RBSE Solutions Class 11 Maths Chapter 11 Straight Line Exercise 11.1

Question 1. Find the equation of straight line which is parallel to x-axis and
(i) lie at a distance of 5 unit from origin (above origin)
(ii) lie at a distance of 3 unit from origin (below origin)
Answer:
(i) When the line is parallel to the x-axis and lies 5 units above the origin, its equation is \( y = 5 \). This means every point on this line has a y-coordinate of 5. The line would be represented as AB.
(ii) When the line is parallel to the x-axis and lies 3 units below the origin, its equation is \( y = -3 \). This shows that all points on this line have a y-coordinate of -3. The line would be represented as PQ.

🎯 Exam Tip: Remember that lines parallel to the x-axis have the form \( y = c \), where \( c \) is the y-intercept. If the line is above the x-axis, \( c \) is positive; if below, \( c \) is negative.

Question 2. Find the equations of those straight lines which are parallel to x-axis and lie at a distance:
(i) \( a + b \)
(ii) \( a^2 - b^2 \)
(iii) \( b \cos \theta \)
Answer: Since the lines are parallel to the x-axis, their equations will be in the form \( y = c \) or \( y = -c \), depending on whether they are above or below the origin. We consider both positive and negative distances for each case.
(i) Equation of line AB is \( y = \pm (a+b) \). This includes two lines: one above and one below the x-axis.
(ii) Equation of line PQ is \( y = \pm (a^2 - b^2) \). This also represents two lines, one with a positive y-intercept and one with a negative y-intercept.
(iii) Equation of line RS is \( y = \pm (b \cos \theta) \). This similarly describes two lines, reflecting the distance on both sides of the x-axis.
In simple words: For lines parallel to the x-axis, the equation is always \( y \) equals some number. If the distance from the origin is 'd', then the lines are \( y = d \) (above) and \( y = -d \) (below).

🎯 Exam Tip: When a problem specifies "at a distance" without direction, always consider both positive and negative values for that distance, leading to two possible lines.

Question 3. Find the equation of those straight lines parallel to y-axis which are at a distance of:
(i) 5 unit from origin (positive x-axis)
(ii) 3 unit from origin (negative x-axis)
(iii) \( \frac{2}{5} \) unit from origin (positive x-axis)
Answer: Since the lines are parallel to the y-axis, their equations will be in the form \( x = c \) or \( x = -c \), depending on whether they are to the right or left of the origin.
(i) Equation of line AB is \( x = 5 \). This line is 5 units to the right of the y-axis.
(ii) Equation of line PQ is \( x = -3 \). This line is 3 units to the left of the y-axis.
(iii) Equation of line PQ is \( x = \frac{2}{5} \). This can also be written as \( 5x - 2 = 0 \). This line is \( \frac{2}{5} \) units to the right of the y-axis.
In simple words: For lines parallel to the y-axis, the equation is always \( x \) equals some number. If the distance from the origin is 'd', then the line is \( x = d \) (to the right) or \( x = -d \) (to the left).

🎯 Exam Tip: Lines parallel to the y-axis have the form \( x = k \), where \( k \) is the x-intercept. If the line is to the right of the y-axis, \( k \) is positive; if to the left, \( k \) is negative.

Question 4. Find the equation of those straight lines which are parallel to y-axis and at a distance of:
(i) \( \sqrt{7} \)
(ii) \( -\sqrt{3} + 2 \)
(iii) \( p + q \)
Answer: For lines parallel to the y-axis, the equation is \( x = \pm \text{distance} \). We consider both positive and negative distances.
(i) Equation of line AB is \( x = \pm \sqrt{7} \). These are two lines, one to the right and one to the left of the y-axis.
(ii) Equation of line MN is \( x = \pm (-\sqrt{3} + 2) \). This simplifies to \( x = \pm (2 - \sqrt{3}) \). Since \( 2 - \sqrt{3} \) is positive, these are two distinct lines on either side of the y-axis. The value \( 2-\sqrt{3} \) is a specific distance.
(iii) Equation of line RS is \( x = \pm (p+q) \). This means there are two lines, one at distance \( (p+q) \) to the right of y-axis and one at \( -(p+q) \) to the left of y-axis. These equations represent lines where all points have the same x-coordinate.
In simple words: Lines that go up and down (parallel to the y-axis) have equations like \( x = \text{number} \). If the problem gives a distance, you write \( x \) equals that distance and also \( x \) equals the negative of that distance, because a line can be on either side.

🎯 Exam Tip: Be careful with expressions involving square roots; calculate their approximate value to understand the line's position relative to the origin.

Question 6. Find the equations of lines passing through point (3,4) and parallel to both axis. Also find the equation of line parallel to these lines at a distance of 8 unit.
Answer:
Given point is \( (3, 4) \).
(i) Equation of line AB parallel to x-axis and passing through \( (3,4) \):
Since the line is parallel to the x-axis, its y-coordinate will be constant. As it passes through \( (3,4) \), the y-coordinate is 4.
So, the equation is \( y = 4 \).
Now, find the equation of a line parallel to \( y = 4 \) at a distance of 8 units.
These lines would be \( y = 4 \pm 8 \).
So, \( y = 12 \) or \( y = -4 \). This can be written as \( y+4=0 \).
(ii) Equation of line PQ parallel to y-axis and passing through \( (3,4) \):
Since the line is parallel to the y-axis, its x-coordinate will be constant. As it passes through \( (3,4) \), the x-coordinate is 3.
So, the equation is \( x = 3 \).
Now, find the equation of a line parallel to \( x = 3 \) at a distance of 8 units.
These lines would be \( x = 3 \pm 8 \).
So, \( x = 11 \) or \( x = -5 \). This can be written as \( x+5=0 \).
In simple words: If a line goes through a point and is parallel to the x-axis, its equation is \( y \) equals the y-value of the point. If it's parallel to the y-axis, its equation is \( x \) equals the x-value of the point. To find lines parallel to these at a specific distance, add and subtract that distance from the constant value in the equation.

🎯 Exam Tip: Visualizing the lines on a coordinate plane can help confirm if you're using the correct axis for parallelism and distance calculations.

Question 7. Write the coordinates of intersection points of \( x = \pm 4 \) and \( y = \pm 3 \) and find the area of rectangle so formed.
Answer:
Given the lines are \( x = 4, x = -4, y = 3, y = -3 \).
These lines form a rectangle. The intersection points are the vertices of this rectangle.
1. Point A: Intersection of \( y = 3 \) and \( x = 4 \). Coordinates: \( (4, 3) \).
2. Point B: Intersection of \( y = 3 \) and \( x = -4 \). Coordinates: \( (-4, 3) \).
3. Point C: Intersection of \( y = -3 \) and \( x = -4 \). Coordinates: \( (-4, -3) \).
4. Point D: Intersection of \( y = -3 \) and \( x = 4 \). Coordinates: \( (4, -3) \).
The coordinates of the vertices are \( (4, 3), (-4, 3), (-4, -3), (4, -3) \).
To find the area of the rectangle ABCD, we need its length and breadth.
Length of AB \( = \sqrt{(-4-4)^2 + (3-3)^2} = \sqrt{(-8)^2 + 0^2} = \sqrt{64} = 8 \) units.
Breadth of BC \( = \sqrt{(-4-(-4))^2 + (-3-3)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6 \) units.
Area of rectangle ABCD \( = \text{length} \times \text{breadth} = 8 \times 6 = 48 \) square units.
In simple words: The lines \( x = 4, x = -4, y = 3, y = -3 \) make a box shape. The corners of this box are \( (4,3), (-4,3), (-4,-3), (4,-3) \). We find how long and wide the box is by using these points, and then multiply length by width to get the area.

🎯 Exam Tip: Always sketch the given lines to correctly identify the vertices and calculate distances for length and breadth, rather than relying solely on formulas.

Question 8. Find the equations of those lines which pass through origin and
(i) make an angle of \( -135^\circ \) with x-axis.
(ii) have an angle of inclination \( 90^\circ - 60^\circ \).
(iii) make an angle of \( 45^\circ \) with x-axis and have x-intercept of -5.
Answer: The equation of a line passing through the origin is \( y = mx \), where \( m \) is the slope.
(i) Here, the angle \( \alpha = -135^\circ \). The slope \( m = \tan(-135^\circ) = \tan(180^\circ - 45^\circ) = -\tan(45^\circ) = -1 \).
So, the equation of the line is \( y = -1 \cdot x \), which simplifies to \( y = -x \) or \( x + y = 0 \).
(ii) Here, the angle of inclination is \( 90^\circ - 60^\circ = 30^\circ \). The slope \( m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
So, the equation of the line is \( y = \frac{1}{\sqrt{3}} x \).
Multiply both sides by \( \sqrt{3} \): \( \sqrt{3}y = x \).
Rearranging, we get \( x - \sqrt{3}y = 0 \). This equation represents a straight line.
(iii) The angle with the x-axis is \( 45^\circ \), so the slope \( m = \tan(45^\circ) = 1 \).
The x-intercept is -5. This means the line passes through the point \( (-5, 0) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \):
\( y - 0 = 1(x - (-5)) \)
\( y = x + 5 \)
Rearranging, we get \( x - y + 5 = 0 \). This is the required equation.
In simple words: For lines going through the origin, we use \( y = mx \). 'm' is found using \( \tan(\text{angle}) \). If it doesn't go through the origin but we know a point and its angle, we use a different formula.

🎯 Exam Tip: Remember to use the correct formula based on the given information: \( y=mx \) for lines through origin, \( y-y_1 = m(x-x_1) \) for point-slope, and \( \frac{x}{a} + \frac{y}{b} = 1 \) for intercept form.

Question 9. Find the equations of those lines which cut the following intercepts at x-axis and y-axis.
(i) 5, 3
(ii) \( -2, 3 \)
Answer: The equation of a line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept.
(i) Given x-intercept \( a = 5 \) and y-intercept \( b = 3 \).
Substitute these values into the intercept form:
\( \frac{x}{5} + \frac{y}{3} = 1 \)
To remove the denominators, find the least common multiple of 5 and 3, which is 15. Multiply the entire equation by 15:
\( 15 \left( \frac{x}{5} \right) + 15 \left( \frac{y}{3} \right) = 15 \cdot 1 \)
\( 3x + 5y = 15 \)
Rearranging, we get \( 3x + 5y - 15 = 0 \). This is the general form of the line's equation.
(ii) Given x-intercept \( a = -2 \) and y-intercept \( b = 3 \).
Substitute these values into the intercept form:
\( \frac{x}{-2} + \frac{y}{3} = 1 \)
Multiply the entire equation by -6 (the least common multiple of -2 and 3) to clear the denominators:
\( -6 \left( \frac{x}{-2} \right) + -6 \left( \frac{y}{3} \right) = -6 \cdot 1 \)
\( 3x - 2y = -6 \)
Rearranging, we get \( 3x - 2y + 6 = 0 \). This is the general form of the line's equation.
In simple words: When you know where a line crosses the x-axis (the x-intercept) and where it crosses the y-axis (the y-intercept), you can write its equation using a special form. Then, you can clear fractions and move everything to one side to get a simpler equation.

🎯 Exam Tip: Always simplify the intercept form equation to the general form \( Ax + By + C = 0 \) unless specifically asked to leave it in intercept form.

Question 10. Find the equation of the line which passes through (2, 3) and cuts equal intercepts on both axis.
Answer:
Let the x-intercept be \( a \) and the y-intercept be \( b \).
Since the line cuts equal intercepts on both axes, we have \( a = b \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substitute \( b = a \) into the equation:
\( \frac{x}{a} + \frac{y}{a} = 1 \)
This simplifies to \( x + y = a \).
The line passes through the point \( (2, 3) \). We can substitute these coordinates into the equation to find the value of \( a \).
\( 2 + 3 = a \)
\( a = 5 \)
Now, substitute the value of \( a \) back into the equation \( x + y = a \):
\( x + y = 5 \)
This is the required equation of the line.
In simple words: If a line crosses the x-axis and y-axis at the same distance from the origin (like 5 and 5), its equation looks like \( x + y = \text{distance} \). If we know a point the line goes through, we can find that distance.

🎯 Exam Tip: When dealing with "equal intercepts," remember this implies \( a=b \), which quickly simplifies the intercept form to \( x+y=a \).

Question 11. Find the equation of straight line which passes through point (1,2) and cuts intercepts on x-axis which is twice the intercepts on y-axis.
Answer:
Let the y-intercept be \( n \).
According to the problem, the x-intercept is twice the y-intercept. So, the x-intercept is \( 2n \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Here, \( a = 2n \) and \( b = n \). Substitute these values:
\( \frac{x}{2n} + \frac{y}{n} = 1 \)
To clear the denominators, multiply the entire equation by \( 2n \):
\( 2n \left( \frac{x}{2n} \right) + 2n \left( \frac{y}{n} \right) = 2n \cdot 1 \)
\( x + 2y = 2n \)
The line passes through the point \( (1, 2) \). Substitute these coordinates into the equation:
\( 1 + 2(2) = 2n \)
\( 1 + 4 = 2n \)
\( 5 = 2n \)
So, \( n = \frac{5}{2} \).
Now, substitute the value of \( n \) back into the equation \( x + 2y = 2n \):
\( x + 2y = 2 \left( \frac{5}{2} \right) \)
\( x + 2y = 5 \)
This is the required equation of the line. This problem combines the concept of intercepts with the condition of passing through a given point.
In simple words: We are looking for a line where the x-intercept is double the y-intercept. We write the equation using a variable for the intercepts, and then use the point the line passes through to find the exact values for those intercepts. Finally, we put these values back into the equation.

🎯 Exam Tip: Clearly define your variables (e.g., let the y-intercept be 'n') to avoid confusion, and make sure to substitute the point's coordinates correctly into the equation to find the unknown variable.

Question 12. Find the equation of the lines which passes through point (- 3, – 5) and intercepts cut by line between both axis bisect this point.
Answer:
Let the equation of the line in intercept form be \( \frac{x}{a} + \frac{y}{b} = 1 \).
This line cuts intercepts \( a \) on the x-axis (point \( (a, 0) \)) and \( b \) on the y-axis (point \( (0, b) \)).
The problem states that the point \( (-3, -5) \) bisects the segment formed by these intercepts. This means \( (-3, -5) \) is the midpoint of the segment connecting \( (a, 0) \) and \( (0, b) \).
Using the midpoint formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \):
\( (-3, -5) = \left( \frac{a + 0}{2}, \frac{0 + b}{2} \right) \)
\( (-3, -5) = \left( \frac{a}{2}, \frac{b}{2} \right) \)
Equating the coordinates:
For the x-coordinate: \( -3 = \frac{a}{2} \implies a = -6 \)
For the y-coordinate: \( -5 = \frac{b}{2} \implies b = -10 \)
Now, substitute these intercept values \( a = -6 \) and \( b = -10 \) into the intercept form equation:
\( \frac{x}{-6} + \frac{y}{-10} = 1 \)
To eliminate the denominators, find the least common multiple of -6 and -10, which is 30. Multiply the entire equation by 30:
\( 30 \left( \frac{x}{-6} \right) + 30 \left( \frac{y}{-10} \right) = 30 \cdot 1 \)
\( -5x - 3y = 30 \)
Rearranging into the standard form \( Ax + By + C = 0 \):
\( 5x + 3y + 30 = 0 \)
This is the required equation of the line.
In simple words: The line cuts the x and y axes, making a small piece of line. The given point is exactly in the middle of this piece. By using the midpoint formula, we can figure out where the line crosses the x and y axes. Once we have these crossing points, we can write the full equation of the line.

🎯 Exam Tip: Remember that if a point bisects a segment between intercepts, it's the midpoint of the segment. This is a key insight for determining the intercepts \( a \) and \( b \).

Question 13. Find the equation of two lines which passes through point (4, -3) and sum of intercepts cut by axis is 5 unit.
Answer:
Let the equation of the line in intercept form be \( \frac{x}{a} + \frac{y}{b} = 1 \).
Given that the sum of the intercepts is 5 units, so \( a + b = 5 \).
From this, we can express \( b \) in terms of \( a \): \( b = 5 - a \).
Substitute \( b = 5 - a \) into the intercept form equation:
\( \frac{x}{a} + \frac{y}{5 - a} = 1 \)
The line passes through the point \( (4, -3) \). Substitute these coordinates into the equation:
\( \frac{4}{a} + \frac{-3}{5 - a} = 1 \)
To solve for \( a \), find a common denominator, which is \( a(5 - a) \):
\( \frac{4(5 - a) - 3a}{a(5 - a)} = 1 \)
\( 20 - 4a - 3a = a(5 - a) \)
\( 20 - 7a = 5a - a^2 \)
Rearrange into a quadratic equation:
\( a^2 - 7a - 5a + 20 = 0 \)
\( a^2 - 12a + 20 = 0 \)
Factor the quadratic equation:
\( a^2 - 10a - 2a + 20 = 0 \)
\( a(a - 10) - 2(a - 10) = 0 \)
\( (a - 2)(a - 10) = 0 \)
This gives two possible values for \( a \):
Case 1: \( a - 2 = 0 \implies a = 2 \)
If \( a = 2 \), then \( b = 5 - a = 5 - 2 = 3 \).
The equation of the line is \( \frac{x}{2} + \frac{y}{3} = 1 \).
Multiply by 6 to clear denominators: \( 3x + 2y = 6 \), or \( 3x + 2y - 6 = 0 \).
Case 2: \( a - 10 = 0 \implies a = 10 \)
If \( a = 10 \), then \( b = 5 - a = 5 - 10 = -5 \).
The equation of the line is \( \frac{x}{10} + \frac{y}{-5} = 1 \).
Multiply by 10 to clear denominators: \( x - 2y = 10 \), or \( x - 2y - 10 = 0 \).
Thus, there are two lines that satisfy the given conditions: \( 3x + 2y - 6 = 0 \) and \( x - 2y - 10 = 0 \). These lines demonstrate how different intercept values can still meet the sum condition and pass through the specific point.
In simple words: We're looking for lines that go through a certain point and whose x and y intercepts add up to 5. We use algebra to find two possible pairs of intercepts that fit these rules. Each pair gives us an equation for a straight line.

🎯 Exam Tip: When a problem yields a quadratic equation for an intercept, remember that there will often be two valid solutions, leading to two distinct lines.

Question 14. Prove that equation of line at which axis reciprocals of intercepts are a and b is \( ax + by = 1 \).
Answer:
Let the x-intercept of the line be \( A \) and the y-intercept be \( B \).
The intercept form of the equation of a line is given by:
\( \frac{x}{A} + \frac{y}{B} = 1 \)
According to the problem, the reciprocals of the intercepts are \( a \) and \( b \).
This means: \( a = \frac{1}{A} \)
And: \( b = \frac{1}{B} \)
From these relationships, we can express \( A \) and \( B \) in terms of \( a \) and \( b \):
\( A = \frac{1}{a} \)
\( B = \frac{1}{b} \)
Now, substitute these expressions for \( A \) and \( B \) back into the intercept form equation:
\( \frac{x}{\frac{1}{a}} + \frac{y}{\frac{1}{b}} = 1 \)
When you divide by a fraction, it's the same as multiplying by its reciprocal:
\( x \cdot a + y \cdot b = 1 \)
\( ax + by = 1 \)
This proves that if the reciprocals of the intercepts are \( a \) and \( b \), the equation of the line is indeed \( ax + by = 1 \). This form is useful when the reciprocals of intercepts are known directly.
In simple words: If you take the numbers where a line crosses the axes, and then flip them upside down (find their reciprocals), let those flipped numbers be 'a' and 'b'. Then the equation of that line will always be \( ax + by = 1 \).

🎯 Exam Tip: This question demonstrates the reciprocal intercept form. Understanding the relationship between intercepts and their reciprocals is crucial for questions involving this unique form of a linear equation.

Question 16. The perpendicular drawn from origin to a straight line makes an angle of 30° with y-axis and its length is 2 units. Find the equation of this line.
Answer: The perpendicular distance from the origin to the line is given as \( p = 2 \) units.
The perpendicular (normal) makes an angle of \( 30^\circ \) with the y-axis. This means the angle \( \alpha \) it makes with the positive x-axis is \( 90^\circ + 30^\circ = 120^\circ \). This angle is in the second quadrant.
Now, we find the cosine and sine of this angle:
\( \cos \alpha = \cos 120^\circ = -\frac{1}{2} \)
\( \sin \alpha = \sin 120^\circ = \frac{\sqrt{3}}{2} \)
The equation of a straight line in the normal form is \( x \cos \alpha + y \sin \alpha = p \).
Substitute the values of \( p \), \( \cos \alpha \), and \( \sin \alpha \):
\( x \left(-\frac{1}{2}\right) + y \left(\frac{\sqrt{3}}{2}\right) = 2 \)
To remove the denominators, multiply the entire equation by 2:
\( -x + \sqrt{3}y = 4 \)
Rearrange the terms to get the standard form:
\( x - \sqrt{3}y + 4 = 0 \)
This is the required equation of the straight line. The normal form helps find the line's equation when its distance from the origin and the angle of the normal are known.
In simple words: We know how far the line is from the center (origin) and the angle of the shortest line to it. Using these, we can write down the equation of the straight line. The perpendicular makes an angle of 120 degrees with the positive x-axis.

🎯 Exam Tip: Remember the normal form of a line is \( x \cos \alpha + y \sin \alpha = p \). Carefully determine \( \alpha \) (angle with the positive x-axis) and \( p \) (perpendicular distance from the origin).

Question 17. Find the length of that part of line \( x \sin a + y \cos a = \sin 2a \) which cuts axis at mid point. Also, find the coordinate of mid point of this part.
Answer: The given equation of the line is \( x \sin a + y \cos a = \sin 2a \).
We can rewrite \( \sin 2a \) as \( 2 \sin a \cos a \).
So, the equation becomes: \( x \sin a + y \cos a = 2 \sin a \cos a \)
To find the x and y intercepts, we convert this into the intercept form \( \frac{x}{X} + \frac{y}{Y} = 1 \).
Divide the entire equation by \( 2 \sin a \cos a \):
\( \frac{x \sin a}{2 \sin a \cos a} + \frac{y \cos a}{2 \sin a \cos a} = \frac{2 \sin a \cos a}{2 \sin a \cos a} \)
\( \implies \frac{x}{2 \cos a} + \frac{y}{2 \sin a} = 1 \)
From this form, the x-intercept is \( X = 2 \cos a \) and the y-intercept is \( Y = 2 \sin a \).
The line cuts the x-axis at point A \( (2 \cos a, 0) \) and the y-axis at point B \( (0, 2 \sin a) \).
Now, we find the length of the line segment AB using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
Length of AB \( = \sqrt{(0 - 2 \cos a)^2 + (2 \sin a - 0)^2} \)
\( = \sqrt{(-2 \cos a)^2 + (2 \sin a)^2} \)
\( = \sqrt{4 \cos^2 a + 4 \sin^2 a} \)
Factor out 4:
\( = \sqrt{4(\cos^2 a + \sin^2 a)} \)
Using the identity \( \cos^2 a + \sin^2 a = 1 \):
\( = \sqrt{4 \times 1} \)
\( = \sqrt{4} = 2 \) units.
Next, we find the coordinates of the midpoint M of the line segment AB using the midpoint formula \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \):
Midpoint M \( = \left(\frac{2 \cos a + 0}{2}, \frac{0 + 2 \sin a}{2}\right) \)
\( = (\cos a, \sin a) \)
The length of the line segment is 2 units, and its midpoint coordinates are \( (\cos a, \sin a) \). This method clearly shows how the intercepts lead to both the length and the midpoint.
In simple words: First, we change the line equation to find where it crosses the x-axis and y-axis. Then, we use these crossing points to find how long the line segment is between them and also find the exact middle point of that segment.

🎯 Exam Tip: To find intercepts, always convert the line equation to the form \( \frac{x}{a} + \frac{y}{b} = 1 \), where 'a' is the x-intercept and 'b' is the y-intercept.

Question 18. Find the equation of straight line for which \( p = 3 \) and \( \cos \alpha = \frac {\sqrt { 3 }}{2} \), where \( p \) is the length of perpendicular drawn from origin to the line and \( \alpha \) is the angle formed by perpendicular with x-axis.
Answer: We are given the perpendicular distance from the origin to the line as \( p = 3 \) units.
The angle \( \alpha \) that the perpendicular makes with the x-axis is such that \( \cos \alpha = \frac{\sqrt{3}}{2} \).
Since \( \cos \alpha \) is positive, \( \alpha \) can be in the first quadrant or the fourth quadrant.
Case 1: \( \alpha \) is in the first quadrant.
If \( \cos \alpha = \frac{\sqrt{3}}{2} \), then \( \alpha = 30^\circ \).
Then \( \sin \alpha = \sin 30^\circ = \frac{1}{2} \).
The equation of the line in normal form is \( x \cos \alpha + y \sin \alpha = p \).
Substitute the values:
\( x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 3 \)
Multiply by 2 to clear denominators:
\( \sqrt{3}x + y = 6 \)
This is the first possible equation of the straight line.

Case 2: \( \alpha \) is in the fourth quadrant.
If \( \cos \alpha = \frac{\sqrt{3}}{2} \), then \( \alpha = 360^\circ - 30^\circ = 330^\circ \).
Then \( \sin \alpha = \sin 330^\circ = -\frac{1}{2} \).
The equation of the line in normal form is \( x \cos \alpha + y \sin \alpha = p \).
Substitute the values:
\( x \left(\frac{\sqrt{3}}{2}\right) + y \left(-\frac{1}{2}\right) = 3 \)
Multiply by 2 to clear denominators:
\( \sqrt{3}x - y = 6 \)
This is the second possible equation of the straight line. Both equations are valid because the given cosine value allows for two different angles for the normal. The specific quadrant of \( \alpha \) affects the sign of \( \sin \alpha \).
In simple words: We are given how far the line is from the origin and the cosine of the angle its shortest perpendicular line makes with the x-axis. Since cosine can be positive in two quadrants, we find two possible angles. Each angle gives a different equation for the straight line.

🎯 Exam Tip: When \( \cos \alpha \) is given, remember that \( \alpha \) can be in two different quadrants. Always calculate \( \sin \alpha \) for both possibilities to find all valid equations for the line.