RBSE Solutions Class 11 Maths Chapter 11 Straight Line Exercise 11.2

Question 1. Convert the following equations in slope form and intercept form and find the value of constants used in standard form.
(i) \( 7x - 13y = 15 \)
(ii) \( 5x + 6y + 8 = 0 \)
Answer:
(i) Given equation is \( 7x - 13y = 15 \).
To convert to slope form (\( y = mx + c \)):
\( -13y = -7x + 15 \)
\( y = \frac{-7x + 15}{-13} \)
\( \implies y = \frac{7}{13}x - \frac{15}{13} \)
Comparing with \( y = mx + c \), we find:
\( m = \frac{7}{13} \)
\( c = -\frac{15}{13} \)
To convert to intercept form (\( \frac{x}{a} + \frac{y}{b} = 1 \)):
\( 7x - 13y = 15 \)
Divide the entire equation by 15:
\( \frac{7x}{15} - \frac{13y}{15} = \frac{15}{15} \)
\( \implies \frac{x}{15/7} + \frac{y}{-15/13} = 1 \)
Comparing with \( \frac{x}{a} + \frac{y}{b} = 1 \), we get:
\( a = \frac{15}{7} \)
\( b = -\frac{15}{13} \)

(ii) Given equation is \( 5x + 6y + 8 = 0 \).
To convert to slope form (\( y = mx + c \)):
\( 6y = -5x - 8 \)
\( \implies y = -\frac{5}{6}x - \frac{8}{6} \)
\( \implies y = -\frac{5}{6}x - \frac{4}{3} \)
Comparing with \( y = mx + c \), we find:
\( m = -\frac{5}{6} \)
\( c = -\frac{4}{3} \)
To convert to intercept form (\( \frac{x}{a} + \frac{y}{b} = 1 \)):
\( 5x + 6y = -8 \)
Divide the entire equation by -8:
\( \frac{5x}{-8} + \frac{6y}{-8} = \frac{-8}{-8} \)
\( \implies \frac{x}{-8/5} + \frac{y}{-8/6} = 1 \)
\( \implies \frac{x}{-8/5} + \frac{y}{-4/3} = 1 \)
Comparing with \( \frac{x}{a} + \frac{y}{b} = 1 \), we get:
\( a = -\frac{8}{5} \)
\( b = -\frac{4}{3} \)
In simple words: We change each equation into two standard forms: slope-intercept form (to find slope 'm' and y-intercept 'c') and intercept form (to find x-intercept 'a' and y-intercept 'b'). We do this by rearranging the terms and dividing by constants until the equation matches the standard look.

🎯 Exam Tip: Remember to simplify fractions for 'c', 'a', and 'b' to their lowest terms. Double-check your signs, as a small error can change the entire line's position.

Question 3. Find the tangent of the following lines with +ve x-axis.
(i) \( \sqrt{3}x-y+2=0 \)
(ii) \( x + \sqrt{3}y - 2\sqrt{3} = 0 \)
Answer:
(i) Given equation: \( \sqrt{3}x - y + 2 = 0 \).
To find the tangent (slope), convert to slope-intercept form \( y = mx + c \):
\( y = \sqrt{3}x + 2 \)
Comparing with \( y = mx + c \), we get \( m = \sqrt{3} \).
Since \( m = \tan \theta \), where \( \theta \) is the angle with the positive x-axis:
\( \tan \theta = \sqrt{3} \)
\( \implies \theta = 60^\circ \).
Thus, the tangent of the required angle is \( 60^\circ \). This means the line makes a 60-degree angle with the x-axis.

(ii) Given equation: \( x + \sqrt{3}y - 2\sqrt{3} = 0 \).
To find the tangent (slope), convert to slope-intercept form \( y = mx + c \):
\( \sqrt{3}y = -x + 2\sqrt{3} \)
\( y = -\frac{1}{\sqrt{3}}x + \frac{2\sqrt{3}}{\sqrt{3}} \)
\( \implies y = -\frac{1}{\sqrt{3}}x + 2 \)
Comparing with \( y = mx + c \), we get \( m = -\frac{1}{\sqrt{3}} \).
Since \( m = \tan \theta \):
\( \tan \theta = -\frac{1}{\sqrt{3}} \)
\( \implies \theta = 150^\circ \).
Thus, the tangent of the required angle is \( 150^\circ \). This line leans backwards, forming an obtuse angle with the x-axis.
In simple words: We rewrite each line equation to find its slope. The slope is the "tangent" of the angle the line makes with the positive x-axis. We then find that angle using the calculated tangent value.

🎯 Exam Tip: Remember that \( \tan \theta \) is positive for angles between 0° and 90° and negative for angles between 90° and 180°. Use this to quickly check if your angle matches the sign of the slope.

Question 4. Prove that \( (x_1, y_1) \) will be coordinate of mid point of section cut by line
Answer: Let the line cut the coordinate axes at points \( A(0, 2y_1) \) on the y-axis and \( B(2x_1, 0) \) on the x-axis. We want to prove that \( (x_1, y_1) \) is the midpoint of the line segment AB.
The coordinates of the midpoint of a line segment joining \( (x_a, y_a) \) and \( (x_b, y_b) \) are given by \( \left( \frac{x_a + x_b}{2}, \frac{y_a + y_b}{2} \right) \).
Substituting the coordinates of A and B:
Midpoint \( = \left( \frac{0 + 2x_1}{2}, \frac{2y_1 + 0}{2} \right) \)
\( \implies \) Midpoint \( = \left( \frac{2x_1}{2}, \frac{2y_1}{2} \right) \)
\( \implies \) Midpoint \( = (x_1, y_1) \).
Therefore, it is proven that \( (x_1, y_1) \) is the coordinate of the midpoint of the section cut by the line between the axes at these specific points. This formula is a key way to find the center of any line segment.
In simple words: If a line crosses the 'x' axis at \( (2x_1, 0) \) and the 'y' axis at \( (0, 2y_1) \), then the point exactly in the middle of these two crossings is \( (x_1, y_1) \). We use the midpoint formula to show this.

🎯 Exam Tip: Always clearly define your two endpoints before applying the midpoint formula to avoid confusion. Labeling \( (x_1, y_1) \) and \( (x_2, y_2) \) explicitly helps prevent errors.

Question 5. Find the length of intercept cut between the axis from straight line \( 3x + 4y = 6 \) and also find its mid point.
Answer: Given equation of the straight line: \( 3x + 4y = 6 \).
First, convert the equation into intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \) to find the x and y intercepts.
Divide the entire equation by 6:
\( \frac{3x}{6} + \frac{4y}{6} = \frac{6}{6} \)
\( \implies \frac{x}{2} + \frac{y}{3/2} = 1 \)
So, the x-intercept is \( a = 2 \), meaning the line crosses the x-axis at point \( A(2, 0) \).
The y-intercept is \( b = \frac{3}{2} \), meaning the line crosses the y-axis at point \( B(0, 3/2) \).

Next, find the length of the intercept cut between the axes. This is the distance between points \( A(2, 0) \) and \( B(0, 3/2) \).
Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( d = \sqrt{(0-2)^2 + (\frac{3}{2}-0)^2} \)
\( \implies d = \sqrt{(-2)^2 + (\frac{3}{2})^2} \)
\( \implies d = \sqrt{4 + \frac{9}{4}} \)
\( \implies d = \sqrt{\frac{16}{4} + \frac{9}{4}} \)
\( \implies d = \sqrt{\frac{25}{4}} \)
\( \implies d = \frac{5}{2} \) units.

Finally, find the midpoint of the intercept AB. The midpoint formula for \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
Midpoint \( = \left( \frac{2 + 0}{2}, \frac{0 + 3/2}{2} \right) \)
\( \implies \) Midpoint \( = \left( \frac{2}{2}, \frac{3/2}{2} \right) \)
\( \implies \) Midpoint \( = (1, \frac{3}{4}) \).
The length of the line segment between the axes is 2.5 units, and its middle point is \( (1, 0.75) \).
In simple words: First, we change the line's equation to find where it cuts the 'x' and 'y' axes. Then, we use the distance formula to find how long this cut-off piece of the line is. After that, we find the middle point of this piece using the midpoint formula.

🎯 Exam Tip: Always convert the equation to intercept form first to correctly identify the points where the line crosses the axes. This makes calculating both length and midpoint straightforward.

Question 6. Find the values of a and b where equations \( 5x - 4y = 20 \) and \( ax - by + 1 = 0 \) represent same straight line.
Answer: Given that the two equations represent the same straight line:
Equation 1: \( 5x - 4y = 20 \)
Equation 2: \( ax - by + 1 = 0 \)
For two lines \( A_1x + B_1y + C_1 = 0 \) and \( A_2x + B_2y + C_2 = 0 \) to be the same, their coefficients must be proportional:
\( \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \)
Rewrite the equations in the standard form \( Ax + By + C = 0 \):
Equation 1: \( 5x - 4y - 20 = 0 \)
Equation 2: \( ax - by + 1 = 0 \)
Now, compare the coefficients:
\( \frac{5}{a} = \frac{-4}{-b} = \frac{-20}{1} \)
From \( \frac{5}{a} = -20 \):
\( 5 = -20a \)
\( \implies a = \frac{5}{-20} = -\frac{1}{4} \)
From \( \frac{-4}{-b} = -20 \):
\( \frac{4}{b} = -20 \)
\( 4 = -20b \)
\( \implies b = \frac{4}{-20} = -\frac{1}{5} \)
Thus, the values are \( a = -\frac{1}{4} \) and \( b = -\frac{1}{5} \). For two lines to be identical, all their corresponding parts must be in the same ratio.
In simple words: If two different equations describe the exact same straight line, it means their numbers (coefficients) are all in proportion to each other. We use this rule to find the unknown 'a' and 'b' values.

🎯 Exam Tip: When comparing coefficients of two identical lines, ensure both equations are in the same standard form (e.g., \( Ax + By + C = 0 \)) before setting up the proportions.

Question 9. \( \frac {x}{a} + \frac {y}{b} = 1 \) and \( 2x - 3y = 5 \) represent same line, then find the values of a and b.
Answer: Given that the two equations represent the same straight line:
Equation 1: \( \frac{x}{a} + \frac{y}{b} = 1 \)
Equation 2: \( 2x - 3y = 5 \)
Rewrite both equations in the standard form \( Ax + By + C = 0 \):
Equation 1: \( \frac{bx + ay}{ab} = 1 \)
\( \implies bx + ay = ab \)
\( \implies bx + ay - ab = 0 \)
Equation 2: \( 2x - 3y - 5 = 0 \)
For the lines to be the same, their coefficients must be proportional:
\( \frac{b}{2} = \frac{a}{-3} = \frac{-ab}{-5} \)
From \( \frac{a}{-3} = \frac{-ab}{-5} \):
\( \frac{a}{-3} = \frac{ab}{5} \)
Since \( a \) cannot be zero (otherwise, Eq 1 is undefined or trivial), we can divide by \( a \):
\( \frac{1}{-3} = \frac{b}{5} \)
\( \implies -3b = 5 \)
\( \implies b = -\frac{5}{3} \)
From \( \frac{b}{2} = \frac{a}{-3} \):
\( -3b = 2a \)
Substitute the value of \( b = -\frac{5}{3} \):
\( -3 \left(-\frac{5}{3}\right) = 2a \)
\( \implies 5 = 2a \)
\( \implies a = \frac{5}{2} \)
Thus, the values are \( a = \frac{5}{2} \) and \( b = -\frac{5}{3} \). These values ensure the two different-looking equations actually describe the exact same line on a graph.
In simple words: When two lines are exactly the same, their numerical parts (coefficients) are proportional. We use this rule to solve for 'a' and 'b' by setting up fractions of their corresponding numbers.

🎯 Exam Tip: Convert both equations into the general form \( Ax + By + C = 0 \) before comparing coefficients. Be careful with signs, especially when moving terms across the equals sign.

Question 10. If straight line \( y = mx + c \) and \( x \cos \alpha + 7 \sin \alpha = p \) represent same line, then find slope of line with x axis, and length of intercept cut from y – axis.
Answer: Given that the two equations represent the same straight line:
Equation 1: \( y = mx + c \)
Equation 2: \( x \cos \alpha + y \sin \alpha = p \) (Note: We assume \( 7 \sin \alpha \) in the question is a typo and should be \( y \sin \alpha \), which is consistent with the standard form of the equation of a line.)
To find the slope and y-intercept, convert Equation 2 into the slope-intercept form (\( y = M x + C \)):
\( y \sin \alpha = -x \cos \alpha + p \)
Divide by \( \sin \alpha \) (assuming \( \sin \alpha \ne 0 \)):
\( y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha} \)
\( \implies y = -(\cot \alpha) x + p \operatorname{cosec} \alpha \)
Comparing this with Equation 1, \( y = mx + c \):
The slope of the line with the x-axis is \( m \), so \( m = -\cot \alpha \).
The length of the intercept cut from the y-axis is \( c \), so \( c = p \operatorname{cosec} \alpha \).
The angle \( \theta \) that the line makes with the positive x-axis is found using \( m = \tan \theta \).
\( \tan \theta = -\cot \alpha \)
\( \implies \tan \theta = \tan (90^\circ + \alpha) \) (since \( -\cot \alpha = \tan(90^\circ + \alpha) \))
\( \implies \theta = 90^\circ + \alpha \).
The slope of the line is \( -\cot \alpha \), and it creates an angle of \( (90^\circ + \alpha) \) with the x-axis. The line also cuts the y-axis at a distance of \( p \operatorname{cosec} \alpha \).
In simple words: When two different-looking equations describe the same line, we can compare their parts. We convert the second equation into the "y = mx + c" format to directly find the slope ('m') and the y-intercept ('c'). The slope tells us the tangent of the angle the line makes with the x-axis.

🎯 Exam Tip: When given two forms of the same line equation, always convert them to a common form (like slope-intercept or general form) to easily compare coefficients or extract desired values.

Question 12. Find the equation of line passing through the following two given points :
(i) \( (3, 4) \) and \( (5, 6) \)
(ii) \( (0, -a) \) and \( (b, 0) \)
(iii) \( (a, b) \) and \( (a + b, a- b) \)
(iv) \( (at_1, a/t_1) \) and \( (at_2, a/t_2) \)
(v) \( (a \sec \alpha, b \tan \alpha) \) and \( (a \sec \beta, b \tan \beta) \)
Answer: The equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \) or \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \).

(i) Points: \( (x_1, y_1) = (3, 4) \) and \( (x_2, y_2) = (5, 6) \)
\( \frac{y - 4}{x - 3} = \frac{6 - 4}{5 - 3} \)
\( \implies \frac{y - 4}{x - 3} = \frac{2}{2} \)
\( \implies \frac{y - 4}{x - 3} = 1 \)
\( \implies y - 4 = x - 3 \)
\( \implies y - x = 4 - 3 \)
\( \implies y - x = 1 \).

(ii) Points: \( (x_1, y_1) = (0, -a) \) and \( (x_2, y_2) = (b, 0) \)
\( \frac{y - (-a)}{x - 0} = \frac{0 - (-a)}{b - 0} \)
\( \implies \frac{y + a}{x} = \frac{a}{b} \)
\( \implies b(y + a) = ax \)
\( \implies by + ab = ax \)
\( \implies ax - by - ab = 0 \). (This is also the intercept form \( \frac{x}{b} + \frac{y}{-a} = 1 \)).

(iii) Points: \( (x_1, y_1) = (a, b) \) and \( (x_2, y_2) = (a+b, a-b) \)
Slope \( m = \frac{(a-b) - b}{(a+b) - a} = \frac{a - 2b}{b} \)
Using \( y - y_1 = m(x - x_1) \):
\( y - b = \frac{a - 2b}{b} (x - a) \)
\( \implies b(y - b) = (a - 2b)(x - a) \)
\( \implies by - b^2 = (a - 2b)x - a(a - 2b) \)
\( \implies by - b^2 = (a - 2b)x - a^2 + 2ab \)
\( \implies (a - 2b)x - by + b^2 - a^2 + 2ab = 0 \).

(iv) Points: \( (x_1, y_1) = (at_1, a/t_1) \) and \( (x_2, y_2) = (at_2, a/t_2) \)
Slope \( m = \frac{a/t_2 - a/t_1}{at_2 - at_1} = \frac{a(1/t_2 - 1/t_1)}{a(t_2 - t_1)} = \frac{(t_1 - t_2)/(t_1t_2)}{t_2 - t_1} = -\frac{1}{t_1t_2} \)
Using \( y - y_1 = m(x - x_1) \):
\( y - \frac{a}{t_1} = -\frac{1}{t_1t_2} (x - at_1) \)
Multiply both sides by \( t_1t_2 \):
\( t_1t_2 y - at_2 = -(x - at_1) \)
\( \implies t_1t_2 y - at_2 = -x + at_1 \)
\( \implies x + t_1t_2 y = at_1 + at_2 \)
\( \implies x + t_1t_2 y = a(t_1 + t_2) \).

(v) Points: \( (x_1, y_1) = (a \sec \alpha, b \tan \alpha) \) and \( (x_2, y_2) = (a \sec \beta, b \tan \beta) \)
Using the two-point form of the line equation:
\( \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \)
\( \frac{y - b \tan \alpha}{x - a \sec \alpha} = \frac{b \tan \beta - b \tan \alpha}{a \sec \beta - a \sec \alpha} \)
Convert tan and sec to sin and cos:
\( \frac{y - b \frac{\sin \alpha}{\cos \alpha}}{x - a \frac{1}{\cos \alpha}} = \frac{b (\frac{\sin \beta}{\cos \beta} - \frac{\sin \alpha}{\cos \alpha})}{a (\frac{1}{\cos \beta} - \frac{1}{\cos \alpha})} \)
Simplify the complex fractions:
\( \frac{\frac{y \cos \alpha - b \sin \alpha}{\cos \alpha}}{\frac{x \cos \alpha - a}{\cos \alpha}} = \frac{b \frac{\sin \beta \cos \alpha - \sin \alpha \cos \beta}{\cos \beta \cos \alpha}}{a \frac{\cos \alpha - \cos \beta}{\cos \beta \cos \alpha}} \)
\( \implies \frac{y \cos \alpha - b \sin \alpha}{x \cos \alpha - a} = \frac{b (\sin \beta \cos \alpha - \sin \alpha \cos \beta)}{a (\cos \alpha - \cos \beta)} \)
Using the sine difference formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \):
\( \implies \frac{y \cos \alpha - b \sin \alpha}{x \cos \alpha - a} = \frac{b \sin(\beta - \alpha)}{a (\cos \alpha - \cos \beta)} \)
Cross-multiply to get the final equation:
\( a(\cos \alpha - \cos \beta)(y \cos \alpha - b \sin \alpha) = b \sin(\beta - \alpha)(x \cos \alpha - a) \).
This process applies the fundamental formula for a straight line through two points, adapting it for each pair of given coordinates. It's a key method for defining lines mathematically.
In simple words: For each pair of points, we use a special formula that finds the equation of the straight line connecting them. We plug in the given 'x' and 'y' values from both points and then simplify the equation to its final form.

🎯 Exam Tip: Always write down the general two-point formula first. Be methodical with algebraic simplifications and keep track of signs, especially when dealing with variables or trigonometric functions in coordinates.