Samacheer Kalvi Class 10 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.2

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1. What is the slope of a line whose inclination with positive direction of x-axis is
(i) 90°
(ii) 0°
Answer:
(i) Here \( \theta = 90^\circ \)
Slope \( m = \tan \theta \)
Slope \( m = \tan 90^\circ \)
Slope \( m \) is undefined.
When a line is perpendicular to the x-axis, its slope cannot be defined because it's a vertical line.
(ii) Here \( \theta = 0^\circ \)
Slope \( m = \tan \theta \)
Slope \( m = \tan 0^\circ \)
Slope \( m = 0 \)
A line parallel to the x-axis has an inclination of 0 degrees and therefore a slope of 0.
In simple words: The slope tells us how steep a line is. If the line goes straight up (90 degrees), its steepness is undefined. If the line is flat (0 degrees), its steepness is zero.

🎯 Exam Tip: Remember that the slope of a vertical line (inclination 90°) is undefined, and the slope of a horizontal line (inclination 0° or 180°) is 0. These are fundamental rules for slopes.

Question 2. What is the inclination of a line whose slope is
(i) 0
(ii) 1
Answer:
(i) Given, slope \( m = 0 \)
We know that \( m = \tan \theta \)
So, \( \tan \theta = 0 \)
\( \implies \theta = 0^\circ \)
When a line has a slope of zero, it means the line is completely flat and parallel to the x-axis.
(ii) Given, slope \( m = 1 \)
We know that \( m = \tan \theta \)
So, \( \tan \theta = 1 \)
\( \implies \theta = 45^\circ \)
A slope of 1 indicates that the line rises at a perfect 45-degree angle, making it equally steep in both horizontal and vertical directions.
In simple words: We are given how steep a line is (its slope) and need to find its angle (inclination). If the slope is 0, the angle is 0 degrees. If the slope is 1, the angle is 45 degrees.

🎯 Exam Tip: To find the inclination, always use the inverse tangent function (\( \arctan \)) on your calculator if the slope is not a standard value like 0, 1, \( \frac{1}{\sqrt{3}} \), \( \sqrt{3} \), or undefined.

Question 3. Find the slope of a line joining the points
(i) \( (5, \sqrt{5}) \) with the origin
(ii) \( (\sin \theta, -\cos \theta) \) and \( (-\sin \theta, \cos \theta) \)
Answer:
(i) The given points are \( (5, \sqrt{5}) \) and the origin \( (0, 0) \).
The formula for slope is \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
Slope of the line \( = \frac{0 - \sqrt{5}}{0 - 5} = \frac{-\sqrt{5}}{-5} = \frac{\sqrt{5}}{5} \)
We can simplify this by multiplying the numerator and denominator by \( \sqrt{5} \) to get \( \frac{\sqrt{5} \cdot \sqrt{5}}{5 \cdot \sqrt{5}} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}} \).
(ii) The given points are \( (\sin \theta, -\cos \theta) \) and \( (-\sin \theta, \cos \theta) \).
Slope of the line \( = \frac{\cos \theta - (-\cos \theta)}{-\sin \theta - \sin \theta} \)
\( = \frac{\cos \theta + \cos \theta}{-\sin \theta - \sin \theta} = \frac{2 \cos \theta}{-2 \sin \theta} \)
\( = -\frac{\cos \theta}{\sin \theta} = -\cot \theta \)
In simple words: To find the slope between two points, subtract the y-coordinates and divide by the difference of the x-coordinates. For trigonometric points, use the same rule and simplify the expression.

🎯 Exam Tip: Always remember the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For simplifying trigonometric expressions, recall basic identities like \( \frac{\cos \theta}{\sin \theta} = \cot \theta \).

Question 4. What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Answer:
First, find the mid-point P of the segment joining \( (4,2) \) and \( (-6,4) \).
Mid-point formula is \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)
Mid-point P \( = \left(\frac{4 + (-6)}{2}, \frac{2 + 4}{2}\right) = \left(\frac{4 - 6}{2}, \frac{6}{2}\right) = \left(\frac{-2}{2}, \frac{6}{2}\right) = (-1, 3) \)
Now, find the slope of the line joining A(5,1) and P(-1,3).
Slope of AP \( = \frac{3 - 1}{-1 - 5} = \frac{2}{-6} = -\frac{1}{3} \)
Let the slope of line AP be \( m_1 = -\frac{1}{3} \).
We need to find the slope of a line perpendicular to AP. Let this be \( m_2 \).
For perpendicular lines, the product of their slopes is -1: \( m_1 \times m_2 = -1 \).
So, \( \left(-\frac{1}{3}\right) \times m_2 = -1 \)
\( \implies m_2 = -1 \times (-3) = 3 \)
The slope of the line perpendicular to AP is 3. This means if a line cuts AP at 90 degrees, its steepness will be 3.
In simple words: First, find the middle point (P) between two given points. Then, find the steepness (slope) of the line connecting A to this middle point P. Finally, find the steepness of a line that forms a perfect corner (90 degrees) with the line AP.

🎯 Exam Tip: Always remember the two key formulas: the midpoint formula and the condition for perpendicular lines (\( m_1 m_2 = -1 \)). Make sure to calculate the midpoint correctly before finding the slope.

Question 5. Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Answer:
Let the given points be A(-3, -4), B(7, 2), and C(12, 5).
For points to be collinear, the slope of AB must be equal to the slope of BC (or AC).
Slope of AB \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-4)}{7 - (-3)} = \frac{2 + 4}{7 + 3} = \frac{6}{10} = \frac{3}{5} \)
Slope of BC \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 2}{12 - 7} = \frac{3}{5} \)
Since the slope of AB is equal to the slope of BC \( \left(\frac{3}{5}\right) \), the points A, B, and C are collinear. This means all three points lie on the same straight line.
In simple words: To check if three points are in a straight line, find the steepness (slope) between the first two points and then between the second and third points. If these steepnesses are the same, the points are in a straight line.

🎯 Exam Tip: When proving collinearity, calculate the slope for two different pairs of points that share a common point. If the slopes are equal, the points are collinear. This method is effective and direct.

Question 6. If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Answer:
Let the given points be A(3, -1), B(a, 3), and C(1, -3).
Since the points are collinear, the slope of AB must be equal to the slope of BC.
Slope of AB \( = \frac{3 - (-1)}{a - 3} = \frac{3 + 1}{a - 3} = \frac{4}{a - 3} \)
Slope of BC \( = \frac{-3 - 3}{1 - a} = \frac{-6}{1 - a} \)
Set the slopes equal to each other:
\( \frac{4}{a - 3} = \frac{-6}{1 - a} \)
Cross-multiply:
\( 4(1 - a) = -6(a - 3) \)
\( 4 - 4a = -6a + 18 \)
Move terms with 'a' to one side and constants to the other:
\( -4a + 6a = 18 - 4 \)
\( 2a = 14 \)
\( \implies a = \frac{14}{2} = 7 \)
So, the value of 'a' is 7. This value ensures that all three points lie perfectly on the same line.
In simple words: We know three points are on a straight line, but one point has a missing number ('a'). We use the fact that the steepness between any two pairs of points on a straight line must be the same. By making the steepness formulas equal, we can find the missing number 'a'.

🎯 Exam Tip: When points are collinear and an unknown variable is involved, equate the slopes of any two segments formed by these points. Carefully handle the algebra, especially with negative signs and cross-multiplication.

Question 7. The line through the points (-2, a) and (9,3) has slope \( -\frac{1}{2} \). Find the value of a.
Answer:
Let the given points be \( (x_1, y_1) = (-2, a) \) and \( (x_2, y_2) = (9, 3) \).
The slope of the line is given as \( m = -\frac{1}{2} \).
Using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\( -\frac{1}{2} = \frac{3 - a}{9 - (-2)} \)
\( -\frac{1}{2} = \frac{3 - a}{9 + 2} \)
\( -\frac{1}{2} = \frac{3 - a}{11} \)
Cross-multiply:
\( -1 \times 11 = 2 \times (3 - a) \)
\( -11 = 6 - 2a \)
Move the constant to the left side:
\( -11 - 6 = -2a \)
\( -17 = -2a \)
Divide by -2:
\( a = \frac{-17}{-2} = \frac{17}{2} \)
The value of 'a' is \( \frac{17}{2} \). This means the y-coordinate of the first point is 8.5, which gives the line its specific slope.
In simple words: We are given two points and the steepness of the line between them. One of the points has a missing 'y' number ('a'). We use the steepness formula to set up an equation and solve for 'a'.

🎯 Exam Tip: Always substitute the coordinates and the given slope into the slope formula carefully. Be extra mindful of negative signs, especially when subtracting negative numbers, to avoid calculation errors.

Question 8. The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Answer:
First, find the slope of the line joining \( (-2, 6) \) and \( (4, 8) \). Let this be \( m_1 \).
\( m_1 = \frac{8 - 6}{4 - (-2)} = \frac{2}{4 + 2} = \frac{2}{6} = \frac{1}{3} \)
Next, find the slope of the line joining \( (8, 12) \) and \( (x, 24) \). Let this be \( m_2 \).
\( m_2 = \frac{24 - 12}{x - 8} = \frac{12}{x - 8} \)
Since the two lines are perpendicular, the product of their slopes is -1.
\( m_1 \times m_2 = -1 \)
\( \frac{1}{3} \times \frac{12}{x - 8} = -1 \)
\( \frac{12}{3(x - 8)} = -1 \)
\( \frac{4}{x - 8} = -1 \)
Multiply both sides by \( (x - 8) \):
\( 4 = -1(x - 8) \)
\( 4 = -x + 8 \)
Move \( -x \) to the left side and 4 to the right side:
\( x = 8 - 4 \)
\( x = 4 \)
The value of x is 4. This ensures that the second line is perpendicular to the first, creating a 90-degree angle between them.
In simple words: We have two lines. We find the steepness of both lines. Because they cross at a perfect right angle, their steepnesses, when multiplied, must equal -1. Using this rule, we can find the missing number 'x' for the second line.

🎯 Exam Tip: The condition for perpendicular lines (\( m_1 m_2 = -1 \)) is crucial here. Be careful when simplifying the fractions and solving the linear equation for 'x'.

Question 9. Show that the given points form a right-angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4), B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Answer:
(i) Given points are A(1, -4), B(2, -3), and C(4, -7).
To show it's a right-angled triangle using slopes, we check if any two sides are perpendicular.
Slope of AB \( = m_{AB} = \frac{-3 - (-4)}{2 - 1} = \frac{-3 + 4}{1} = \frac{1}{1} = 1 \)
Slope of BC \( = m_{BC} = \frac{-7 - (-3)}{4 - 2} = \frac{-7 + 3}{2} = \frac{-4}{2} = -2 \)
Slope of AC \( = m_{AC} = \frac{-7 - (-4)}{4 - 1} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1 \)
Check for perpendicularity: \( m_{AB} \times m_{AC} = 1 \times (-1) = -1 \).
Since \( m_{AB} \times m_{AC} = -1 \), line AB is perpendicular to line AC. This means the angle at A is \( 90^\circ \).
Thus, \( \triangle ABC \) is a right-angled triangle.
**Verification using Distance Formula (Pythagoras Theorem):**
Distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
AB \( = \sqrt{(2 - 1)^2 + (-3 - (-4))^2} = \sqrt{(1)^2 + (-3 + 4)^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \)
BC \( = \sqrt{(4 - 2)^2 + (-7 - (-3))^2} = \sqrt{(2)^2 + (-7 + 3)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \)
AC \( = \sqrt{(4 - 1)^2 + (-7 - (-4))^2} = \sqrt{(3)^2 + (-7 + 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \)
According to Pythagoras theorem, in a right-angled triangle, \( \text{hypotenuse}^2 = \text{side}_1^2 + \text{side}_2^2 \).
Here, BC is the longest side, so it should be the hypotenuse.
\( BC^2 = (\sqrt{20})^2 = 20 \)
\( AB^2 + AC^2 = (\sqrt{2})^2 + (\sqrt{18})^2 = 2 + 18 = 20 \)
Since \( BC^2 = AB^2 + AC^2 \) (i.e., \( 20 = 20 \)), Pythagoras theorem is verified. The triangle indeed has a right angle.

(ii) Given points are L(0, 5), M(9, 12), and N(3, 14).
Slope of LM \( = m_{LM} = \frac{12 - 5}{9 - 0} = \frac{7}{9} \)
Slope of MN \( = m_{MN} = \frac{14 - 12}{3 - 9} = \frac{2}{-6} = -\frac{1}{3} \)
Slope of LN \( = m_{LN} = \frac{14 - 5}{3 - 0} = \frac{9}{3} = 3 \)
Check for perpendicularity:
\( m_{MN} \times m_{LN} = \left(-\frac{1}{3}\right) \times 3 = -1 \).
Since \( m_{MN} \times m_{LN} = -1 \), line MN is perpendicular to line LN. This means the angle at N is \( 90^\circ \).
Thus, \( \triangle LMN \) is a right-angled triangle.
**Verification using Distance Formula (Pythagoras Theorem):**
LM \( = \sqrt{(9 - 0)^2 + (12 - 5)^2} = \sqrt{9^2 + 7^2} = \sqrt{81 + 49} = \sqrt{130} \)
MN \( = \sqrt{(3 - 9)^2 + (14 - 12)^2} = \sqrt{(-6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} \)
LN \( = \sqrt{(3 - 0)^2 + (14 - 5)^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} \)
According to Pythagoras theorem, \( \text{hypotenuse}^2 = \text{side}_1^2 + \text{side}_2^2 \).
Here, LM is the longest side, so it should be the hypotenuse.
\( LM^2 = (\sqrt{130})^2 = 130 \)
\( MN^2 + LN^2 = (\sqrt{40})^2 + (\sqrt{90})^2 = 40 + 90 = 130 \)
Since \( LM^2 = MN^2 + LN^2 \) (i.e., \( 130 = 130 \)), Pythagoras theorem is verified. The triangle is indeed a right-angled triangle.
In simple words: To show a triangle has a right angle, we can check two things. First, find the steepness of each side; if two sides have slopes that multiply to -1, they form a right angle. Second, use the distance formula to find the length of each side. Then, check if the square of the longest side is equal to the sum of the squares of the other two sides. If both checks work, it's a right-angled triangle.

🎯 Exam Tip: You can prove a right-angled triangle using either slopes (if two slopes multiply to -1) or the distance formula (Pythagoras theorem). For full marks, often both methods are expected, or one explicitly asked for. Ensure calculations for both slopes and distances are accurate.

Question 10. Show that the given points A(2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) form a parallelogram.
Answer:
Let the vertices of the quadrilateral be A(2.5, 3.5), B(10, -4), C(2.5, -2.5), and D(-5, 5).
A quadrilateral is a parallelogram if its opposite sides are parallel. We can check this by comparing the slopes of opposite sides.
Slope of AB \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 3.5}{10 - 2.5} = \frac{-7.5}{7.5} = -1 \)
Slope of CD \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-2.5)}{-5 - 2.5} = \frac{5 + 2.5}{-7.5} = \frac{7.5}{-7.5} = -1 \)
Since Slope of AB = Slope of CD, AB is parallel to CD.

Slope of BC \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2.5 - (-4)}{2.5 - 10} = \frac{-2.5 + 4}{-7.5} = \frac{1.5}{-7.5} = -\frac{15}{75} = -\frac{1}{5} \)
Slope of AD \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3.5}{-5 - 2.5} = \frac{1.5}{-7.5} = -\frac{15}{75} = -\frac{1}{5} \)
Since Slope of BC = Slope of AD, BC is parallel to AD.
Because both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram. This means its opposite sides are equal in length and run in the same direction.
In simple words: To show that a shape with four corners (a quadrilateral) is a parallelogram, we need to prove that its opposite sides are parallel. We do this by calculating the steepness (slope) of each side. If the steepness of side AB is the same as side CD, and the steepness of side BC is the same as side AD, then it's a parallelogram.

🎯 Exam Tip: The most straightforward way to prove a parallelogram using coordinates is to show that both pairs of opposite sides have equal slopes. This immediately confirms they are parallel.

Question 11. If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Answer:
Let the vertices of the parallelogram be A(2, 2), B(-2, -3), C(1, -3), and D(x, y).
In a parallelogram, opposite sides are parallel, so their slopes must be equal.
Slope of AB \( = \frac{-3 - 2}{-2 - 2} = \frac{-5}{-4} = \frac{5}{4} \)
Slope of BC \( = \frac{-3 - (-3)}{1 - (-2)} = \frac{-3 + 3}{1 + 2} = \frac{0}{3} = 0 \)
Slope of CD \( = \frac{y - (-3)}{x - 1} = \frac{y + 3}{x - 1} \)
Slope of AD \( = \frac{y - 2}{x - 2} \)

Since ABCD is a parallelogram, Slope of AB = Slope of CD.
\( \frac{5}{4} = \frac{y + 3}{x - 1} \)
Cross-multiply:
\( 5(x - 1) = 4(y + 3) \)
\( 5x - 5 = 4y + 12 \)
\( 5x - 4y = 12 + 5 \)
\( 5x - 4y = 17 \) (Equation 1)

Also, Slope of BC = Slope of AD.
\( 0 = \frac{y - 2}{x - 2} \)
For a fraction to be zero, its numerator must be zero (as long as the denominator is not zero).
\( y - 2 = 0 \)
\( \implies y = 2 \)
Now substitute the value of \( y = 2 \) into Equation 1:
\( 5x - 4(2) = 17 \)
\( 5x - 8 = 17 \)
\( 5x = 17 + 8 \)
\( 5x = 25 \)
\( \implies x = \frac{25}{5} = 5 \)
So, the value of x is 5 and the value of y is 2. This makes point D(5,2), completing the parallelogram.
In simple words: We are given three points of a parallelogram and need to find the fourth point (x, y). We use the property that opposite sides of a parallelogram have the same steepness (slope). By setting the slopes of opposite sides equal, we get equations that help us find 'x' and 'y'.

🎯 Exam Tip: When finding unknown coordinates of a parallelogram, always remember that opposite sides have equal slopes. Setting up two equations (one for each pair of opposite sides) and solving them simultaneously is the standard approach.

Question 12. Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Answer:
Let the vertices of the quadrilateral be A(3, -4), B(9, -4), C(5, -7), and D(7, -7).
A quadrilateral is a trapezium if at least one pair of opposite sides is parallel.
Slope of AB \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - (-4)}{9 - 3} = \frac{-4 + 4}{6} = \frac{0}{6} = 0 \)
Slope of CD \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - (-7)}{7 - 5} = \frac{-7 + 7}{2} = \frac{0}{2} = 0 \)
Since Slope of AB = Slope of CD (both are 0), AB is parallel to CD.

Now, let's check the other pair of opposite sides.
Slope of BC \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - (-4)}{5 - 9} = \frac{-7 + 4}{-4} = \frac{-3}{-4} = \frac{3}{4} \)
Slope of AD \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - (-4)}{7 - 3} = \frac{-7 + 4}{4} = \frac{-3}{4} = -\frac{3}{4} \)
Since Slope of BC \( \left(\frac{3}{4}\right) \) is not equal to Slope of AD \( \left(-\frac{3}{4}\right) \), BC is not parallel to AD.
Because exactly one pair of opposite sides (AB and CD) is parallel, the quadrilateral ABCD is a trapezium. This means two of its sides are parallel, while the other two are not.
In simple words: To prove a four-sided shape is a trapezium, we need to show that at least one pair of its opposite sides runs in the same direction (is parallel). We calculate the steepness (slope) of each side. If one pair of opposite sides has the same steepness, then it's a trapezium.

🎯 Exam Tip: The key definition of a trapezium is having at least one pair of parallel sides. Therefore, calculating the slopes of both pairs of opposite sides and identifying one pair with equal slopes is sufficient proof.

Question 13. A quadrilateral has vertices at A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Answer:
Let the vertices of the quadrilateral be A(-4, -2), B(5, -1), C(6, 5), and D(-7, 6).
First, find the mid-points of each side:
Mid-point of AB (E) \( = \left(\frac{-4 + 5}{2}, \frac{-2 + (-1)}{2}\right) = \left(\frac{1}{2}, \frac{-3}{2}\right) \)
Mid-point of BC (F) \( = \left(\frac{5 + 6}{2}, \frac{-1 + 5}{2}\right) = \left(\frac{11}{2}, \frac{4}{2}\right) = \left(\frac{11}{2}, 2\right) \)
Mid-point of CD (G) \( = \left(\frac{6 + (-7)}{2}, \frac{5 + 6}{2}\right) = \left(\frac{-1}{2}, \frac{11}{2}\right) \)
Mid-point of AD (H) \( = \left(\frac{-4 + (-7)}{2}, \frac{-2 + 6}{2}\right) = \left(\frac{-11}{2}, \frac{4}{2}\right) = \left(\frac{-11}{2}, 2\right) \)
Now, we have a new quadrilateral EFGH with vertices E\( \left(\frac{1}{2}, -\frac{3}{2}\right) \), F\( \left(\frac{11}{2}, 2\right) \), G\( \left(-\frac{1}{2}, \frac{11}{2}\right) \), and H\( \left(-\frac{11}{2}, 2\right) \).
To show EFGH is a parallelogram, we check if its opposite sides are parallel (i.e., have equal slopes).
Slope of EF \( = \frac{2 - (-\frac{3}{2})}{\frac{11}{2} - \frac{1}{2}} = \frac{\frac{4}{2} + \frac{3}{2}}{\frac{10}{2}} = \frac{\frac{7}{2}}{5} = \frac{7}{10} \)
Slope of GH \( = \frac{2 - \frac{11}{2}}{-\frac{11}{2} - (-\frac{1}{2})} = \frac{\frac{4}{2} - \frac{11}{2}}{-\frac{11}{2} + \frac{1}{2}} = \frac{-\frac{7}{2}}{-\frac{10}{2}} = \frac{7}{10} \)
Since Slope of EF = Slope of GH \( \left(\frac{7}{10}\right) \), EF is parallel to GH. (1)

Slope of FG \( = \frac{\frac{11}{2} - 2}{-\frac{1}{2} - \frac{11}{2}} = \frac{\frac{11}{2} - \frac{4}{2}}{-\frac{12}{2}} = \frac{\frac{7}{2}}{-6} = -\frac{7}{12} \)
Slope of EH \( = \frac{2 - (-\frac{3}{2})}{-\frac{11}{2} - \frac{1}{2}} = \frac{\frac{4}{2} + \frac{3}{2}}{-\frac{12}{2}} = \frac{\frac{7}{2}}{-6} = -\frac{7}{12} \)
Since Slope of FG = Slope of EH \( \left(-\frac{7}{12}\right) \), FG is parallel to EH. (2)
From (1) and (2), both pairs of opposite sides of quadrilateral EFGH are parallel. Therefore, EFGH is a parallelogram. This demonstrates a geometric property that connects any quadrilateral to a parallelogram formed by its midpoints.
In simple words: We start with a four-sided shape. First, we find the exact middle point of each of its four sides. These four middle points become the corners of a new, inner shape. Then, we check if this new inner shape is a parallelogram by seeing if its opposite sides are parallel (have the same steepness). It turns out that they always are, forming a parallelogram.

🎯 Exam Tip: This question proves a well-known theorem in coordinate geometry. To solve it, systematically find all four midpoints using the midpoint formula, then calculate the slopes of all four sides of the newly formed quadrilateral. Ensure accurate fraction arithmetic throughout.

Question 14. PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Answer:
Let the coordinates of P be (a, b).
In a rhombus, the diagonals bisect each other at right angles. This means M is the midpoint of both PR and QS, and PM is perpendicular to SM.
Coordinates of S = (1, 1)
Coordinates of M = (2, -1)
Coordinates of P = (a, b)

First, let's use the fact that PM is perpendicular to SM.
Slope of SM \( = \frac{-1 - 1}{2 - 1} = \frac{-2}{1} = -2 \)
Since PM \( \perp \) SM, the slope of PM \( = -\frac{1}{\text{Slope of SM}} = -\frac{1}{-2} = \frac{1}{2} \)
Now, use the coordinates of P(a,b) and M(2,-1) to write the slope of PM:
Slope of PM \( = \frac{-1 - b}{2 - a} \)
Equating the slopes:
\( \frac{-1 - b}{2 - a} = \frac{1}{2} \)
Cross-multiply:
\( 2(-1 - b) = 1(2 - a) \)
\( -2 - 2b = 2 - a \)
\( a - 2b = 4 \) (Equation 1)

Next, use the condition QS = 2PR, which implies SM = 2PM (since M is the midpoint of both, QS = 2SM and PR = 2PM).
Distance SM \( = \sqrt{(2 - 1)^2 + (-1 - 1)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \)
Distance PM \( = \sqrt{(a - 2)^2 + (b - (-1))^2} = \sqrt{(a - 2)^2 + (b + 1)^2} \)
Given SM = 2PM:
\( \sqrt{5} = 2\sqrt{(a - 2)^2 + (b + 1)^2} \)
Square both sides:
\( 5 = 4((a - 2)^2 + (b + 1)^2) \) (Equation 2)

From Equation 1, \( a = 4 + 2b \). Substitute this into Equation 2:
\( 5 = 4((4 + 2b - 2)^2 + (b + 1)^2) \)
\( 5 = 4((2 + 2b)^2 + (b + 1)^2) \)
\( 5 = 4( (2(1 + b))^2 + (b + 1)^2 ) \)
\( 5 = 4( 4(1 + b)^2 + (b + 1)^2 ) \)
\( 5 = 4( 5(b + 1)^2 ) \)
\( 5 = 20(b + 1)^2 \)
Divide by 20:
\( \frac{5}{20} = (b + 1)^2 \)
\( \frac{1}{4} = (b + 1)^2 \)
Take the square root of both sides:
\( b + 1 = \pm\sqrt{\frac{1}{4}} \)
\( b + 1 = \pm\frac{1}{2} \)

Case 1: \( b + 1 = \frac{1}{2} \)
\( b = \frac{1}{2} - 1 = -\frac{1}{2} \)
Substitute \( b = -\frac{1}{2} \) into \( a = 4 + 2b \):
\( a = 4 + 2\left(-\frac{1}{2}\right) = 4 - 1 = 3 \)
So, \( P = \left(3, -\frac{1}{2}\right) \)

Case 2: \( b + 1 = -\frac{1}{2} \)
\( b = -\frac{1}{2} - 1 = -\frac{3}{2} \)
Substitute \( b = -\frac{3}{2} \) into \( a = 4 + 2b \):
\( a = 4 + 2\left(-\frac{3}{2}\right) = 4 - 3 = 1 \)
So, \( P = \left(1, -\frac{3}{2}\right) \)

There are two possible coordinates for P: \( \left(3, -\frac{1}{2}\right) \) or \( \left(1, -\frac{3}{2}\right) \). These two points complete the rhombus based on the given conditions and ensure its specific diagonal properties.
In simple words: We know a rhombus has special diagonal properties: they cut each other in half at right angles. Also, one diagonal is twice the length of the other. We use these rules with the given points (S and M) and a missing point (P) to set up equations. We find the steepness of SM and then PM (which is at a right angle). We also use the distance formula for SM and PM, linking them with the "QS = 2PR" rule. Solving these equations helps us find the 'x' and 'y' values for point P.

🎯 Exam Tip: For rhombus problems, remember two key properties: diagonals bisect each other at right angles (\( m_1 m_2 = -1 \) for the slopes of the diagonals if they are sides of the triangles, or the two perpendicular segments at intersection), and the midpoint of one diagonal is the midpoint of the other. The length relationship QS = 2PR translates to SM = 2PM since M is the midpoint. This problem requires solving a system of equations, one from the slope condition and one from the distance condition.

Question 1. What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Answer:
(i) Here, the inclination \( \theta = 90^\circ \). The slope (m) of a line is given by \( \text{m} = \tan \theta \). Since the tangent of 90 degrees is undefined, the slope of the line is undefined. This means the line is a vertical line.
(ii) Here, the inclination \( \theta = 0^\circ \). The slope (m) of a line is given by \( \text{m} = \tan \theta \). The tangent of 0 degrees is 0, so the slope of the line is 0. This indicates the line is a horizontal line.
In simple words: If a line is perfectly straight up and down (vertical), its slope is impossible to define. If a line is perfectly flat (horizontal), its slope is zero.

🎯 Exam Tip: Remember that vertical lines (like x=c) have an undefined slope, while horizontal lines (like y=c) have a slope of 0. Always link the angle of inclination directly to the tangent function.

Question 2. What is the inclination of a line whose slope is
(i) 0
(ii) 1
Answer:
(i) Given that the slope \( m = 0 \). We know that \( m = \tan \theta \).
So, \( \tan \theta = 0 \)
\( \implies \theta = 0^\circ \). The inclination of the line is 0 degrees. This means the line is perfectly horizontal.
(ii) Given that the slope \( m = 1 \). We know that \( m = \tan \theta \).
So, \( \tan \theta = 1 \)
\( \implies \theta = 45^\circ \). The inclination of the line is 45 degrees. This angle means the line rises at a 45-degree angle from the positive x-axis.
In simple words: If the slope is 0, the line is flat. If the slope is 1, the line goes up at a 45-degree angle.

🎯 Exam Tip: When finding the inclination from the slope, be careful with negative slopes. A negative slope means the angle of inclination is obtuse (between 90° and 180°), which often requires using the reference angle.

Question 3. Find the slope of a line joining the points
(i) \( (5,\sqrt { 5 }) \) with the origin
(ii) \( (\sin \theta, -\cos \theta) \) and \( (-\sin \theta, \cos \theta) \)
Answer:
(i) The given points are \( (x_1, y_1) = (5, \sqrt{5}) \) and the origin \( (x_2, y_2) = (0, 0) \).
The slope of a line is calculated using the formula \( m = \frac{y_2-y_1}{x_2-x_1} \).
Substituting the values, we get: \( m = \frac{0-\sqrt{5}}{0-5} = \frac{-\sqrt{5}}{-5} = \frac{\sqrt{5}}{5} \).
This can be simplified further: \( m = \frac{\sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{1}{\sqrt{5}} \). The slope is a positive value, indicating an upward-sloping line.
(ii) The given points are \( (x_1, y_1) = (\sin \theta, -\cos \theta) \) and \( (x_2, y_2) = (-\sin \theta, \cos \theta) \).
Using the slope formula: \( m = \frac{y_2-y_1}{x_2-x_1} \).
Substituting the values, we get: \( m = \frac{\cos \theta - (-\cos \theta)}{-\sin \theta - \sin \theta} = \frac{\cos \theta + \cos \theta}{-\sin \theta - \sin \theta} = \frac{2 \cos \theta}{-2 \sin \theta} \).
Simplifying the expression, we get: \( m = -\frac{\cos \theta}{\sin \theta} = -\cot \theta \). The slope depends on the angle \( \theta \).
In simple words: For part (i), we use the coordinates of the point and the origin (0,0) to find the slope. For part (ii), we use trigonometry values in the slope formula and simplify it.

🎯 Exam Tip: Remember the slope formula \( m = \frac{y_2-y_1}{x_2-x_1} \). Be careful with signs when substituting coordinates, especially with negative values or trigonometric expressions. Simplify the final fraction if possible.

Question 4. What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Answer:
First, we need to find the coordinates of point P. P is the mid-point of the segment joining X(4,2) and Y(-6,4).
The midpoint formula is \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
So, \( P = \left(\frac{4+(-6)}{2}, \frac{2+4}{2}\right) = \left(\frac{4-6}{2}, \frac{6}{2}\right) = \left(\frac{-2}{2}, \frac{6}{2}\right) = (-1, 3) \).
Now we have point A(5,1) and point P(-1,3). Let's find the slope of the line AP, let's call it \( m_1 \).
\( m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{3-1}{-1-5} = \frac{2}{-6} = -\frac{1}{3} \).
We need to find the slope of a line perpendicular to line AP. If two lines are perpendicular, the product of their slopes is -1. Let the slope of the perpendicular line be \( m_2 \).
\( m_1 \times m_2 = -1 \)
\( -\frac{1}{3} \times m_2 = -1 \)
\( \implies m_2 = -1 \times (-3) = 3 \).
Therefore, the slope of the line perpendicular to line AP is 3. This indicates a line that rises steeply.
In simple words: First, find the middle point (P) of the two given points. Then find the slope of the line connecting A to P. Finally, find the slope of a line that would cross this line at a perfect right angle.

🎯 Exam Tip: Always remember the two key formulas: the midpoint formula and the slope formula. For perpendicular lines, the product of their slopes is -1. For parallel lines, their slopes are equal.

Question 5. Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Answer:
Let the given points be \( A(-3, -4) \), \( B(7, 2) \), and \( C(12, 5) \).
For points to be collinear (lie on the same straight line), the slope between any two pairs of points must be equal.
First, calculate the slope of line segment AB:
\( \text{Slope of AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{2-(-4)}{7-(-3)} = \frac{2+4}{7+3} = \frac{6}{10} = \frac{3}{5} \).
Next, calculate the slope of line segment BC:
\( \text{Slope of BC} = \frac{y_2-y_1}{x_2-x_1} = \frac{5-2}{12-7} = \frac{3}{5} \).
Since the slope of AB is equal to the slope of BC (both are \( \frac{3}{5} \)), the points A, B, and C are collinear. If the slopes were different, the points would form a triangle instead of a straight line.
In simple words: To check if points are on the same line, calculate the "steepness" (slope) between the first two points and then between the next two points. If the steepness is the same, they are on one straight line.

🎯 Exam Tip: To prove collinearity using slopes, always calculate the slopes of two different segments formed by the three points (e.g., AB and BC, or AB and AC). If they are equal, the points are collinear.

Question 6. If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Answer:
Let the given points be \( A(3, -1) \), \( B(a, 3) \), and \( C(1, -3) \).
Since the points are collinear, the slope of AB must be equal to the slope of BC.
First, find the slope of AB:
\( \text{Slope of AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{3-(-1)}{a-3} = \frac{3+1}{a-3} = \frac{4}{a-3} \).
Next, find the slope of BC:
\( \text{Slope of BC} = \frac{y_2-y_1}{x_2-x_1} = \frac{-3-3}{1-a} = \frac{-6}{1-a} \). (Note: The source used (3+3)/(a-1) which is for C and B coordinates swapped and an error, I will use C(1,-3) and B(a,3)) Wait, the source calculation for `Slope of BC = \frac { 3+3 }{ a-1 }` implies B=(a,3) and C=(1,-3) and they did (y_b - y_c) / (x_b - x_c) = (3 - (-3)) / (a - 1) = (3+3)/(a-1) = 6/(a-1). I will follow the source's calculation method to remain consistent with Iron Rule 6. So, using B(a,3) and C(1,-3) as `(x_1, y_1)` and `(x_2, y_2)` respectively (reversed from my initial thought, but matching source calculation structure): \( \text{Slope of BC} = \frac{y_c-y_b}{x_c-x_b} = \frac{-3-3}{1-a} = \frac{-6}{1-a} \). If I follow the source exactly: `Slope of BC = \frac { 3 - (-3) }{ a-1 } = \frac { 6 }{ a-1 }`. This implies B(a,3) and C(1,-3) with formula \( (y_B-y_C) / (x_B-x_C) \). Let's use the points in the given order for AB and BC: \( \text{Slope of AB} = \frac{3-(-1)}{a-3} = \frac{4}{a-3} \). \( \text{Slope of BC} = \frac{-3-3}{1-a} = \frac{-6}{1-a} \). Now set the slopes equal: \( \frac{4}{a-3} = \frac{-6}{1-a} \)
\( 4(1-a) = -6(a-3) \)
\( 4 - 4a = -6a + 18 \)
\( -4a + 6a = 18 - 4 \)
\( 2a = 14 \)
\( \implies a = \frac{14}{2} = 7 \). Thus, the value of 'a' is 7. This value ensures that all three points lie on the same straight line.
In simple words: Because all three points are on the same line, the steepness from the first point to the second must be the same as the steepness from the second point to the third. We set up an equation with 'a' and solve for it.

🎯 Exam Tip: When points are collinear, the slope calculated using any two pairs of points will be the same. This property is crucial for solving problems where an unknown coordinate needs to be found to satisfy collinearity.

Question 7. The line through the points (-2, a) and (9,3) has slope \( -\frac { 1 }{ 2 } \) Find the value of a.
Answer:
The given points are \( (x_1, y_1) = (-2, a) \) and \( (x_2, y_2) = (9, 3) \).
The given slope is \( m = -\frac{1}{2} \).
Using the slope formula: \( m = \frac{y_2-y_1}{x_2-x_1} \).
Substitute the known values:
\( -\frac{1}{2} = \frac{3-a}{9-(-2)} \)
\( -\frac{1}{2} = \frac{3-a}{9+2} \)
\( \implies -\frac{1}{2} = \frac{3-a}{11} \).
Now, cross-multiply to solve for 'a':
\( -1 \times 11 = 2 \times (3-a) \)
\( -11 = 6 - 2a \)
\( 2a = 6 + 11 \)
\( 2a = 17 \)
\( \implies a = \frac{17}{2} \).
So, the value of 'a' is \( \frac{17}{2} \). This means the y-coordinate of the first point is 8.5.
In simple words: We are given two points and the slope of the line that goes through them. We use the slope formula, put in all the numbers we know, and then solve to find the missing 'a' value.

🎯 Exam Tip: Double-check your algebraic steps, especially when dealing with negative numbers and fractions. A common mistake is an error in cross-multiplication or sign manipulation.

Question 8. The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Answer:
Let the first line pass through \( A(-2, 6) \) and \( B(4, 8) \). Its slope is \( m_1 \).
\( m_1 = \frac{8-6}{4-(-2)} = \frac{2}{4+2} = \frac{2}{6} = \frac{1}{3} \).
Let the second line pass through \( C(8, 12) \) and \( D(x, 24) \). Its slope is \( m_2 \).
\( m_2 = \frac{24-12}{x-8} = \frac{12}{x-8} \).
Since the two lines are perpendicular, the product of their slopes is -1:
\( m_1 \times m_2 = -1 \)
\( \frac{1}{3} \times \frac{12}{x-8} = -1 \)
\( \implies \frac{12}{3(x-8)} = -1 \)
\( \frac{4}{x-8} = -1 \).
Now, cross-multiply:
\( 4 = -1(x-8) \)
\( 4 = -x + 8 \)
\( x = 8 - 4 \)
\( \implies x = 4 \).
So, the value of 'x' is 4. This means the second point on the second line is (4, 24).
In simple words: First, find the steepness of the first line. Then, find an expression for the steepness of the second line using 'x'. Because the lines cross at a right angle, multiply their steepnesses and set the answer to -1. Then, solve to find 'x'.

🎯 Exam Tip: Remember to apply the correct condition for perpendicular lines (\( m_1 m_2 = -1 \)). Make sure to handle the algebraic rearrangement carefully, especially when the unknown is in the denominator.

Question 9. Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4), B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Answer:
(i) Given points are \( A(1, -4) \), \( B(2, -3) \), and \( C(4, -7) \).
To show it's a right-angled triangle, we check if the product of slopes of any two sides is -1.
Slope of AB \( (m_{AB}) = \frac{-3-(-4)}{2-1} = \frac{-3+4}{1} = \frac{1}{1} = 1 \).
Slope of BC \( (m_{BC}) = \frac{-7-(-3)}{4-2} = \frac{-7+3}{2} = \frac{-4}{2} = -2 \).
Slope of AC \( (m_{AC}) = \frac{-7-(-4)}{4-1} = \frac{-7+4}{3} = \frac{-3}{3} = -1 \).
Check for perpendicularity:
\( m_{AB} \times m_{AC} = 1 \times (-1) = -1 \).
Since the product of slopes of AB and AC is -1, AB is perpendicular to AC. This means \( \angle A = 90^\circ \).
Therefore, \( \triangle \text{ABC} \) is a right-angled triangle.
Verification using Pythagoras theorem (Distance Formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)):
\( \text{AB} = \sqrt{(2-1)^2+(-3-(-4))^2} = \sqrt{1^2+(-3+4)^2} = \sqrt{1^2+1^2} = \sqrt{1+1} = \sqrt{2} \).
\( \text{BC} = \sqrt{(4-2)^2+(-7-(-3))^2} = \sqrt{2^2+(-7+3)^2} = \sqrt{2^2+(-4)^2} = \sqrt{4+16} = \sqrt{20} \).
\( \text{AC} = \sqrt{(4-1)^2+(-7-(-4))^2} = \sqrt{3^2+(-7+4)^2} = \sqrt{3^2+(-3)^2} = \sqrt{9+9} = \sqrt{18} \).
According to Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides. Here, BC is the hypotenuse.
\( \text{BC}^2 = \text{AB}^2 + \text{AC}^2 \)
\( (\sqrt{20})^2 = (\sqrt{2})^2 + (\sqrt{18})^2 \)
\( 20 = 2 + 18 \)
\( 20 = 20 \)
\( \implies \text{Pythagoras theorem is verified} \). This confirms the triangle is indeed right-angled.
(ii) Given points are \( L(0, 5) \), \( M(9, 12) \), and \( N(3, 14) \).
Slope of LM \( (m_{LM}) = \frac{12-5}{9-0} = \frac{7}{9} \).
Slope of MN \( (m_{MN}) = \frac{14-12}{3-9} = \frac{2}{-6} = -\frac{1}{3} \).
Slope of LN \( (m_{LN}) = \frac{14-5}{3-0} = \frac{9}{3} = 3 \).
Check for perpendicularity:
\( m_{MN} \times m_{LN} = -\frac{1}{3} \times 3 = -1 \).
Since the product of slopes of MN and LN is -1, MN is perpendicular to LN. This means \( \angle N = 90^\circ \).
Therefore, \( \triangle \text{LMN} \) is a right-angled triangle.
Verification using Pythagoras theorem:
\( \text{LM} = \sqrt{(9-0)^2+(12-5)^2} = \sqrt{9^2+7^2} = \sqrt{81+49} = \sqrt{130} \).
\( \text{MN} = \sqrt{(3-9)^2+(14-12)^2} = \sqrt{(-6)^2+2^2} = \sqrt{36+4} = \sqrt{40} \).
\( \text{LN} = \sqrt{(3-0)^2+(14-5)^2} = \sqrt{3^2+9^2} = \sqrt{9+81} = \sqrt{90} \).
According to Pythagoras theorem, LM is the hypotenuse.
\( \text{LM}^2 = \text{MN}^2 + \text{LN}^2 \)
\( (\sqrt{130})^2 = (\sqrt{40})^2 + (\sqrt{90})^2 \)
\( 130 = 40 + 90 \)
\( 130 = 130 \)
\( \implies \text{Pythagoras theorem is verified} \). This also confirms the triangle is right-angled.
In simple words: For each set of points, first find the steepness (slope) of each side. If any two sides have slopes that multiply to -1, then those sides form a right angle. Then, use the distance formula to find the length of each side. Check if the square of the longest side equals the sum of the squares of the other two sides (Pythagoras theorem).

🎯 Exam Tip: To prove a right-angled triangle, you can either show that two sides are perpendicular (product of slopes is -1) or verify the Pythagorean theorem using distance formula. Both methods lead to the same conclusion and can act as a cross-check.

Question 10. Show that the given points form a parallelogram: A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Answer:
To show that the points \( A(2.5, 3.5) \), \( B(10, -4) \), \( C(2.5, -2.5) \), and \( D(-5, 5) \) form a parallelogram, we need to show that opposite sides are parallel. This means their slopes must be equal.
Slope of AB: \( m_{AB} = \frac{-4-3.5}{10-2.5} = \frac{-7.5}{7.5} = -1 \).
Slope of CD: \( m_{CD} = \frac{5-(-2.5)}{-5-2.5} = \frac{5+2.5}{-5-2.5} = \frac{7.5}{-7.5} = -1 \).
Since \( m_{AB} = m_{CD} = -1 \), line AB is parallel to line CD \( (\text{AB} \parallel \text{CD}) \).
Slope of BC: \( m_{BC} = \frac{-2.5-(-4)}{2.5-10} = \frac{-2.5+4}{-7.5} = \frac{1.5}{-7.5} = -\frac{15}{75} = -\frac{1}{5} \).
Slope of AD: \( m_{AD} = \frac{5-3.5}{-5-2.5} = \frac{1.5}{-7.5} = -\frac{15}{75} = -\frac{1}{5} \).
Since \( m_{BC} = m_{AD} = -\frac{1}{5} \), line BC is parallel to line AD \( (\text{BC} \parallel \text{AD}) \).
Because both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram. A parallelogram is a fundamental shape in geometry with many useful properties.
In simple words: To prove it's a parallelogram, calculate the steepness (slope) of all four sides. If the opposite sides have the same steepness, then the shape is a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram, the most common methods involve showing 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 diagonals bisect each other (using midpoint formula).

Question 11. If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Answer:
Let the vertices of the parallelogram be \( A(2, 2) \), \( B(-2, -3) \), \( C(1, -3) \), and \( D(x, y) \).
In a parallelogram, opposite sides are parallel, which means their slopes are equal.
Slope of AB \( (m_{AB}) = \frac{-3-2}{-2-2} = \frac{-5}{-4} = \frac{5}{4} \).
Slope of BC \( (m_{BC}) = \frac{-3-(-3)}{1-(-2)} = \frac{-3+3}{1+2} = \frac{0}{3} = 0 \).
Slope of CD \( (m_{CD}) = \frac{y-(-3)}{x-1} = \frac{y+3}{x-1} \).
Slope of AD \( (m_{AD}) = \frac{y-2}{x-2} \).
Since ABCD is a parallelogram, \( \text{AB} \parallel \text{CD} \) and \( \text{BC} \parallel \text{AD} \).
Therefore, \( m_{AB} = m_{CD} \):
\( \frac{5}{4} = \frac{y+3}{x-1} \)
\( 5(x-1) = 4(y+3) \)
\( 5x-5 = 4y+12 \)
\( 5x-4y = 12+5 \)
\( 5x-4y = 17 \quad \text{(Equation 1)} \).
Also, \( m_{BC} = m_{AD} \):
\( 0 = \frac{y-2}{x-2} \).
For this fraction to be zero, the numerator must be zero:
\( y-2 = 0 \)
\( y = 2 \).
Now, substitute the value of \( y=2 \) into Equation 1:
\( 5x-4(2) = 17 \)
\( 5x-8 = 17 \)
\( \implies 5x = 17+8 \)
\( 5x = 25 \)
\( \implies x = \frac{25}{5} = 5 \).
Therefore, the value of \( x = 5 \) and \( y = 2 \). This completes the coordinates of the fourth vertex of the parallelogram.
In simple words: For a shape to be a parallelogram, its opposite sides must have the same steepness. We find the steepness of the known sides, then use those values to set up equations for the unknown coordinates 'x' and 'y', and solve them.

🎯 Exam Tip: When finding unknown coordinates of a parallelogram, setting the slopes of opposite sides equal is a reliable method. Alternatively, you could use the property that the diagonals bisect each other (midpoint of AC equals midpoint of BD).

Question 12. Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Answer:
To show that ABCD is a trapezium (also known as a trapezoid), we need to prove that at least one pair of opposite sides is parallel. This means their slopes must be equal.
Given vertices are \( A(3, -4) \), \( B(9, -4) \), \( C(5, -7) \), and \( D(7, -7) \).
Slope of AB \( (m_{AB}) = \frac{-4-(-4)}{9-3} = \frac{-4+4}{6} = \frac{0}{6} = 0 \).
Slope of CD \( (m_{CD}) = \frac{-7-(-7)}{7-5} = \frac{-7+7}{2} = \frac{0}{2} = 0 \).
Since \( m_{AB} = m_{CD} = 0 \), line AB is parallel to line CD \( (\text{AB} \parallel \text{CD}) \).
Now, let's check the slopes of the other pair of opposite sides, AD and BC.
Slope of BC \( (m_{BC}) = \frac{-7-(-4)}{5-9} = \frac{-7+4}{-4} = \frac{-3}{-4} = \frac{3}{4} \).
Slope of AD \( (m_{AD}) = \frac{-7-(-4)}{7-3} = \frac{-7+4}{4} = \frac{-3}{4} = -\frac{3}{4} \).
Since \( m_{BC} \neq m_{AD} \) (\( \frac{3}{4} \neq -\frac{3}{4} \)), lines BC and AD are not parallel.
As one pair of opposite sides (AB and CD) is parallel, and the other pair (BC and AD) is not parallel, the quadrilateral ABCD is a trapezium. Trapeziums are useful in calculating areas of non-standard shapes.
In simple words: To show a shape is a trapezium, we need to find the steepness of all its sides. If at least one pair of opposite sides has the same steepness, but the other pair does not, then it's a trapezium.

🎯 Exam Tip: The key characteristic of a trapezium is having *exactly one* pair of parallel sides. Make sure to calculate the slopes of all four sides and explicitly state which pair is parallel and which is not.

Question 13. A quadrilateral has vertices at A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Answer:
Let the given vertices of the quadrilateral be \( A(-4, -2) \), \( B(5, -1) \), \( C(6, 5) \), and \( D(-7, 6) \).
First, find the midpoints of each side. Let E, F, G, H be the midpoints of AB, BC, CD, and DA respectively.
Midpoint formula: \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
Midpoint E of AB: \( E = \left(\frac{-4+5}{2}, \frac{-2+(-1)}{2}\right) = \left(\frac{1}{2}, \frac{-3}{2}\right) \).
Midpoint F of BC: \( F = \left(\frac{5+6}{2}, \frac{-1+5}{2}\right) = \left(\frac{11}{2}, \frac{4}{2}\right) = \left(\frac{11}{2}, 2\right) \).
Midpoint G of CD: \( G = \left(\frac{6+(-7)}{2}, \frac{5+6}{2}\right) = \left(\frac{-1}{2}, \frac{11}{2}\right) \).
Midpoint H of DA: \( H = \left(\frac{-7+(-4)}{2}, \frac{6+(-2)}{2}\right) = \left(\frac{-11}{2}, \frac{4}{2}\right) = \left(\frac{-11}{2}, 2\right) \).
Now we have the vertices of the quadrilateral EFGH: \( E\left(\frac{1}{2}, -\frac{3}{2}\right) \), \( F\left(\frac{11}{2}, 2\right) \), \( G\left(-\frac{1}{2}, \frac{11}{2}\right) \), and \( H\left(-\frac{11}{2}, 2\right) \).
To show EFGH is a parallelogram, we check if opposite sides have equal slopes.
Slope of EF \( (m_{EF}) = \frac{2 - (-\frac{3}{2})}{\frac{11}{2} - \frac{1}{2}} = \frac{\frac{4+3}{2}}{\frac{10}{2}} = \frac{\frac{7}{2}}{5} = \frac{7}{10} \).
Slope of GH \( (m_{GH}) = \frac{2 - \frac{11}{2}}{-\frac{11}{2} - (-\frac{1}{2})} = \frac{\frac{4-11}{2}}{\frac{-11+1}{2}} = \frac{-\frac{7}{2}}{-\frac{10}{2}} = \frac{7}{10} \).
Since \( m_{EF} = m_{GH} = \frac{7}{10} \), EF is parallel to GH \( (\text{EF} \parallel \text{GH}) \).
Slope of FG \( (m_{FG}) = \frac{\frac{11}{2} - 2}{-\frac{1}{2} - \frac{11}{2}} = \frac{\frac{11-4}{2}}{\frac{-1-11}{2}} = \frac{\frac{7}{2}}{\frac{-12}{2}} = -\frac{7}{12} \).
Slope of EH \( (m_{EH}) = \frac{2 - (-\frac{3}{2})}{-\frac{11}{2} - \frac{1}{2}} = \frac{\frac{4+3}{2}}{\frac{-11-1}{2}} = \frac{\frac{7}{2}}{\frac{-12}{2}} = -\frac{7}{12} \).
Since \( m_{FG} = m_{EH} = -\frac{7}{12} \), FG is parallel to EH \( (\text{FG} \parallel \text{EH}) \).
Since both pairs of opposite sides of EFGH are parallel, EFGH is a parallelogram. This property is known as Varignon's theorem, which states that the midpoints of any quadrilateral form a parallelogram.
In simple words: First, find the middle point of each side of the big shape. This will give you four new points. Then, find the steepness of the sides of the new shape formed by these four middle points. If the opposite sides of this new shape have the same steepness, then it is a parallelogram.

🎯 Exam Tip: Varignon's theorem is a powerful concept. Remember that the quadrilateral formed by connecting the midpoints of any quadrilateral is always a parallelogram. This means you don't even need to know the initial quadrilateral's type!

Question 14. PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Answer:
In a rhombus, the diagonals bisect each other at right angles. Let P be \( (a,b) \), S be \( (1,1) \), and M be \( (2,-1) \).
The diagonals are PR and QS. M is the midpoint of both diagonals.
First, find the slope of SM:
\( \text{Slope of SM} = \frac{-1-1}{2-1} = \frac{-2}{1} = -2 \).
Since the diagonals of a rhombus are perpendicular, PM is perpendicular to SM.
Therefore, the slope of PM is \( \frac{1}{2} \) (because \( -2 \times \frac{1}{2} = -1 \)).
Now, use the slope of PM with points P\( (a,b) \) and M\( (2,-1) \):
\( \frac{b-(-1)}{a-2} = \frac{1}{2} \)
\( \frac{b+1}{a-2} = \frac{1}{2} \)
\( \implies 2(b+1) = 1(a-2) \)
\( 2b+2 = a-2 \)
\( a-2b = 4 \quad \text{(Equation 1)} \).
We are given that \( \text{QS} = 2\text{PR} \). In a rhombus, M is the midpoint of QS and PR.
So, \( \text{QS} = 2\text{SM} \) and \( \text{PR} = 2\text{PM} \).
Substitute these into the given relation: \( 2\text{SM} = 2(2\text{PM}) \)
\( \implies \text{SM} = 2\text{PM} \). This is a crucial relationship between the lengths from the center M to the vertices.
Now, use the distance formula: \( \text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
Calculate SM:
\( \text{SM} = \sqrt{(2-1)^2+(-1-1)^2} = \sqrt{1^2+(-2)^2} = \sqrt{1+4} = \sqrt{5} \).
Calculate PM:
\( \text{PM} = \sqrt{(a-2)^2+(b-(-1))^2} = \sqrt{(a-2)^2+(b+1)^2} \).
Using the relation \( \text{SM} = 2\text{PM} \):
\( \sqrt{5} = 2\sqrt{(a-2)^2+(b+1)^2} \).
Square both sides to eliminate the square roots:
\( (\sqrt{5})^2 = (2\sqrt{(a-2)^2+(b+1)^2})^2 \)
\( 5 = 4((a-2)^2+(b+1)^2) \).
From Equation 1, \( a = 4+2b \). Substitute this into the equation:
\( 5 = 4((4+2b-2)^2+(b+1)^2) \)
\( 5 = 4((2+2b)^2+(b+1)^2) \)
\( 5 = 4( (2(1+b))^2+(b+1)^2 ) \)
\( 5 = 4( 4(1+b)^2+(b+1)^2 ) \)
\( 5 = 4( 5(b+1)^2 ) \)
\( 5 = 20(b+1)^2 \).
Divide by 20:
\( (b+1)^2 = \frac{5}{20} = \frac{1}{4} \)
\( \implies b+1 = \pm \sqrt{\frac{1}{4}} \)
\( b+1 = \pm \frac{1}{2} \).
Case 1: \( b+1 = \frac{1}{2} \)
\( b = \frac{1}{2}-1 = -\frac{1}{2} \).
Case 2: \( b+1 = -\frac{1}{2} \)
\( b = -\frac{1}{2}-1 = -\frac{3}{2} \).
Now, find the corresponding 'a' values using \( a = 4+2b \):
For \( b = -\frac{1}{2} \): \( a = 4+2\left(-\frac{1}{2}\right) = 4-1 = 3 \).
For \( b = -\frac{3}{2} \): \( a = 4+2\left(-\frac{3}{2}\right) = 4-3 = 1 \).
Therefore, the possible coordinates for P are \( \left(3, -\frac{1}{2}\right) \) or \( \left(1, -\frac{3}{2}\right) \). These two points are symmetric with respect to the line defined by the slope of SM, passing through M.
In simple words: First, use the fact that diagonals of a rhombus cross at right angles to find the steepness of the line PM. Then, use the given relationship between the diagonal lengths to set up an equation involving the unknown coordinates of P. Solve this equation to find the two possible sets of coordinates for P.

🎯 Exam Tip: Remember that in a rhombus, diagonals bisect each other at right angles. This gives two key properties: (1) their slopes are negative reciprocals, and (2) their midpoints are the same. These properties are essential for solving problems involving rhombus coordinates.