertical line is undefined (often referred to as infinite). These are special cases for perpendicular slopes.

Question 8. Find the slope of a line perpendicular to the line which passes through the pair of the following points :
(a) (0, 8) and (-5, 2);
(b) (1, -11) and (5, 2);
(c) (-k, h) and (b, -f);
(d) (x1, y1) and (x2, y2)
Answer:
(a) For points \( (0, 8) \) and \( (-5, 2) \):
First, find the slope of the line joining these points, \( m \).
\( m = \frac{y_2-y_1}{x_2-x_1} = \frac{2 - 8}{-5 - 0} = \frac{-6}{-5} = \frac{6}{5} \).
The slope of the line perpendicular to this line \( = \frac{-1}{m} = \frac{-1}{\frac{6}{5}} = \frac{-5}{6} \).
(b) For points \( (1, -11) \) and \( (5, 2) \):
First, find the slope of the line joining these points, \( m \).
\( m = \frac{y_2-y_1}{x_2-x_1} = \frac{2 - (-11)}{5 - 1} = \frac{2 + 11}{4} = \frac{13}{4} \).
The slope of the line perpendicular to this line \( = \frac{-1}{m} = \frac{-1}{\frac{13}{4}} = \frac{-4}{13} \).
(c) For points \( (-k, h) \) and \( (b, -f) \):
First, find the slope of the line joining these points, \( m \).
\( m = \frac{y_2-y_1}{x_2-x_1} = \frac{-f - h}{b - (-k)} = \frac{-(f + h)}{b + k} \).
The slope of the line perpendicular to this line \( = \frac{-1}{m} = \frac{-1}{\frac{-(f + h)}{b + k}} = \frac{b + k}{f + h} \).
(d) For points \( (x_1, y_1) \) and \( (x_2, y_2) \):
First, find the slope of the line joining these points, \( m \).
\( m = \frac{y_2-y_1}{x_2-x_1} \).
The slope of the line perpendicular to this line \( = \frac{-1}{m} = \frac{-1}{\frac{y_2-y_1}{x_2-x_1}} = \frac{-(x_2-x_1)}{y_2-y_1} \).
In simple words: For each pair of points, first calculate the slope of the line connecting them. Then, to find the slope of a line perpendicular to it, simply flip this calculated slope and change its sign. This works for both numbers and algebraic expressions.

🎯 Exam Tip: When dealing with algebraic coordinates, keep track of all the variables and signs carefully. The formula \( \frac{-1}{m} \) is universal for perpendicular slopes, but algebraic manipulation needs precision.

Question 9. In rectangle ABCD, the slope of AB = \( \frac{5}{6} \). State the slope of
(a) BC
(b) CD
(c) DA
Answer:
In a rectangle ABCD, opposite sides are parallel, and adjacent sides are perpendicular.
Given, the slope of AB \( = \frac{5}{6} \).
(a) For side BC:
Side BC is perpendicular to side AB. If the slope of AB is \( m_{AB} \), then the slope of BC, \( m_{BC} \), is \( \frac{-1}{m_{AB}} \).
So, slope of BC \( = \frac{-1}{\frac{5}{6}} = \frac{-6}{5} \).
(b) For side CD:
Side CD is parallel to side AB. Parallel lines have equal slopes.
So, slope of CD \( = \text{slope of AB} = \frac{5}{6} \).
(c) For side DA:
Side DA is parallel to side BC. Parallel lines have equal slopes.
So, slope of DA \( = \text{slope of BC} = \frac{-6}{5} \). Alternatively, DA is perpendicular to AB, so its slope is also \( \frac{-1}{\frac{5}{6}} = \frac{-6}{5} \).
In simple words: In a rectangle, opposite sides run in the same direction, so they have the same slope. Sides next to each other are at right angles, so their slopes are negative reciprocals of each other (meaning you flip the fraction and change its sign).

🎯 Exam Tip: Always draw a quick sketch of the rectangle (or any polygon) and label its vertices to visualize which sides are parallel and which are perpendicular. This helps in correctly applying the slope rules.

Question 10. In parallelogram ABCD, slope of AB = - 2, slope of BC = \( \frac{3}{5} \). State the slope of
(a) AD
(b) CD
(c) the altitude of AD
(d) the altitude of CD
Answer:
In a parallelogram ABCD, opposite sides are parallel.
Given, slope of AB \( = -2 \).
Given, slope of BC \( = \frac{3}{5} \).
(a) For side AD:
Side AD is parallel to side BC. Therefore, their slopes are equal.
Slope of AD \( = \text{slope of BC} = \frac{3}{5} \).
(b) For side CD:
Side CD is parallel to side AB. Therefore, their slopes are equal.
Slope of CD \( = \text{slope of AB} = -2 \).
(c) For the altitude of AD:
An altitude to a side is perpendicular to that side. So, the altitude of AD is perpendicular to AD.
Slope of AD \( = \frac{3}{5} \).
Slope of altitude of AD \( = \frac{-1}{\text{slope of AD}} = \frac{-1}{\frac{3}{5}} = \frac{-5}{3} \).
(d) For the altitude of CD:
The altitude of CD is perpendicular to CD.
Slope of CD \( = -2 \).
Slope of altitude of CD \( = \frac{-1}{\text{slope of CD}} = \frac{-1}{-2} = \frac{1}{2} \).
In simple words: In a parallelogram, sides opposite each other have the same slope. A line that is an "altitude" to a side means it meets that side at a right angle. So, to find the slope of an altitude, you take the slope of the side it's perpendicular to, flip it, and change its sign.

🎯 Exam Tip: Make sure to distinguish between the slope of a side and the slope of an altitude to that side. One uses parallelism, the other uses perpendicularity. This is a common source of confusion in problems.

Question 11. The vertices of a ∆ABC are A (1, 1), B (7, 3) and C (3, 6). State the slope of the altitude to
(a) AB
(b) BC
(c) AC.
Answer:
The vertices of \( \triangle ABC \) are A(1, 1), B(7, 3), and C(3, 6).
(a) Slope of the altitude to AB:
First, find the slope of side AB:
\( \text{Slope of AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{3 - 1}{7 - 1} = \frac{2}{6} = \frac{1}{3} \).
The altitude to AB is perpendicular to AB. So, its slope \( = \frac{-1}{\text{Slope of AB}} = \frac{-1}{\frac{1}{3}} = -3 \).
(b) Slope of the altitude to BC:
First, find the slope of side BC:
\( \text{Slope of BC} = \frac{y_2-y_1}{x_2-x_1} = \frac{6 - 3}{3 - 7} = \frac{3}{-4} = \frac{-3}{4} \).
The altitude to BC is perpendicular to BC. So, its slope \( = \frac{-1}{\text{Slope of BC}} = \frac{-1}{\frac{-3}{4}} = \frac{4}{3} \).
(c) Slope of the altitude to AC:
First, find the slope of side AC:
\( \text{Slope of AC} = \frac{y_2-y_1}{x_2-x_1} = \frac{6 - 1}{3 - 1} = \frac{5}{2} \).
The altitude to AC is perpendicular to AC. So, its slope \( = \frac{-1}{\text{Slope of AC}} = \frac{-1}{\frac{5}{2}} = \frac{-2}{5} \).
In simple words: An altitude from one corner of a triangle to the opposite side always meets that side at a perfect right angle. To find the slope of such an altitude, we first find the slope of the side it connects to, and then calculate its negative reciprocal.

🎯 Exam Tip: Remember the formula for slope and the relationship between slopes of perpendicular lines. It's crucial to correctly identify the two points for each side to calculate its slope accurately before finding the perpendicular slope.

Question 12. The line joining (-5, 7) and (0, -2) is perpendicular to the line joining (1, -3) and (4, x). Find x.
Answer:
Let \( m_1 \) be the slope of the line joining \( (-5, 7) \) and \( (0, -2) \).
\( m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{-2 - 7}{0 - (-5)} = \frac{-9}{0 + 5} = \frac{-9}{5} \).
Let \( m_2 \) be the slope of the line joining \( (1, -3) \) and \( (4, x) \).
\( m_2 = \frac{y_2-y_1}{x_2-x_1} = \frac{x - (-3)}{4 - 1} = \frac{x + 3}{3} \).
Since the two lines are perpendicular to each other, the product of their slopes must be -1.
\( m_1 \times m_2 = -1 \)
\( \implies \frac{-9}{5} \times \frac{x + 3}{3} = -1 \)
Simplify the left side:
\( \implies \frac{-3(x + 3)}{5} = -1 \)
Multiply both sides by 5:
\( \implies -3(x + 3) = -5 \)
Divide both sides by -3:
\( \implies x + 3 = \frac{-5}{-3} = \frac{5}{3} \)
Subtract 3 from both sides:
\( \implies x = \frac{5}{3} - 3 \)
To subtract, find a common denominator:
\( x = \frac{5}{3} - \frac{9}{3} \)
\( x = \frac{5 - 9}{3} = \frac{-4}{3} \).
In simple words: We found the slope of the first line and the slope of the second line (which had an unknown 'x'). Since the lines are perpendicular, we knew their slopes multiply to give -1. We then solved this equation to find the value of 'x'.

🎯 Exam Tip: When using the perpendicular slope rule, remember that \( m_1 \cdot m_2 = -1 \). Be careful with algebraic fractions and signs when solving for an unknown variable like x.

Question 13. The vertices of a quad. PMQS are P (0, 0), M (3, 2), Q (7, 7) and S (4, 5). Show that PMQS is a parallelogram.
Answer:
To show that a quadrilateral is a parallelogram, we can prove that its opposite sides have equal slopes (meaning they are parallel).
The vertices are P(0, 0), M(3, 2), Q(7, 7), and S(4, 5).
1. Slope of side PM \( (m_1) \):
\( m_{PM} = \frac{2 - 0}{3 - 0} = \frac{2}{3} \).
2. Slope of side MQ \( (m_2) \):
\( m_{MQ} = \frac{7 - 2}{7 - 3} = \frac{5}{4} \).
3. Slope of side QS \( (m_3) \):
\( m_{QS} = \frac{5 - 7}{4 - 7} = \frac{-2}{-3} = \frac{2}{3} \).
4. Slope of side SP \( (m_4) \):
\( m_{SP} = \frac{5 - 0}{4 - 0} = \frac{5}{4} \).
Comparing the slopes:
Slope of PM \( (m_{PM}) = \frac{2}{3} \) and Slope of QS \( (m_{QS}) = \frac{2}{3} \). Since \( m_{PM} = m_{QS} \), side PM is parallel to side QS.
Slope of MQ \( (m_{MQ}) = \frac{5}{4} \) and Slope of SP \( (m_{SP}) = \frac{5}{4} \). Since \( m_{MQ} = m_{SP} \), side MQ is parallel to side SP.
Since both pairs of opposite sides are parallel, the quadrilateral PMQS is a parallelogram.
In simple words: A parallelogram is a shape where opposite sides are parallel. We checked this by calculating the 'steepness' (slope) of all four sides. Since the slopes of PM and QS were the same, and the slopes of MQ and SP were the same, we proved it's a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram, you can either show that both pairs of opposite sides are parallel (using slopes), or that both pairs of opposite sides are equal in length (using distance formula), or that its diagonals bisect each other (using midpoint formula).

Question 14. Show that P (a, b), Q (a + 3, b + 4),R (a – 1, b + 7), S (a – 4, b + 3) are the vertices of a square. What is the area of the square?
Answer:
To show that the vertices P, Q, R, S form a square, we need to prove that all four sides are equal in length and that the diagonals are also equal in length. We will use the distance formula: \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
1. Length of PQ:
\( PQ = \sqrt{((a+3)-a)^2 + ((b+4)-b)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
2. Length of QR:
\( QR = \sqrt{((a-1)-(a+3))^2 + ((b+7)-(b+4))^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
3. Length of RS:
\( RS = \sqrt{((a-4)-(a-1))^2 + ((b+3)-(b+7))^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
4. Length of SP:
\( SP = \sqrt{(a-(a-4))^2 + (b-(b+3))^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Since PQ = QR = RS = SP = 5 units, all four sides are equal.
Now, let's check the lengths of the diagonals:
1. Length of diagonal PR:
\( PR = \sqrt{((a-1)-a)^2 + ((b+7)-b)^2} = \sqrt{(-1)^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} \) units.
2. Length of diagonal QS:
\( QS = \sqrt{((a-4)-(a+3))^2 + ((b+3)-(b+4))^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} \) units.
Since PR = QS = \( \sqrt{50} \) units, the diagonals are equal.
Because all sides are equal and both diagonals are equal, the quadrilateral PMQS is a square.
Area of the square \( = (\text{side})^2 = (5)^2 = 25 \) square units.
In simple words: To prove it's a square, we calculated the length of all four sides and both diagonals using the distance formula. Since all sides were equal (5 units) and both diagonals were equal ( \( \sqrt{50} \) units), it's a square. The area is simply the side length multiplied by itself.

🎯 Exam Tip: To prove a quadrilateral is a square, it's generally best to show two things: (1) all four sides are equal in length, and (2) both diagonals are equal in length. Alternatively, you could show all sides are equal and one pair of adjacent sides are perpendicular (by checking their slopes).

Question 15. Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
Answer:
To show that the points form a right-angled triangle without using the Pythagoras theorem, we can use the concept of slopes. If two sides of a triangle are perpendicular, then their slopes' product will be -1.
The vertices are A(0, 4), B(1, 2), and C(3, 3).
1. Slope of AB \( (m_{AB}) \):
\( m_{AB} = \frac{2 - 4}{1 - 0} = \frac{-2}{1} = -2 \).
2. Slope of BC \( (m_{BC}) \):
\( m_{BC} = \frac{3 - 2}{3 - 1} = \frac{1}{2} \).
3. Slope of AC \( (m_{AC}) \):
\( m_{AC} = \frac{3 - 4}{3 - 0} = \frac{-1}{3} \).
Now, let's check the product of slopes for each pair of sides:
\( m_{AB} \times m_{BC} = -2 \times \frac{1}{2} = -1 \).
Since the product of the slopes of AB and BC is -1, it means that side AB is perpendicular to side BC. Therefore, \( \triangle ABC \) is a right-angled triangle with the right angle at B.
In simple words: We calculated the 'steepness' (slope) of each side of the triangle. When we multiplied the slopes of sides AB and BC, we got -1. This special number means those two sides meet at a perfect right angle, proving it's a right-angled triangle.

🎯 Exam Tip: The quickest way to show a right-angled triangle using coordinate geometry (without Pythagoras) is by checking the slopes of all three pairs of sides. If any pair has a product of slopes equal to -1, then those sides are perpendicular, and the triangle is right-angled.

Question 16. If the points (x, 1), (1, 2) and (0, y + 1) are collinear, show that \( \frac{1}{x} + \frac{1}{y} = 1 \).
Answer:
If three points are collinear (lie on the same straight line), then the slope between any two pairs of these points must be equal.
Let A = (x, 1), B = (1, 2), and C = (0, y + 1).
1. Slope of AB \( (m_{AB}) \):
\( m_{AB} = \frac{2 - 1}{1 - x} = \frac{1}{1 - x} \).
2. Slope of BC \( (m_{BC}) \):
\( m_{BC} = \frac{(y+1) - 2}{0 - 1} = \frac{y - 1}{-1} = -(y - 1) \).
Since A, B, and C are collinear, \( m_{AB} = m_{BC} \).
\( \implies \frac{1}{1 - x} = -(y - 1) \)
\( \implies 1 = -(1 - x)(y - 1) \)
\( \implies 1 = -(y - 1 - xy + x) \)
\( \implies 1 = -y + 1 + xy - x \)
Rearrange the terms to get \( x+y = xy \):
\( x + y - 1 = -y + 1 + xy - x \) is incorrect. Let's re-distribute.
\( 1 = -(y - 1 - xy + x) \)
\( 1 = -y + 1 + xy - x \)
Add y and x to both sides:
\( x + y = 1 + xy - 1 \)
\( x + y = xy \).
Now, divide the entire equation by \( xy \) (assuming \( x \neq 0 \) and \( y \neq 0 \)):
\( \frac{x}{xy} + \frac{y}{xy} = \frac{xy}{xy} \)
\( \implies \frac{1}{y} + \frac{1}{x} = 1 \).
This shows the desired relationship.
In simple words: When three points are on the same line, the slope calculated between the first two points will be the same as the slope between the next two points. We set these slopes equal to each other and solved the equation, then did some algebra to get the final simple equation that was asked.

🎯 Exam Tip: The condition for collinearity (equal slopes) is a powerful tool. When simplifying algebraic expressions, especially with fractions and negative signs, take extra care to avoid calculation errors. Remember that dividing by \( xy \) implies \( x \neq 0 \) and \( y \neq 0 \).