OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (C)

 

Question 1. write down the equation of the straight line cutting off intercepts a and b from the axes where
(i) a = - 2, b = 3
(ii) a = 5, b = -6
(iii) a = \( \frac { k }{ m } \), b = k
Answer:
(i) Let the equation of a straight line with intercepts 'a' and 'b' on the axes be given by the formula: \( \frac { x }{ a } + \frac { y }{ b } = 1 \). This formula helps us find the line when we know where it crosses the x-axis and y-axis.
Given a = -2 and b = 3.
Substitute these values into the formula:
\( \frac{x}{-2} + \frac{y}{3} = 1 \)
To remove the fractions, find a common multiple for -2 and 3, which is 6. Multiply the entire equation by 6:
\( 3x + (-2)y = 6 \)
\( 3x - 2y = 6 \)
To write the equation in the standard form (Ax + By + C = 0), move the constant term to the left side:
\( 3x - 2y - 6 = 0 \)
This is the required equation of the line.
(ii) The equation of a straight line with intercepts 'a' and 'b' on the axes is: \( \frac { x }{ a } + \frac { y }{ b } = 1 \). This is a general form to easily set up the line's equation when intercepts are known.
Given a = 5 and b = -6.
Substitute these values into the formula:
\( \frac { x }{ 5 } + \frac { y }{ -6 } = 1 \)
To clear the denominators, find the least common multiple of 5 and -6, which is 30. Multiply the entire equation by 30:
\( 6x + (-5)y = 30 \)
\( 6x - 5y = 30 \)
To get the equation in standard form (Ax + By + C = 0), move the constant term to the left side:
\( 6x - 5y - 30 = 0 \)
This is the required equation of the line.
(iii) The equation of a straight line that cuts off intercepts 'a' and 'b' on the axes is: \( \frac { x }{ a } + \frac { y }{ b } = 1 \). This formula provides a straightforward way to express the line's path.
Given \( a = -\frac { k }{ m } \) and \( b = k \).
Substitute these values into the formula:
\( \frac{x}{-\frac{k}{m}} + \frac{y}{k} = 1 \)
Simplify the first term: \( \frac{x}{-\frac{k}{m}} = \frac{xm}{-k} = -\frac{mx}{k} \).
So the equation becomes:
\( -\frac{mx}{k} + \frac{y}{k} = 1 \)
To remove the denominator 'k', multiply the entire equation by k:
\( -mx + y = k \)
To express it in the standard form (Ax + By + C = 0), move all terms to one side:
\( mx - y + k = 0 \)
This is the required equation of the line.
In simple words: When you know where a straight line crosses the x-axis and y-axis (these are called intercepts 'a' and 'b'), you can write its equation using the formula \( \frac{x}{a} + \frac{y}{b} = 1 \). Just put in the numbers for 'a' and 'b' and then clear any fractions to get the final simple equation.

🎯 Exam Tip: Always remember the intercept form of a straight line equation: \( \frac{x}{a} + \frac{y}{b} = 1 \). This is crucial for solving problems where intercepts are given directly or can be easily found. Simplify the equation to the general form \(Ax + By + C = 0\) at the end.

 

Question 2. Determine the x-intercept 'a' with the y intercept 'b' of the following lines. Sketch each.
(i) 3x + 5y - 15 = 0,
(ii) x - y - 7 = 0.
Answer:
(i) Given equation of the line is \( 3x + 5y - 15 = 0 \).
To find the intercepts, we need to rewrite the equation in the intercept form, which is \( \frac{x}{a} + \frac{y}{b} = 1 \).
First, move the constant term to the right side of the equation:
\( 3x + 5y = 15 \)
Now, divide the entire equation by the constant term (15) to make the right side equal to 1:
\( \frac{3x}{15} + \frac{5y}{15} = \frac{15}{15} \)
Simplify the fractions:
\( \frac{x}{5} + \frac{y}{3} = 1 \)
By comparing this with the standard intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), we can identify the intercepts.
So, the x-intercept is \( a = 5 \) and the y-intercept is \( b = 3 \). These points tell us where the line crosses the axes.

(ii) Given equation of the line is \( x - y - 7 = 0 \).
To find the intercepts, we need to convert this equation into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \).
First, move the constant term to the right side:
\( x - y = 7 \)
Now, divide the entire equation by 7 to make the right side equal to 1:
\( \frac{x}{7} - \frac{y}{7} = \frac{7}{7} \)
Rewrite the second term to match the standard form \( \frac{y}{b} \):
\( \frac{x}{7} + \frac{y}{-7} = 1 \)
By comparing this with the standard intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), we can find the intercepts.
So, the x-intercept is \( a = 7 \) and the y-intercept is \( b = -7 \). These are the points where the line crosses the axes.

In simple words: To find where a line crosses the x-axis (x-intercept) and y-axis (y-intercept), change its equation to the form \( \frac{x}{a} + \frac{y}{b} = 1 \). The number under 'x' is 'a' (x-intercept), and the number under 'y' is 'b' (y-intercept). Then, draw a graph by marking these two points and connecting them with a straight line.

🎯 Exam Tip: When sketching, label the axes, the origin, the intercepts, and the equation of the line clearly. Make sure the intercepts are correctly identified, especially with negative signs.

 

Question 3. Find the equation of the line which makes equal intercepts on the axes and passes through the point (2, 3).
Answer: Since the line makes equal intercepts on the axes, let's call the length of both intercepts 'a'. This means the line crosses the x-axis at (a, 0) and the y-axis at (0, a).
The equation of a line with equal intercepts 'a' is given by:
\( \frac{x}{a} + \frac{y}{a} = 1 \)
Multiply the entire equation by 'a' to remove the denominators:
\( x + y = a \) ... (1)
We are given that this line passes through the point (2, 3). This means if we substitute x=2 and y=3 into the equation, it should hold true.
Substitute x = 2 and y = 3 into equation (1):
\( 2 + 3 = a \)
\( \implies a = 5 \)
Now that we have the value of 'a', substitute it back into equation (1) to find the required equation of the line:
\( x + y = 5 \)
This is the final equation of the straight line.
In simple words: If a line crosses the x-axis and y-axis at the same distance from the center (equal intercepts), its equation is \( x + y = a \). If you know a point the line goes through, put those numbers into \( x \) and \( y \) to find what 'a' is. Once you have 'a', you have the full equation.

🎯 Exam Tip: The condition "equal intercepts" simplifies the intercept form to \( x+y=a \). Always use the given point to find the unknown intercept value 'a' by substituting its coordinates into the simplified equation.

 

Question 4. Write down the equation of the line which makes an intercept of 2a on the x-axis and 3a on the y-axis. Given that the line passes through the point (14, -9), find the numerical value of a.
Answer: Let the x-intercept be \( 2a \) and the y-intercept be \( 3a \).
The general equation of a line in intercept form is \( \frac{x}{X\text{-intercept}} + \frac{y}{Y\text{-intercept}} = 1 \).
Substitute the given intercepts into the formula:
\( \frac{x}{2a} + \frac{y}{3a} = 1 \)
To simplify, find a common denominator for \( 2a \) and \( 3a \), which is \( 6a \). Multiply the entire equation by \( 6a \):
\( 3x + 2y = 6a \) ... (1)
We are given that the line passes through the point (14, -9). This means that if we substitute \( x = 14 \) and \( y = -9 \) into equation (1), the equation must hold true.
Substitute these coordinates into equation (1):
\( 3 \times 14 + 2 \times (-9) = 6a \)
\( 42 - 18 = 6a \)
\( 24 = 6a \)
To find the numerical value of 'a', divide both sides by 6:
\( \implies a = \frac{24}{6} \)
\( \implies a = 4 \)
So, the numerical value of 'a' is 4.
In simple words: If a line crosses the axes at specific points (like 2a on x and 3a on y), you can write its equation. Then, if the line goes through another known point, put those numbers into the equation to find the value of 'a'.

🎯 Exam Tip: Always set up the equation using the intercept form first. When a line passes through a point, it means the coordinates of that point satisfy the line's equation, allowing you to solve for any unknown variables.

 

Question 5. Find the equation of the straight line which passes through the point (5, 6) and has intercept on the axes equal in magnitude but opposite in sign.
Answer: Let the equation of the straight line in intercept form be given by \( \frac{x}{a} + \frac{y}{b} = 1 \). This form is useful when dealing with intercepts directly.
We are told that the intercepts on the axes are equal in magnitude but opposite in sign. This means if the x-intercept is 'a', then the y-intercept 'b' must be '-a'.
So, \( b = -a \).
Substitute \( b = -a \) into the intercept form equation:
\( \frac{x}{a} + \frac{y}{-a} = 1 \)
This can be written as:
\( \frac{x}{a} - \frac{y}{a} = 1 \)
To remove the denominator 'a', multiply the entire equation by 'a':
\( x - y = a \) ... (2)
We are given that this line passes through the point (5, 6). This means that the coordinates \( x = 5 \) and \( y = 6 \) must satisfy equation (2).
Substitute \( x = 5 \) and \( y = 6 \) into equation (2):
\( 5 - 6 = a \)
\( \implies a = -1 \)
Now, substitute the value of \( a = -1 \) back into equation (2) to get the required equation of the line:
\( x - y = -1 \)
This is the final equation of the straight line.
In simple words: If a line crosses the x-axis and y-axis at points that are the same distance from zero but on opposite sides (like 5 and -5), its equation becomes \( x - y = a \). If you know a point that the line goes through, you can use its numbers to find what 'a' is, and then you have the complete equation for the line.

🎯 Exam Tip: The phrase "equal in magnitude but opposite in sign" means if one intercept is 'a', the other is '-a'. This is a key translation from words to algebraic setup. Always clearly state the relationship between the intercepts before substituting into the general form.

 

Question 6. A straight line passes through (2, 3) and the portion of the line intercepted between the this point. Find its equation.
Answer: Let the equation of the line in intercept form be \( \frac{x}{a} + \frac{y}{b} = 1 \) ... (1). This form helps relate the line to its intersections with the axes.
This line meets the coordinate axes at two points: A(a, 0) on the x-axis and B(0, b) on the y-axis.
We are given that the line passes through (2, 3), and this point (2, 3) is the midpoint of the portion of the line intercepted between the axes. This means (2, 3) is the midpoint of the line segment AB.
The coordinates of the midpoint P of a line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) are given by \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
Here, \((x_1, y_1) = A(a, 0)\) and \((x_2, y_2) = B(0, b)\). The midpoint P is (2, 3).
So, we can set up two equations:
\( \frac{a+0}{2} = 2 \)
\( \implies \frac{a}{2} = 2 \)
\( \implies a = 4 \)
And,
\( \frac{0+b}{2} = 3 \)
\( \implies \frac{b}{2} = 3 \)
\( \implies b = 6 \)
Now we have the intercepts \( a = 4 \) and \( b = 6 \). Substitute these values back into the intercept form equation (1):
\( \frac{x}{4} + \frac{y}{6} = 1 \)
To simplify this equation and remove denominators, find the least common multiple of 4 and 6, which is 12. Multiply the entire equation by 12:
\( 3x + 2y = 12 \)
To write it in the standard form (Ax + By + C = 0), move the constant term to the left side:
\( 3x + 2y - 12 = 0 \)
This is the required equation of the line.

In simple words: If a line passes through a point that is exactly in the middle of where the line crosses the x-axis and y-axis, you can use the midpoint formula. First, write the line equation using intercepts. Then, use the given midpoint to find the actual values of those intercepts. Finally, put those numbers back into the line equation.

🎯 Exam Tip: When a point is described as the "midpoint" of a line segment between the axes, it implies that it's the midpoint of the x-intercept and y-intercept. This information is key to setting up and solving for the unknown intercepts 'a' and 'b'.

 

Question 7. Show that the three points (5, 1), (1, -1) and (11, 4) lie on a straight line. Further find
(i) its intercepts on the axes;
(ii) the length of the portion of the line intercepted between the axes;
(iii) the slope of the line.
Answer: First, let's find the equation of the line passing through two of the given points, say (5, 1) and (1, -1). We'll use the two-point form of a straight line equation: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \). This formula helps us find the line equation when two points are known.
Let \((x_1, y_1) = (5, 1)\) and \((x_2, y_2) = (1, -1)\).
Substitute these values into the formula:
\( y - 1 = \frac{-1 - 1}{1 - 5} (x - 5) \)
\( y - 1 = \frac{-2}{-4} (x - 5) \)
\( y - 1 = \frac{1}{2} (x - 5) \)
Multiply both sides by 2 to clear the fraction:
\( 2(y - 1) = 1(x - 5) \)
\( 2y - 2 = x - 5 \)
Rearrange the terms to get the standard form \( Ax + By + C = 0 \):
\( x - 2y - 3 = 0 \) ... (1)
Now, we need to check if the third point (11, 4) lies on this line. Substitute \( x = 11 \) and \( y = 4 \) into equation (1):
\( 11 - 2(4) - 3 = 0 \)
\( 11 - 8 - 3 = 0 \)
\( 3 - 3 = 0 \)
\( 0 = 0 \)
Since the equation holds true, the point (11, 4) lies on the line. Therefore, all three given points (5, 1), (1, -1), and (11, 4) lie on the same straight line.

(i) To find the intercepts on the axes, we need to rewrite equation (1) in the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \).
Start with the line equation: \( x - 2y - 3 = 0 \)
Move the constant term to the right side:
\( x - 2y = 3 \)
Divide the entire equation by 3 to make the right side equal to 1:
\( \frac{x}{3} - \frac{2y}{3} = \frac{3}{3} \)
\( \frac{x}{3} + \frac{y}{-\frac{3}{2}} = 1 \)
Comparing this with \( \frac{x}{a} + \frac{y}{b} = 1 \), we find:
x-intercept \( a = 3 \)
y-intercept \( b = -\frac{3}{2} \)
These points are \( A(3, 0) \) on the x-axis and \( B(0, -\frac{3}{2}) \) on the y-axis.

(ii) The length of the portion of the line intercepted between the axes is the distance between the x-intercept \( A(3, 0) \) and the y-intercept \( B(0, -\frac{3}{2}) \). We can use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). The distance formula helps calculate the length between two points.
Length \( AB = \sqrt{(0 - 3)^2 + (-\frac{3}{2} - 0)^2} \)
\( = \sqrt{(-3)^2 + (-\frac{3}{2})^2} \)
\( = \sqrt{9 + \frac{9}{4}} \)
To add the terms under the square root, find a common denominator:
\( = \sqrt{\frac{36}{4} + \frac{9}{4}} \)
\( = \sqrt{\frac{45}{4}} \)
Simplify the square root:
\( = \frac{\sqrt{45}}{\sqrt{4}} = \frac{\sqrt{9 \times 5}}{2} = \frac{3\sqrt{5}}{2} \)
So, the length of the intercepted portion is \( \frac{3\sqrt{5}}{2} \) units.

(iii) To find the slope of the line, we can use the equation \( x - 2y - 3 = 0 \) and rewrite it in the slope-intercept form \( y = mx + c \), where 'm' is the slope.
Start with: \( x - 2y - 3 = 0 \)
Isolate the 'y' term:
\( -2y = -x + 3 \)
Divide the entire equation by -2:
\( y = \frac{-x}{-2} + \frac{3}{-2} \)
\( y = \frac{1}{2}x - \frac{3}{2} \)
Comparing this with \( y = mx + c \), we find that the slope \( m = \frac{1}{2} \).
Alternatively, we could have used the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) with any two of the given points. Using (5, 1) and (1, -1):
\( m = \frac{-1 - 1}{1 - 5} = \frac{-2}{-4} = \frac{1}{2} \).
In simple words: First, check if all three points are on the same line by finding the equation of the line using two points, then see if the third point fits. After that, change the line's equation to find where it crosses the x and y axes (intercepts). Use these intercept points to calculate the distance between them. Finally, change the equation into \( y = mx + c \) form to easily see the slope 'm', which tells you how steep the line is.

🎯 Exam Tip: To show three points are collinear, find the equation of the line using any two points and then verify if the third point satisfies that equation. Always use the distance formula for length and convert to \( y=mx+c \) for slope.

 

Question 8. Find the equation of the straight line which passes through the point (3, -2) and cuts off positive intercepts on the x and y-axes which are in the ratio 4: 3.
Answer: Let the equation of the straight line in intercept form be \( \frac{x}{a} + \frac{y}{b} = 1 \) ... (1). This is the base formula for lines given intercepts.
We are given that the intercepts on the x and y-axes are positive, so \( a > 0 \) and \( b > 0 \).
Also, the intercepts are in the ratio 4:3. This means that \( \frac{a}{b} = \frac{4}{3} \).
We can express 'a' and 'b' in terms of a common factor. Let the x-intercept be \( a = 4k \) and the y-intercept be \( b = 3k \) for some positive constant \( k \).
Substitute these into equation (1):
\( \frac{x}{4k} + \frac{y}{3k} = 1 \)
To clear the denominators, find the least common multiple of \( 4k \) and \( 3k \), which is \( 12k \). Multiply the entire equation by \( 12k \):
\( 3x + 4y = 12k \) ... (2)
We are given that the line passes through the point (3, -2). This means that if we substitute \( x = 3 \) and \( y = -2 \) into equation (2), it must hold true.
Substitute \( x = 3 \) and \( y = -2 \) into equation (2):
\( 3(3) + 4(-2) = 12k \)
\( 9 - 8 = 12k \)
\( 1 = 12k \)
\( \implies k = \frac{1}{12} \)
Now that we have the value of \( k \), we can substitute it back into equation (2) to find the required equation of the line:
\( 3x + 4y = 12 \left(\frac{1}{12}\right) \)
\( 3x + 4y = 1 \)
This is the required equation of the line.

In simple words: If a line crosses the x and y axes at positive points that have a certain ratio (like 4:3), you can write these intercepts as '4k' and '3k'. Then use the intercept form of the line equation. Since you also know a specific point the line passes through, substitute its coordinates into the equation to find the value of 'k'. Once 'k' is known, you have the full equation of the line.

🎯 Exam Tip: When intercepts are given in a ratio (e.g., m:n), represent them as 'mk' and 'nk'. This introduces a single unknown 'k' that can be solved using the point the line passes through. Always ensure the intercepts meet any "positive" or "negative" conditions.

 

Question 9. Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle \( \alpha \), given by the equation \( \tan \alpha = \frac { 5 }{ 12 } \), with the positive direction of the axis of x.
Answer: We are given the distance of the line from the origin, \( p = 3 \) units. This is the length of the perpendicular from the origin to the line.
We are also given that the perpendicular from the origin to the line makes an angle \( \alpha \) with the positive x-axis, and \( \tan \alpha = \frac { 5 }{ 12 } \).
We can form a right-angled triangle with the angle \( \alpha \). Since \( \tan \alpha = \frac{\text{opposite}}{\text{adjacent}} \), we have the opposite side as 5 and the adjacent side as 12.
Using the Pythagorean theorem (\( \text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2 \)), the hypotenuse is \( \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).
Now we can find \( \sin \alpha \) and \( \cos \alpha \):
\( \sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13} \)
\( \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13} \)
The normal form of the equation of a straight line is \( x \cos \alpha + y \sin \alpha = p \). This form is used when the perpendicular distance from the origin and its angle are known.
Substitute the values of \( p \), \( \cos \alpha \), and \( \sin \alpha \) into the normal form:
\( x \left(\frac{12}{13}\right) + y \left(\frac{5}{13}\right) = 3 \)
Multiply the entire equation by 13 to remove the denominators:
\( 12x + 5y = 39 \)
This is the required equation of the straight line.

In simple words: When you know how far a line is from the origin (point 0,0) and the angle this shortest distance makes with the x-axis, you can find the line's equation. First, use the tangent of the angle to find its sine and cosine values. Then, put these values into the special "normal form" equation: \( x \cos \alpha + y \sin \alpha = p \). Simplify to get the final line equation.

🎯 Exam Tip: The normal form \( x \cos \alpha + y \sin \alpha = p \) is ideal for problems involving the perpendicular distance from the origin and the angle of this perpendicular. Remember to correctly determine \( \sin \alpha \) and \( \cos \alpha \) from \( \tan \alpha \), possibly by drawing a right-angled triangle.

 

Question 10. n the position of the straight line x cos 30° + y sin 30° = 2 in relation to the co-ordinate axes, indicating clearly which angle is 30° and which length is 2 units. Find
(i) the equation of the straight line parallel to that given and passing through the point (4, 3);
(ii) the length of the perpendicular from the origin on to this line;
(iii) the distance between the two parallel straight lines.
Answer: The given equation of the straight line is \( x \cos 30^\circ + y \sin 30^\circ = 2 \). This is already in the normal form \( x \cos \alpha + y \sin \alpha = p \).
Here, \( \alpha = 30^\circ \) and \( p = 2 \). This means the perpendicular from the origin to the line has a length of 2 units and makes an angle of 30° with the positive x-axis.
Let's find 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 values into the equation:
\( x \frac{\sqrt{3}}{2} + y \frac{1}{2} = 2 \)
Multiply the entire equation by 2 to clear the denominators:
\( \sqrt{3}x + y = 4 \) ... (1)
To find the x and y-intercepts of this line, set y=0 for x-intercept and x=0 for y-intercept:
If \( y = 0 \), then \( \sqrt{3}x = 4 \implies x = \frac{4}{\sqrt{3}} \). So, \( A(\frac{4}{\sqrt{3}}, 0) \) is the x-intercept.
If \( x = 0 \), then \( y = 4 \). So, \( B(0, 4) \) is the y-intercept.

(i) To find the equation of a straight line parallel to the given line, we first need to find the slope of the given line. From equation (1), \( \sqrt{3}x + y = 4 \), we can write it in slope-intercept form \( y = mx + c \).
\( y = -\sqrt{3}x + 4 \)
The slope of this line is \( m_1 = -\sqrt{3} \).
A line parallel to this line will have the same slope. So, the slope of the new line is \( m_2 = -\sqrt{3} \).
The new line passes through the point (4, 3). We can use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \). This form is great for finding an equation when you have a point and the slope.
Substitute \( (x_1, y_1) = (4, 3) \) and \( m = -\sqrt{3} \):
\( y - 3 = -\sqrt{3}(x - 4) \)
\( y - 3 = -\sqrt{3}x + 4\sqrt{3} \)
Rearrange the terms to the general form:
\( \sqrt{3}x + y = 3 + 4\sqrt{3} \) ... (2)
This is the required equation of the parallel line.

(ii) To find the length of the perpendicular from the origin (0, 0) to the line \( \sqrt{3}x + y = 3 + 4\sqrt{3} \), we can use the formula for the perpendicular distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\), which is \( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \).
Rewrite equation (2) in the form \( Ax + By + C = 0 \):
\( \sqrt{3}x + y - (3 + 4\sqrt{3}) = 0 \)
Here, \( A = \sqrt{3} \), \( B = 1 \), \( C = -(3 + 4\sqrt{3}) \), and the point is the origin \((x_0, y_0) = (0, 0)\).
Distance \( p_2 = \frac{|\sqrt{3}(0) + 1(0) - (3 + 4\sqrt{3})|}{\sqrt{(\sqrt{3})^2 + 1^2}} \)
\( p_2 = \frac{|-(3 + 4\sqrt{3})|}{\sqrt{3 + 1}} \)
\( p_2 = \frac{3 + 4\sqrt{3}}{\sqrt{4}} \)
\( p_2 = \frac{3 + 4\sqrt{3}}{2} \)
This is the length of the perpendicular from the origin to the second line.

(iii) The distance between two parallel straight lines \( A_1x + B_1y + C_1 = 0 \) and \( A_2x + B_2y + C_2 = 0 \) is given by \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \), provided \( A_1=A_2=A \) and \( B_1=B_2=B \). This formula directly calculates the separation between the lines.
Our two parallel lines are:
Line 1: \( \sqrt{3}x + y - 4 = 0 \) (from \( \sqrt{3}x + y = 4 \), so \( C_1 = -4 \))
Line 2: \( \sqrt{3}x + y - (3 + 4\sqrt{3}) = 0 \) (from \( \sqrt{3}x + y = 3 + 4\sqrt{3} \), so \( C_2 = -(3 + 4\sqrt{3}) \))
Here, \( A = \sqrt{3} \) and \( B = 1 \).
Distance \( d = \frac{|-4 - (-(3 + 4\sqrt{3}))|}{\sqrt{(\sqrt{3})^2 + 1^2}} \)
\( d = \frac{|-4 + 3 + 4\sqrt{3}|}{\sqrt{3 + 1}} \)
\( d = \frac{|-1 + 4\sqrt{3}|}{\sqrt{4}} \)
\( d = \frac{|4\sqrt{3} - 1|}{2} \)
Now, we can approximate the value using \( \sqrt{3} \approx 1.732 \):
\( d = \frac{|4 \times 1.732 - 1|}{2} \)
\( d = \frac{|6.928 - 1|}{2} \)
\( d = \frac{|5.928|}{2} \)
\( d = 2.964 \) units.
In simple words: First, understand the given line equation which shows its distance from the origin (2 units) and its angle (30 degrees). Then, to find a parallel line, use the same slope but make sure it passes through the new point. After that, calculate the shortest distance from the origin to this new parallel line. Finally, use a special formula to find the exact distance between these two parallel lines.

🎯 Exam Tip: For problems involving parallel lines, remember that parallel lines have the same slope. The distance between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is \( \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \). This is a common and important formula to know.

 

Question 11. A straight line \( \frac{x}{a} – \frac{y}{b} = 1 \) passes through the point (8, 6) and cuts off a triangle of area 12 units from the axes of coordinates. Find the equations of the straight line.
Answer: The given equation of the straight line is \( \frac{x}{a} - \frac{y}{b} = 1 \) ... (1).
This line forms a triangle with the coordinate axes. The x-intercept is \( (a, 0) \) and the y-intercept is \( (0, -b) \). Notice the negative sign for the y-intercept because of the \( -\frac{y}{b} \) term in the equation.
The base of the triangle (along the x-axis) is \( |a| \) and the height (along the y-axis) is \( |-b| = |b| \).
The area of the triangle formed by the line and the axes is \( \frac{1}{2} \times \text{base} \times \text{height} \).
Area \( = \frac{1}{2} \times |a| \times |-b| = \frac{1}{2} |ab| \).
We are given that the area of this triangle is 12 units.
\( \frac{1}{2} |ab| = 12 \)
\( |ab| = 24 \)
This implies that \( ab = 24 \) or \( ab = -24 \).
The line passes through the point (8, 6). Substitute \( x = 8 \) and \( y = 6 \) into equation (1):
\( \frac{8}{a} - \frac{6}{b} = 1 \)
To clear the denominators, multiply the entire equation by \( ab \):
\( 8b - 6a = ab \) ... (2)
Now we have a system of two equations:
Case 1: \( ab = 24 \)
From this, \( b = \frac{24}{a} \).
Substitute this into equation (2):
\( 8\left(\frac{24}{a}\right) - 6a = 24 \)
\( \frac{192}{a} - 6a = 24 \)
Multiply by 'a' to clear the denominator:
\( 192 - 6a^2 = 24a \)
Rearrange into a quadratic equation:
\( 6a^2 + 24a - 192 = 0 \)
Divide by 6 to simplify:
\( a^2 + 4a - 32 = 0 \)
Factor the quadratic equation:
\( (a + 8)(a - 4) = 0 \)
So, \( a = -8 \) or \( a = 4 \).
If \( a = -8 \), then \( b = \frac{24}{-8} = -3 \).
If \( a = 4 \), then \( b = \frac{24}{4} = 6 \).

The equations of the straight lines are:
When \( a = 4 \) and \( b = 6 \):
\( \frac{x}{4} - \frac{y}{6} = 1 \)
When \( a = -8 \) and \( b = -3 \):
\( \frac{x}{-8} - \frac{y}{-3} = 1 \) which simplifies to \( -\frac{x}{8} + \frac{y}{3} = 1 \) or \( \frac{x}{8} - \frac{y}{3} = -1 \).
Case 2: \( ab = -24 \)
From this, \( b = -\frac{24}{a} \).
Substitute this into equation (2):
\( 8\left(-\frac{24}{a}\right) - 6a = -24 \)
\( -\frac{192}{a} - 6a = -24 \)
Multiply by 'a' to clear the denominator:
\( -192 - 6a^2 = -24a \)
Rearrange into a quadratic equation:
\( 6a^2 - 24a + 192 = 0 \)
Divide by 6 to simplify:
\( a^2 - 4a + 32 = 0 \)
To solve this quadratic equation, we can use the discriminant \( \Delta = B^2 - 4AC \).
\( \Delta = (-4)^2 - 4(1)(32) = 16 - 128 = -112 \).
Since the discriminant is negative \( (\Delta < 0) \), there are no real solutions for 'a' in this case. This means no line exists under the condition \( ab = -24 \).
So, the two possible equations of the straight line are:
1. \( \frac{x}{4} - \frac{y}{6} = 1 \)
2. \( \frac{x}{-8} - \frac{y}{-3} = 1 \) (or \( \frac{x}{8} - \frac{y}{3} = -1 \))
In simple words: First, write the line's equation using intercepts. Then, use the given area of the triangle formed by the line and the axes to find a relationship between the intercepts 'a' and 'b'. Next, use the point the line passes through to create another relationship. Solve these two relationships to find the possible values for 'a' and 'b'. Sometimes there will be more than one possible line. If a calculation results in no real numbers, that case is not possible.

🎯 Exam Tip: Remember that the area of a triangle formed by a line and the coordinate axes is \( \frac{1}{2} |ab| \), where 'a' and 'b' are the x and y intercepts respectively. You will often get a system of equations (one from the area, one from the point) which might lead to multiple solutions or no real solutions.

 

Question 12. A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of the intercepts on the axes is 60. Calculate the values of a and b and find the equation of the straight line.
Answer: The line passes through \( A(a, 0) \) and \( B(0, b) \). These are the x-intercept and y-intercept, respectively.
The length of the line segment contained between the axes is the distance between \( A(a, 0) \) and \( B(0, b) \). We are given this length as 13 units.
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\( \text{Length } AB = \sqrt{(0 - a)^2 + (b - 0)^2} = 13 \)
\( \sqrt{a^2 + b^2} = 13 \)
Square both sides to remove the square root:
\( a^2 + b^2 = 13^2 \)
\( a^2 + b^2 = 169 \) ... (1)
We are also given that the product of the intercepts on the axes is 60.
So, \( ab = 60 \) ... (2)
Now we have a system of two equations with two unknowns 'a' and 'b'. We can use algebraic identities to solve this.
We know that \( (a+b)^2 = a^2 + b^2 + 2ab \). This identity is useful for relating sums and products of variables.
Substitute the values from (1) and (2) into this identity:
\( (a+b)^2 = 169 + 2(60) \)
\( (a+b)^2 = 169 + 120 \)
\( (a+b)^2 = 289 \)
Take the square root of both sides:
\( a+b = \pm \sqrt{289} \)
\( a+b = \pm 17 \)
This gives us two cases:

Case I: When \( a + b = 17 \)
From this, \( b = 17 - a \).
Substitute this into equation (2) \( ab = 60 \):
\( a(17 - a) = 60 \)
\( 17a - a^2 = 60 \)
Rearrange into a quadratic equation:
\( a^2 - 17a + 60 = 0 \)
Factor the quadratic equation:
\( (a - 5)(a - 12) = 0 \)
So, \( a = 5 \) or \( a = 12 \).
If \( a = 5 \), then \( b = 17 - 5 = 12 \).
If \( a = 12 \), then \( b = 17 - 12 = 5 \).
This gives us two possible sets of intercepts: (5, 12) and (12, 5).
The corresponding equations of the lines (using intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \)) are:
1. For \( a = 5, b = 12 \): \( \frac{x}{5} + \frac{y}{12} = 1 \)
2. For \( a = 12, b = 5 \): \( \frac{x}{12} + \frac{y}{5} = 1 \)

Case II: When \( a + b = -17 \)
From this, \( b = -17 - a \).
Substitute this into equation (2) \( ab = 60 \):
\( a(-17 - a) = 60 \)
\( -17a - a^2 = 60 \)
Rearrange into a quadratic equation:
\( a^2 + 17a + 60 = 0 \)
Factor the quadratic equation:
\( (a + 5)(a + 12) = 0 \)
So, \( a = -5 \) or \( a = -12 \).
If \( a = -5 \), then \( b = -17 - (-5) = -17 + 5 = -12 \).
If \( a = -12 \), then \( b = -17 - (-12) = -17 + 12 = -5 \).
This gives us another two possible sets of intercepts: (-5, -12) and (-12, -5).
The corresponding equations of the lines are:
3. For \( a = -5, b = -12 \): \( \frac{x}{-5} + \frac{y}{-12} = 1 \)
4. For \( a = -12, b = -5 \): \( \frac{x}{-12} + \frac{y}{-5} = 1 \)
Thus, there are four possible equations for the straight line.
In simple words: You are given how long the line segment is between the axes and the multiplication result of its x and y intercepts. Use the distance formula for the length and the given product to set up two equations. Solve these equations to find all possible pairs of 'a' and 'b' (the intercepts). Then, for each pair, write down the equation of the line using the intercept form.

🎯 Exam Tip: Problems involving both the length of an intercept and the product of intercepts require using both the distance formula (for \( a^2+b^2 \)) and the product condition \( ab \). The identity \( (a+b)^2 = a^2+b^2+2ab \) is very helpful here. Remember to consider both positive and negative square roots for \( a+b \), as this leads to multiple possible lines.