Samacheer Kalvi Class 11 Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Exercise 6.2

 

Question 1. Find the equation of the lines passing through the point (1, 1):
(i) with y – intercept – 4
(ii) with slope 3
(iii) and (-2, 3)
(iv) and the perpendicular from the origin makes an angle 60° with x – axis.
Answer:
(i) The equation of a line with slope \( m \) and y-intercept \( b \) is \( y = mx + b \). Given that the y-intercept \( b = -4 \), the equation becomes \( y = mx - 4 \). Since the line passes through the point \( (1, 1) \), we substitute these coordinates into the equation: \( 1 = m(1) - 4 \). Solving for \( m \), we get \( 1 = m - 4 \implies m = 5 \). Thus, the required equation of the line is \( y = 5x - 4 \). This formula helps us quickly find the equation when the slope and y-intercept are known.
(ii) The equation of a line passing through a point \( (x_1, y_1) \) with slope \( m \) is \( y - y_1 = m(x - x_1) \). Given the slope \( m = 3 \) and the point \( (x_1, y_1) = (1, 1) \), we substitute these values: \( y - 1 = 3(x - 1) \). Expanding this, we get \( y - 1 = 3x - 3 \). Rearranging the terms to the standard form \( Ax + By = C \), we get \( 3x - y = 2 \). This method is direct when a point and slope are provided.
(iii) The equation of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \). Given the two points \( (x_1, y_1) = (1, 1) \) and \( (x_2, y_2) = (-2, 3) \), we substitute the coordinates: \( \frac{y - 1}{3 - 1} = \frac{x - 1}{-2 - 1} \). This simplifies to \( \frac{y - 1}{2} = \frac{x - 1}{-3} \). Cross-multiplying, we get \( -3(y - 1) = 2(x - 1) \), which expands to \( -3y + 3 = 2x - 2 \). Rearranging the terms, we get \( 2x + 3y - 2 - 3 = 0 \), resulting in the equation \( 2x + 3y - 5 = 0 \). This formula is very useful for finding a line through any two distinct points.
(iv) The equation of a line in the normal form is \( x \cos \alpha + y \sin \alpha = p \), where \( p \) is the length of the perpendicular from the origin to the line, and \( \alpha \) is the angle this perpendicular makes with the x-axis. Given \( \alpha = 60^\circ \), the equation is \( x \cos 60^\circ + y \sin 60^\circ = p \). Substituting the values for \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we get \( x(\frac{1}{2}) + y(\frac{\sqrt{3}}{2}) = p \), which can be written as \( x + \sqrt{3}y = 2p \). Since the line passes through \( (1, 1) \), we substitute these coordinates: \( 1 + \sqrt{3}(1) = 2p \implies 1 + \sqrt{3} = 2p \). Substituting this value of \( 2p \) back into the equation, we get the required equation: \( x + \sqrt{3}y = 1 + \sqrt{3} \). This form is especially helpful when dealing with distances from the origin.
In simple words: For each part, we use a different formula based on the information given. Whether it's the y-intercept, slope, two points, or the perpendicular from the origin, each method helps us find the unique straight line.

🎯 Exam Tip: Always identify the given information (e.g., slope, points, intercepts, angles) to choose the most appropriate form of the straight line equation (slope-intercept, point-slope, two-point, or normal form).

 

Question 2. If P (r, c) is mid point of a line segment between the axes then show that \( \frac{x}{\mathbf{r}}+\frac{\mathbf{y}}{\mathbf{c}} = 2 \)
Answer: Let's consider a line segment AB that is intercepted between the coordinate axes. Let point A be on the x-axis, so its coordinates are \( (a, 0) \). Let point B be on the y-axis, so its coordinates are \( (0, b) \). We are given that P \( (r, c) \) is the midpoint of this line segment AB.
Using the midpoint formula, the coordinates of P are \( (\frac{a+0}{2}, \frac{0+b}{2}) \). Since P is \( (r, c) \), we have: \( r = \frac{a}{2} \implies a = 2r \)
\( c = \frac{b}{2} \implies b = 2c \) The equation of a line with x-intercept \( a \) and y-intercept \( b \) is given by the intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \). Now, substitute the values of \( a \) and \( b \) in terms of \( r \) and \( c \) into this equation: \( \frac{x}{2r} + \frac{y}{2c} = 1 \) To clear the denominators, multiply the entire equation by 2: \( \frac{2x}{2r} + \frac{2y}{2c} = 2 \times 1 \)
\( \implies \frac{x}{r} + \frac{y}{c} = 2 \) This shows the desired relationship. This property holds true for any line segment whose midpoint is given as (r, c).
In simple words: If a point is exactly in the middle of a line that touches both the x and y axes, we can use its coordinates to write a special equation for that line. The x-part divided by the x-midpoint, plus the y-part divided by the y-midpoint, always equals 2.

🎯 Exam Tip: Remember the midpoint formula and the intercept form of a line. This problem requires you to combine both concepts by expressing the intercepts in terms of the midpoint's coordinates.

 

Question 3. Find the equation of the line passing through the point (1, 5) and also divides the line segment between the coordinate axes in the ratio 3 : 10.
Answer: Let the line divide the coordinate axes such that its x-intercept is \( a \) and its y-intercept is \( b \). We are given that the line segment between the axes is divided in the ratio 3:10. This implies that the x-intercept \( a \) can be represented as \( 3k \) and the y-intercept \( b \) as \( 10k \) for some constant \( k \). The equation of a straight line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \). Substituting \( a = 3k \) and \( b = 10k \), the equation becomes: \( \frac{x}{3k} + \frac{y}{10k} = 1 \) We are told that this line passes through the point P \( (1, 5) \). So, we can substitute \( x = 1 \) and \( y = 5 \) into the equation: \( \frac{1}{3k} + \frac{5}{10k} = 1 \) Simplify the second term: \( \frac{1}{3k} + \frac{1}{2k} = 1 \) To add these fractions, find a common denominator, which is \( 6k \): \( \frac{2}{6k} + \frac{3}{6k} = 1 \)
\( \frac{2 + 3}{6k} = 1 \)
\( \frac{5}{6k} = 1 \) Solving for \( k \): \( 6k = 5 \implies k = \frac{5}{6} \) Now, we can find the actual x-intercept \( a \) and y-intercept \( b \): \( a = 3k = 3(\frac{5}{6}) = \frac{5}{2} \)
\( b = 10k = 10(\frac{5}{6}) = \frac{25}{3} \) Substitute these values back into the intercept form of the line: \( \frac{x}{\frac{5}{2}} + \frac{y}{\frac{25}{3}} = 1 \) This can be rewritten as: \( \frac{2x}{5} + \frac{3y}{25} = 1 \) To eliminate the denominators and get the general form, multiply the entire equation by the least common multiple of 5 and 25, which is 25: \( 25 \left( \frac{2x}{5} \right) + 25 \left( \frac{3y}{25} \right) = 25 \times 1 \)
\( 10x + 3y = 25 \) This is the required equation of the line. Understanding how intercepts relate to a ratio helps in constructing the initial equation.
In simple words: We find numbers for where the line cuts the x and y axes, based on the given ratio. Then, we use the point the line passes through to find the missing value. Finally, we put all these numbers back into the line equation to get the final answer.

🎯 Exam Tip: When a line divides segments between axes in a given ratio, represent the intercepts as multiples of a common variable (e.g., \( nk, mk \)) to establish a relationship that can be solved using the given point.

 

Question 4. If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then show that \( \frac{1}{\mathbf{p}^{2}}=\frac{1}{\mathbf{a}^{2}}+\frac{1}{\mathbf{b}^{2}} \)
Answer: The equation of a line with x-intercept \( a \) and y-intercept \( b \) is given by the intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \) To find the length of the perpendicular from the origin \( (0, 0) \) to this line, we first rewrite the equation in the general form \( Ax + By + C = 0 \): \( \frac{1}{a}x + \frac{1}{b}y - 1 = 0 \) The formula for the perpendicular distance \( p \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is \( p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \). Here, \( (x_0, y_0) = (0, 0) \), \( A = \frac{1}{a} \), \( B = \frac{1}{b} \), and \( C = -1 \). Substituting these values: \[ p = \frac{\left| \frac{1}{a}(0) + \frac{1}{b}(0) - 1 \right|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2}} \] \[ p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \] Since the length \( p \) must be positive, \( |-1| = 1 \): \[ p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \] Now, to show the required relationship, we square both sides of the equation: \[ p^2 = \left( \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right)^2 \] \[ p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} \] Finally, take the reciprocal of both sides: \[ \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \] This proves the relationship. This formula demonstrates a beautiful connection between a line's intercepts and its distance from the origin.
In simple words: We start with the line's equation based on where it crosses the x and y axes. Then we use a special formula to find the distance from the middle point (origin) to this line. When we square and flip both sides of this distance, we get the required connection with the intercepts.

🎯 Exam Tip: Remember the intercept form of a line and the formula for the perpendicular distance from a point to a line. Pay attention to algebraic manipulation, especially squaring and taking reciprocals.

 

Question 5. The normal boiling point of water is 100° C or 212° F and the freezing point of water is 0° C or 32° F.
(i) Find the linear relationship between C and F. Find
(ii) the value of C for 98.6° F and
(iii) the value of F for 38° C.
Answer: Let \( C \) denote the temperature in Celsius and \( F \) denote the temperature in Fahrenheit. We are given two corresponding points: Freezing point: \( (C_1, F_1) = (0^\circ C, 32^\circ F) \) Boiling point: \( (C_2, F_2) = (100^\circ C, 212^\circ F) \) (i) To find the linear relationship between C and F, we use the two-point form of a linear equation: \[ \frac{C - C_1}{C_2 - C_1} = \frac{F - F_1}{F_2 - F_1} \] Substitute the given values: \[ \frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32} \] \[ \frac{C}{100} = \frac{F - 32}{180} \] To simplify this equation, divide both denominators by their greatest common divisor, which is 20: \[ \frac{C}{100 \div 20} = \frac{F - 32}{180 \div 20} \] \[ \frac{C}{5} = \frac{F - 32}{9} \] This is the linear relationship between Celsius and Fahrenheit. This formula allows for easy conversion between the two scales.
(ii) To find the value of \( C \) when \( F = 98.6^\circ F \), we use the derived relationship: \[ \frac{C}{5} = \frac{F - 32}{9} \] Substitute \( F = 98.6 \): \[ \frac{C}{5} = \frac{98.6 - 32}{9} \] \[ \frac{C}{5} = \frac{66.6}{9} \] \[ \frac{C}{5} = 7.4 \] Now, solve for \( C \): \[ C = 5 \times 7.4 \] \[ C = 37^\circ C \] So, 98.6°F is equivalent to 37°C.
(iii) To find the value of \( F \) when \( C = 38^\circ C \), we again use the relationship: \[ \frac{C}{5} = \frac{F - 32}{9} \] Substitute \( C = 38 \): \[ \frac{38}{5} = \frac{F - 32}{9} \] Multiply both sides by 9: \[ 9 \times \frac{38}{5} = F - 32 \] \[ \frac{342}{5} = F - 32 \] \[ 68.4 = F - 32 \] Now, solve for \( F \): \[ F = 68.4 + 32 \] \[ F = 100.4^\circ F \] So, 38°C is equivalent to 100.4°F. These calculations are crucial in many scientific and everyday contexts.
In simple words: We use two known points (freezing and boiling of water) to make a formula that changes Celsius to Fahrenheit and vice-versa. Then, we use this formula to convert other temperatures from one scale to the other.

🎯 Exam Tip: Remember the two key points for water's freezing and boiling in both scales. The two-point form of a linear equation is essential for deriving the conversion formula, and then simple substitution and algebraic steps will yield the required conversions.

 

Question 6. An object was launched from a place P in constant speed to hit a target. At the 15th second, it was 1400 m from the target, and at the 18th second 800 m away. Find
(i) the distance between the place and the target
(ii) The distance covered by it in 15 seconds
(iii) time taken to hit the target.
Answer: Let \( T \) represent time in seconds (along the x-axis) and \( D \) represent the distance from the target in meters (along the y-axis). We are given two data points: At \( T_1 = 15 \) s, \( D_1 = 1400 \) m. This gives the point \( (15, 1400) \). At \( T_2 = 18 \) s, \( D_2 = 800 \) m. This gives the point \( (18, 800) \). First, we find the linear equation connecting \( T \) and \( D \) using the two-point form: \[ \frac{D - D_1}{D_2 - D_1} = \frac{T - T_1}{T_2 - T_1} \] Substitute the points: \[ \frac{D - 1400}{800 - 1400} = \frac{T - 15}{18 - 15} \] \[ \frac{D - 1400}{-600} = \frac{T - 15}{3} \] Cross-multiply: \[ 3(D - 1400) = -600(T - 15) \] Divide both sides by 3: \[ D - 1400 = -200(T - 15) \] \[ D - 1400 = -200T + 3000 \] Rearrange to get the equation of the line: \[ D = -200T + 3000 + 1400 \] \[ D = -200T + 4400 \] This equation describes the object's distance from the target at any given time.
(i) The distance between the place (origin of the motion) and the target is the initial distance from the target when \( T = 0 \). Substitute \( T = 0 \) into the equation: \[ D = -200(0) + 4400 \] \[ D = 4400 \text{ m} \] So, the initial distance from the target was 4400 meters. This tells us how far the target was when the object started moving.
(ii) The distance from the target at 15 seconds: The question states "at the 15th second, it was 1400m from the target". This is already a given fact. Let's verify it with our equation. Substitute \( T = 15 \) into the equation: \[ D = -200(15) + 4400 \] \[ D = -3000 + 4400 \] \[ D = 1400 \text{ m} \] This confirms the distance from the target at 15 seconds is 1400m. The phrasing "distance covered by it in 15 seconds" might be interpreted as total distance traveled, but based on the problem's definition of D, it is the distance *from* the target.
(iii) Time taken to hit the target: Hitting the target means the distance from the target \( D \) is 0. Substitute \( D = 0 \) into the equation: \[ 0 = -200T + 4400 \] Solve for \( T \): \[ 200T = 4400 \] \[ T = \frac{4400}{200} \] \[ T = 22 \text{ seconds} \] It will take 22 seconds for the object to hit the target. This calculation helps predict when the object reaches its destination.
In simple words: We find a rule that connects time and the object's distance from the target. Then we use this rule to find the initial distance from the target, confirm the distance at a specific time, and finally, calculate how long it takes for the object to reach the target (when the distance becomes zero).

🎯 Exam Tip: Clearly define your variables and what they represent (e.g., distance from target, not distance traveled). When solving word problems, read carefully to extract the correct coordinate pairs for linear equation formulation.

 

Question 7. The city in the years 2005 and 2010 are 1, 35, 000 and 1, 45, 000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant).
Answer: Let \( Y \) represent the year (along the x-axis) and \( P \) represent the population of the city (along the y-axis). We are given two data points: In the year 2005, the population was 1,35,000. This gives the point \( (Y_1, P_1) = (2005, 1,35,000) \). In the year 2010, the population was 1,45,000. This gives the point \( (Y_2, P_2) = (2010, 1,45,000) \). We need to find the approximate population in the year 2015, assuming constant growth. This means we can use the two-point form of a linear equation: \[ \frac{P - P_1}{P_2 - P_1} = \frac{Y - Y_1}{Y_2 - Y_1} \] Substitute the given values: \[ \frac{P - 1,35,000}{1,45,000 - 1,35,000} = \frac{Y - 2005}{2010 - 2005} \] \[ \frac{P - 1,35,000}{10,000} = \frac{Y - 2005}{5} \] To solve for \( P \), multiply both sides by 10,000: \[ P - 1,35,000 = \frac{10,000}{5}(Y - 2005) \] \[ P - 1,35,000 = 2000(Y - 2005) \] Now, isolate \( P \): \[ P = 2000(Y - 2005) + 1,35,000 \] To find the population in the year 2015, substitute \( Y = 2015 \) into the equation: \[ P = 2000(2015 - 2005) + 1,35,000 \] \[ P = 2000(10) + 1,35,000 \] \[ P = 20,000 + 1,35,000 \] \[ P = 1,55,000 \] Thus, the approximate population in the year 2015 is 1,55,000. Assuming constant growth simplifies future predictions.
In simple words: We take two points of population over time and use them to create a straight line equation. Then, we use this equation to guess the population for a future year, assuming the city keeps growing at the same steady rate.

🎯 Exam Tip: When dealing with constant rates of change (like population growth), a linear model is appropriate. Always clearly define your variables and their corresponding axes before applying the two-point formula.

 

Question 8. Find the equation of the line if the perpendicular drawn from the origin makes an angle 30° with x - axis and its length is 12.
Answer: The equation of a straight line in normal form is \( x \cos \alpha + y \sin \alpha = p \). In this form, \( p \) is the length of the perpendicular from the origin to the line, and \( \alpha \) is the angle that this perpendicular makes with the positive x-axis. We are given the following information: The angle made by the perpendicular from the origin with the x-axis, \( \alpha = 30^\circ \). The length of this perpendicular from the origin, \( p = 12 \). Now, substitute these values into the normal form equation: \[ x \cos 30^\circ + y \sin 30^\circ = 12 \] Recall the values of \( \cos 30^\circ \) and \( \sin 30^\circ \): \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 30^\circ = \frac{1}{2} \) Substitute these trigonometric values into the equation: \[ x\left(\frac{\sqrt{3}}{2}\right) + y\left(\frac{1}{2}\right) = 12 \] To eliminate the denominators, multiply the entire equation by 2: \[ 2 \times x\left(\frac{\sqrt{3}}{2}\right) + 2 \times y\left(\frac{1}{2}\right) = 2 \times 12 \] \[ \sqrt{3}x + y = 24 \] This is the required equation of the straight line. The normal form is very useful for problems involving the distance from the origin and angles.
In simple words: We use a special formula for lines that involves the distance from the center point (origin) and the angle of the line that is exactly 90 degrees to it. We plug in the given angle and distance, then simplify the math to get the line's equation.

🎯 Exam Tip: Understand the normal form of a line \( x \cos \alpha + y \sin \alpha = p \). Ensure you accurately recall or calculate the trigonometric values for \( \cos \alpha \) and \( \sin \alpha \).

 

Question 9. Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1.
Answer: Let the x-intercept of the line be \( a \) and the y-intercept be \( b \). The intercept form of the equation of a straight line is \( \frac{x}{a} + \frac{y}{b} = 1 \). We are given that the sum of the intercepts is 1: \( a + b = 1 \) From this, we can express \( b \) in terms of \( a \): \( b = 1 - a \) Substitute this expression for \( b \) into the intercept form equation: \[ \frac{x}{a} + \frac{y}{1 - a} = 1 \] We are also given that the line passes through the point \( (8, 3) \). Substitute \( x = 8 \) and \( y = 3 \) into the equation: \[ \frac{8}{a} + \frac{3}{1 - a} = 1 \] To solve for \( a \), multiply the entire equation by the common denominator \( a(1 - a) \): \[ a(1 - a) \left( \frac{8}{a} \right) + a(1 - a) \left( \frac{3}{1 - a} \right) = a(1 - a)(1) \] \[ 8(1 - a) + 3a = a(1 - a) \] Expand both sides: \[ 8 - 8a + 3a = a - a^2 \] \[ 8 - 5a = a - a^2 \] Rearrange the terms to form a quadratic equation: \[ a^2 - 5a - a + 8 = 0 \] \[ a^2 - 6a + 8 = 0 \] Factorize the quadratic equation: \[ (a - 4)(a - 2) = 0 \] This gives two possible values for \( a \): Case 1: \( a - 4 = 0 \implies a = 4 \) Case 2: \( a - 2 = 0 \implies a = 2 \) Now, we find the corresponding values of \( b \) for each case: If \( a = 4 \), then \( b = 1 - a = 1 - 4 = -3 \). The equation of the line is \( \frac{x}{4} + \frac{y}{-3} = 1 \). To clear the denominators, multiply by 12: \( 3x - 4y = 12 \) If \( a = 2 \), then \( b = 1 - a = 1 - 2 = -1 \). The equation of the line is \( \frac{x}{2} + \frac{y}{-1} = 1 \). To clear the denominators, multiply by 2: \( x - 2y = 2 \) Therefore, there are two possible equations for the straight lines that satisfy the given conditions. This problem shows how quadratic equations can arise in geometry.
In simple words: We use the idea that the x and y intercepts add up to 1 to write one intercept using the other. Then, we use the point the line goes through to find the actual values of these intercepts. Since there are two possible answers for the intercept, we get two possible equations for the lines.

🎯 Exam Tip: When a problem involves intercepts and a sum, use the intercept form of the line and express one intercept in terms of the other. Be prepared to solve a quadratic equation, as it often leads to multiple valid solutions.

 

Question 10. Show that the points (1, 3), (2, 1) and \( (\frac{1}{2}, 4) \) are collinear, by using
(i) concept of slope
(ii) using a straight line and
(iii) any other method.
Answer: Let the given points be A\( (1, 3) \), B\( (2, 1) \), and C\( (\frac{1}{2}, 4) \). (i) Using the concept of slope: For three points to be collinear, the slope between any two pairs of points must be the same. Slope of AB \( (m_{AB}) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{2 - 1} = \frac{-2}{1} = -2 \) Slope of BC \( (m_{BC}) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{\frac{1}{2} - 2} = \frac{3}{\frac{1 - 4}{2}} = \frac{3}{\frac{-3}{2}} = 3 \times \left(-\frac{2}{3}\right) = -2 \) Since \( m_{AB} = m_{BC} = -2 \), the points A, B, and C are collinear. This is a fundamental test for collinearity.
(ii) Using a straight line equation: First, find the equation of the line passing through points A\( (1, 3) \) and B\( (2, 1) \). We can use the two-point form: \[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \] Substituting A\( (1, 3) \) and B\( (2, 1) \): \[ \frac{y - 3}{1 - 3} = \frac{x - 1}{2 - 1} \] \[ \frac{y - 3}{-2} = \frac{x - 1}{1} \] Cross-multiply: \( y - 3 = -2(x - 1) \)
\( y - 3 = -2x + 2 \) Rearrange to get the general form of the line equation: \( 2x + y - 3 - 2 = 0 \)
\( 2x + y - 5 = 0 \) Now, check if the third point C\( (\frac{1}{2}, 4) \) lies on this line by substituting its coordinates into the equation: \( 2\left(\frac{1}{2}\right) + 4 - 5 = 1 + 4 - 5 = 5 - 5 = 0 \) Since the equation holds true for point C, point C lies on the line passing through A and B. Therefore, A, B, and C are collinear. This method confirms if all points are on the same path.
(iii) Using the distance formula (any other method): For three points A, B, C to be collinear, the sum of the distances of the two shorter segments must be equal to the distance of the longest segment. That is, if A, B, C are collinear and B is between A and C, then \( AB + BC = AC \). Calculate the distances between each pair of points using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \): Distance AB: \( AB = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \) Distance BC: \( BC = \sqrt{\left(\frac{1}{2} - 2\right)^2 + (4 - 1)^2} = \sqrt{\left(-\frac{3}{2}\right)^2 + (3)^2} = \sqrt{\frac{9}{4} + 9} = \sqrt{\frac{9 + 36}{4}} = \sqrt{\frac{45}{4}} = \frac{\sqrt{9 \times 5}}{\sqrt{4}} = \frac{3\sqrt{5}}{2} \) Distance AC: \( AC = \sqrt{\left(\frac{1}{2} - 1\right)^2 + (4 - 3)^2} = \sqrt{\left(-\frac{1}{2}\right)^2 + (1)^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{1 + 4}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \) Now check the condition for collinearity: \( AB + AC = \sqrt{5} + \frac{\sqrt{5}}{2} = \frac{2\sqrt{5} + \sqrt{5}}{2} = \frac{3\sqrt{5}}{2} \) Since \( AB + AC = BC \) (or \( \sqrt{5} + \frac{\sqrt{5}}{2} = \frac{3\sqrt{5}}{2} \)), the points A, B, and C are collinear. This method is effective but can be more computation-intensive.
In simple words: To show points are on the same line, we can check if the slope between the first two points is the same as the slope between the next two points. Or, we can find the equation of the line passing through the first two points and see if the third point fits that equation. A third way is to measure the distances between all points and check if the sum of the two shorter distances equals the longest distance. All three methods confirm the points lie on one line.

🎯 Exam Tip: For collinearity problems, choose the method that seems simplest for the given coordinates. The slope method is often the quickest if fractions or decimals are not too complex. Always show all steps clearly for whichever method you choose.

 

Question 11. A straight line is passing through the point A (1, 2) with slope \( \frac{5}{12} \). Find the points which are 13 units away from A.
Answer: We are given a point A\( (x_1, y_1) = (1, 2) \) and the slope \( m = \frac{5}{12} \). We need to find points that are 13 units away from A along this line. The slope \( m \) is equal to \( \tan \theta \), where \( \theta \) is the angle the line makes with the positive x-axis. \( \tan \theta = \frac{5}{12} \) From this, we can form a right-angled triangle where the opposite side is 5 and the adjacent side is 12. Using the Pythagorean theorem, the hypotenuse is \( \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). So, \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13} \) and \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13} \). The parametric equation of a line passing through a point \( (x_1, y_1) \) with an angle \( \theta \) and a distance \( r \) from \( (x_1, y_1) \) is: \( x = x_1 + r \cos \theta \)
\( y = y_1 + r \sin \theta \) Here, \( (x_1, y_1) = (1, 2) \) and the distance \( r = \pm 13 \) (since points can be in two directions along the line). Case 1: For \( r = 13 \) \( x = 1 + 13 \left(\frac{12}{13}\right) = 1 + 12 = 13 \)
\( y = 2 + 13 \left(\frac{5}{13}\right) = 2 + 5 = 7 \) So, one point is \( (13, 7) \). Case 2: For \( r = -13 \) \( x = 1 + (-13) \left(\frac{12}{13}\right) = 1 - 12 = -11 \)
\( y = 2 + (-13) \left(\frac{5}{13}\right) = 2 - 5 = -3 \) So, the other point is \( (-11, -3) \). The two points on the line that are 13 units away from A are \( (13, 7) \) and \( (-11, -3) \). This method effectively finds points at a specific distance along a line.
In simple words: First, we use the given slope to find the sine and cosine of the line's angle. Then, using a special formula that includes the starting point, this angle, and the distance, we calculate the new points. We must remember to check both positive and negative distances to find points on both sides.

🎯 Exam Tip: Remember the parametric form of a line \( x = x_1 + r \cos \theta, y = y_1 + r \sin \theta \). When given a slope, form a right triangle to find \( \sin \theta \) and \( \cos \theta \). Always consider both positive and negative values for \( r \) as distance can be in two directions.

 

Question 12. A 150 m long train is moving with a constant velocity of 12.5 m/s.
(i) The equation of motion of the train.
(ii) Time taken to cross a pole.
(iii) The time to cross the bridge of length 850m is?
Answer:
(i) To find the equation of motion for the train, we consider time (in seconds) along the x-axis and distance (in meters) along the y-axis. The train starts at the origin, so its length of 150 m acts as a negative y-intercept, meaning \( b = -150 \). The constant velocity of 12.5 m/s is the slope, \( m = 12.5 \). Using the straight-line equation \( y = mx + b \), the equation of motion is:
\( y = 12.5x - 150 \). This is a linear equation because the train moves at a constant speed.
(ii) To find the time taken to cross a pole, we set the distance covered, \( y \), to 0, because the pole is considered a point. So, we solve for \( x \):
\( 0 = 12.5x - 150 \)
\( 12.5x = 150 \)
\( x = \frac{150}{12.5} \)
\( x = 12 \) seconds. For a point-like obstacle like a pole, the train only needs to cover its own length to pass it.
(iii) To find the time taken to cross a bridge of length 850 m, the train needs to travel a distance equal to its own length plus the bridge's length. This means the total effective distance \( y \) is 850 m. We substitute this into the equation of motion:
\( 850 = 12.5x - 150 \)
\( 12.5x = 850 + 150 \)
\( 12.5x = 1000 \)
\( x = \frac{1000}{12.5} \)
\( x = 80 \) seconds. Understanding that the train's own length adds to the total distance is key for such problems.

🎯 Exam Tip: Remember that for linear motion with constant velocity, the equation \( y = mx + b \) represents distance (y) as a function of time (x), where m is velocity and b is the initial position or offset.

 

Question 13. A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

(a) Draw a graph showing the results.
(b) Find the equation relating the length of the spring to the weight on it.
(c) What is the actual length of the spring?
(d) If the spring stretches to 9 cm long, how much weight should be added?
(e) How long will the spring be when 6 kilograms of weight on it?
Answer:
(a) The graph below shows the relationship between the weight applied (x-axis) and the length of the spring (y-axis). Each point from the table is plotted, and a straight line is drawn through them, illustrating the linear stretching of the spring.

(b) We can find the equation of the straight line by using two points from the table, for example, (2, 3) and (4, 4). The slope \( m \) is calculated as:
\( m = \frac{4 - 3}{4 - 2} = \frac{1}{2} \).
Now, using the point-slope form \( y - y_1 = m(x - x_1) \) with point (2, 3):
\( y - 3 = \frac{1}{2}(x - 2) \)
Multiply by 2:
\( 2(y - 3) = x - 2 \)
\( 2y - 6 = x - 2 \)
Rearrange into the general form:
\( x - 2y + 4 = 0 \). This linear relationship is characteristic of how springs stretch under load, often described by Hooke's Law.
(c) The actual length of the spring is its length when no weight is attached. This means we need to find \( y \) when the weight \( x = 0 \). Substituting \( x = 0 \) into the equation from part (b):
\( 0 - 2y + 4 = 0 \)
\( -2y = -4 \)
\( y = 2 \) cm. This "zero-weight" length is also known as the natural length of the spring.
(d) If the spring stretches to 9 cm long, we need to find the weight \( x \) when \( y = 9 \). Substituting \( y = 9 \) into the equation from part (b):
\( x - 2(9) + 4 = 0 \)
\( x - 18 + 4 = 0 \)
\( x - 14 = 0 \)
\( x = 14 \) kg. This calculation helps us predict how much weight is needed to achieve a specific stretch.
(e) To find how long the spring will be when 6 kg of weight is attached, we set \( x = 6 \). Substituting \( x = 6 \) into the equation from part (b):
\( 6 - 2y + 4 = 0 \)
\( 10 - 2y = 0 \)
\( 2y = 10 \)
\( y = 5 \) cm. This result confirms the spring's behavior aligns with the linear model derived from the given data.

🎯 Exam Tip: When working with linear relationships from data, first derive the equation of the line. Then, use this equation to solve for unknown values or predict outcomes by substituting the known values.

 

Question 14. A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg = 29.5-14.2) If it is consumed at a constant rate, then it lasts for 24 days. Then the new cylinder is replaced
(i) Find the equation relating the quantity of gas in the cylinder to the days,
(ii) Draw the graph for the first 96 days.
Answer:
(i) Let \( x \) represent the number of days of gas consumption and \( y \) represent the quantity of gas remaining in the cylinder (in kg).
Initially, at \( x = 0 \) days, the gas quantity is \( y = 14.2 \) kg. So, the first point is \( (0, 14.2) \).
The gas lasts for 24 days, meaning after \( x = 24 \) days, the gas quantity is \( y = 0 \) kg. So, the second point is \( (24, 0) \).
We can find the equation of the straight line joining these two points using the two-point form:
\( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \)
Substitute the points \( (0, 14.2) \) and \( (24, 0) \):
\( \frac{y - 14.2}{x - 0} = \frac{0 - 14.2}{24 - 0} \)
\( \frac{y - 14.2}{x} = \frac{-14.2}{24} \)
\( y - 14.2 = \frac{-14.2}{24} x \)
\( y = \frac{-14.2}{24} x + 14.2 \)
To simplify the fraction, multiply numerator and denominator by 10:
\( y = \frac{-142}{240} x + 14.2 \)
Divide by 2:
\( y = \frac{-71}{120} x + 14.2 \)
This equation relates the quantity of gas remaining to the number of days passed. It can predict the gas level on any specific day.
(ii) The graph for the first 96 days shows the depletion of gas over time. The gas lasts for 24 days, after which it runs out. Since the cylinder is replaced, the pattern of consumption repeats every 24 days, starting with a full cylinder (14.2 kg) and ending with an empty one (0 kg). The graph shows the initial depletion, and then implies the gas level resets with new cylinders.

🎯 Exam Tip: When drawing graphs from real-world scenarios, ensure your axes are clearly labeled with units and that the plot accurately reflects the given conditions (e.g., constant rate, depletion, or replacement).

 

Question 15. In the shopping mall, there is a hall of cuboid shape with dimension 800 x 800 x 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find
(i) the minimum total length of the escalator
(ii) the height at which the escalator changes its direction
(iii) the slopes of the escalator at the turning points.
Answer:
(i) The hall has dimensions 800 x 800 x 720 units. The escalator path consists of four equal segments (OA, AB, BC, CD). From the problem context and calculations provided, each segment of the escalator has a horizontal run of 800 units and a vertical rise of 180 units (which is 1/4 of the total height 720). We can find the length of one segment (OA) using the Pythagorean theorem:
Length \( OA^2 = (\text{horizontal run})^2 + (\text{vertical rise})^2 \)
\( OA^2 = (800)^2 + (180)^2 \)
\( OA^2 = (40 \times 20)^2 + (9 \times 20)^2 \) (Factoring to simplify calculation)
\( OA^2 = 20^2 (40^2 + 9^2) \)
\( OA^2 = 20^2 (1600 + 81) \)
\( OA^2 = 20^2 \times 1681 \)
\( OA^2 = 20^2 \times 41^2 \) (Since \( 1681 = 41^2 \))
\( OA = \sqrt{20^2 \times 41^2} \)
\( OA = 20 \times 41 = 820 \) units.
Since there are four such steps for the escalator and each has the same length:
Total length of the escalator = \( OA + AB + BC + CD = 4 \times OA \)
Total length = \( 4 \times 820 = 3280 \) units. This calculation ensures the escalator is long enough to cover both the horizontal and vertical distances required for its path.
(ii) The heights at which the escalator changes its direction correspond to the cumulative vertical rises after each segment:
First change (at A): \( EA = \frac{1}{4} \times \text{total height} = \frac{1}{4} \times 720 = 180 \) units.
Second change (at B): \( FB = \frac{1}{2} \times \text{total height} = \frac{1}{2} \times 720 = 360 \) units.
Third change (at C): \( GC = \frac{3}{4} \times \text{total height} = \frac{3}{4} \times 720 = 540 \) units. These heights define the levels where the escalator's path adjusts, ensuring a smooth stepped ascent.
(iii) The slope of the escalator at the turning points is constant because each segment is identical. We can find this slope using the first segment, OA, where the horizontal run is 800 and the vertical rise is 180. The slope is given by \( \tan \theta \), where \( \theta \) is the angle with the horizontal:
\( \tan \theta = \frac{\text{vertical rise}}{\text{horizontal run}} = \frac{EA}{OE} \)
\( \tan \theta = \frac{180}{800} \)
\( \tan \theta = \frac{9}{40} \). A constant slope makes the escalator movement uniform and predictable throughout its path.

🎯 Exam Tip: When calculating the total length of a multi-segment path, ensure you correctly identify the horizontal and vertical components for *each* segment before applying the Pythagorean theorem. For slopes, remember it's always "rise over run".