RBSE Solutions Class 11 Maths Chapter 11 सरल रेखा More Questions

Question 1. उस सरल रेखा का समीकरण जो y-अक्ष के समान्तर तथा y-अक्ष के बायीं ओर 5 इकाई की दूरी पर है
(A) y = 5
(B) x = 5
(C) x = - 5
(D) y = - 5
Answer: (C) x = - 5
In simple words: A line parallel to the y-axis is always in the form \( x = \text{constant} \). Since it is 5 units to the left of the y-axis, the constant must be -5.

🎯 Exam Tip: Remember that lines parallel to the y-axis have the equation \( x = k \) and lines parallel to the x-axis have the equation \( y = k \).

Question 2. उस रेखा का समीकरण जो बिन्दु (3, – 4) से होकर गुजरती है। तथा x-अक्ष के समान्तर है
(A) x = 3
(B) y = – 4
(C) x + 3 = 0
(D)
Answer: (B) y = – 4
In simple words: A line parallel to the x-axis means its y-coordinate stays constant. Since the line passes through (3, -4), its y-coordinate will always be -4. So, the equation is \( y = -4 \).

🎯 Exam Tip: If a line is parallel to the x-axis, its equation is \( y = \text{constant} \). If it's parallel to the y-axis, its equation is \( x = \text{constant} \).

Question 3. (Question text not provided in source)
(A) 1
(B) 0
(C) \( \infty \)
(D) \( \frac{\pi}{2} \)
Answer: (C) \( \infty \)
In simple words: Without the full question, it is hard to say. However, when we get infinity as an answer in geometry, it often relates to slopes of vertical lines or undefined values.

🎯 Exam Tip: Always look for the full question text, especially for MCQs, to understand what is being asked. Sometimes, \( \infty \) (infinity) can be the answer in specific mathematical contexts like finding the slope of a vertical line.

Question 4. समीकरण \( x \cdot \frac{1}{2} + y \frac{\sqrt{3}}{2} = 5 \) द्वारा निरूपित सरल रेखा निम्न रूप में है-
(A) सममित रूप
(B) झुकाव रूप
(C) अन्त:खण्ड रूप
(D) लम्ब रूप
Answer: (D) लम्ब रूप
In simple words: The given equation resembles the normal (perpendicular) form of a straight line, which is \( x \cos \alpha + y \sin \alpha = p \). This form describes a line based on the length of the perpendicular from the origin to the line and the angle it makes with the x-axis.

🎯 Exam Tip: Recognize standard forms of a straight line: slope-intercept (\( y = mx+c \)), intercept form (\( \frac{x}{a} + \frac{y}{b} = 1 \)), and normal/perpendicular form (\( x \cos \alpha + y \sin \alpha = p \)).

Question 5. सरल रेखा \( 3x – 4y = 7 \) के समान्तर और मूल बिन्दु से गुजरने वाले रेखा का समीकरण है-
(A) \( 3x – 4y = 1 \)
(B) \( 3x – 4y = 0 \)
(C) \( 4x - 3y = 1 \)
(D) \( 3y - 4x = 0 \)
Answer: (B) \( 3x – 4y = 0 \)
In simple words: A line parallel to \( 3x – 4y = 7 \) will have the same slope, so its equation will be \( 3x – 4y = k \). Since it passes through the origin (0,0), substituting these values gives \( 3(0) – 4(0) = k \), which means \( k = 0 \). Thus, the equation is \( 3x – 4y = 0 \).

🎯 Exam Tip: Parallel lines have the same slope. If a line passes through the origin, the constant term in its general equation \( Ax + By + C = 0 \) will be zero.

Question 6. मूल बिन्दु से सरल रेखा \( x + \sqrt{3}y = 1 \) पर डाले गए लम्ब की लम्बाई p है तो p का मान है
(A) \( \frac{1}{4} \)
(B) \( \frac{\sqrt{3}}{2} \)
(C) \( \frac{1}{2} \)
(D) 1
Answer: (C) \( \frac{1}{2} \)
In simple words: To find the perpendicular distance from the origin (0,0) to a line \( Ax + By + C = 0 \), use the formula \( \frac{|C|}{\sqrt{A^2 + B^2}} \). Here, the equation is \( x + \sqrt{3}y - 1 = 0 \). So, the distance is \( \frac{|-1|}{\sqrt{1^2 + (\sqrt{3})^2}} = \frac{1}{\sqrt{1+3}} = \frac{1}{\sqrt{4}} = \frac{1}{2} \).

🎯 Exam Tip: The formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \). For the origin (0,0), it simplifies to \( \frac{|C|}{\sqrt{A^2 + B^2}} \).

Question 7. (Question text not provided in source)
(A) \( \frac{5}{3} \)
(B) \( -\frac{5}{3} \)
(C) \( -\frac{3}{5} \)
(D) \( \frac{3}{5} \)
Answer: (A) \( \frac{5}{3} \)
In simple words: Without the question, it's hard to explain why \( \frac{5}{3} \) is the answer. It could be a slope, an intercept, or a distance value from a specific context.

🎯 Exam Tip: Always ensure you have the complete question before attempting to answer, especially for multiple-choice questions, to avoid errors.

Question 8. सरल रेखा \( 3x – 4y + 7 = 0 \) पर लम्ब और बिन्दु (1, -2) में से गुजरने वाली रेखा का समीकरण होगा
(A) \( 4x + 3y - 2 = 0 \)
(B) \( 4x + 3y + 2 = 0 \)
(C) \( 4x - 3y + 2 = 0 \)
(D) \( 4x - 3y - 2 = 0 \)
Answer: (B) \( 4x + 3y + 2 = 0 \)
In simple words: A line perpendicular to \( Ax + By + C = 0 \) has the form \( Bx - Ay + k = 0 \). So, for \( 3x – 4y + 7 = 0 \), the perpendicular line is \( -4x - 3y + k = 0 \) or \( 4x + 3y - k = 0 \). Since it passes through (1, -2), substitute \( x=1 \) and \( y=-2 \) to find \( k \). \( 4(1) + 3(-2) + k = 0 \implies 4 - 6 + k = 0 \implies -2 + k = 0 \implies k = 2 \). So, the equation is \( 4x + 3y + 2 = 0 \).

🎯 Exam Tip: If two lines \( A_1x + B_1y + C_1 = 0 \) and \( A_2x + B_2y + C_2 = 0 \) are perpendicular, then \( A_1A_2 + B_1B_2 = 0 \). A line perpendicular to \( Ax + By + C = 0 \) can be written as \( Bx - Ay + k = 0 \).

Question 9. रेखाओं \( y = -2 \) तथा \( y = x + 2 \) के मध्य का अधिक कोण है-
(A) 145°
(B) 150°
(C) 135°
(D) 120°
Answer: (C) 135°
In simple words: The line \( y = -2 \) is a horizontal line (parallel to the x-axis). The line \( y = x + 2 \) has a slope \( m=1 \), which means it makes an angle of 45° with the positive x-axis. The angle between a horizontal line and a line with a 45° slope is 45°. The *obtuse* angle (अधिक कोण) is \( 180° - 45° = 135° \).

🎯 Exam Tip: The slope \( m \) of a line is equal to \( \tan \theta \), where \( \theta \) is the angle the line makes with the positive x-axis. If \( m=1 \), then \( \theta = 45° \).

Question 10. रेखा \( 3x – 4y - 4 = 0 \) द्वारा x-अक्ष तथा y-अक्ष पर काटे गये अन्त:खण्डों की लम्बाई है-
(A) \( \frac{4}{3} \) और -1
(B) \( -\frac{4}{3} \) और 1
(C) \( \frac{4}{3} \) और -1
(D) \( \frac{4}{3} \) और 1
Answer: (A) \( \frac{4}{3} \) और -1
In simple words: To find the x-intercept, set \( y = 0 \). \( 3x - 4(0) - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \). To find the y-intercept, set \( x = 0 \). \( 3(0) - 4y - 4 = 0 \implies -4y = 4 \implies y = -1 \). The lengths of the intercepts are \( \frac{4}{3} \) and -1.

🎯 Exam Tip: For the general equation of a line \( Ax + By + C = 0 \), the x-intercept is \( -\frac{C}{A} \) and the y-intercept is \( -\frac{C}{B} \). Alternatively, convert to intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept.

Question 12. रेखा के समीकरण \( 2x + \sqrt{3}y - 4 = 0 \) के झुकाव रूप में बदलने पर झुकाव रूप में प्रयुक्त अचर राशि के मान हैं
(A) \( m = 2, c = 4 \)
(B) \( m = \frac{\sqrt{3}}{2}, c = -\frac{4}{\sqrt{3}} \)
(C) \( m = \frac{2}{\sqrt{3}}, c = 2 \)
(D) \( m = -\frac{2}{\sqrt{3}}, c = \frac{4}{\sqrt{3}} \)
Answer: (D) \( m = -\frac{2}{\sqrt{3}}, c = \frac{4}{\sqrt{3}} \)
In simple words: To convert \( 2x + \sqrt{3}y - 4 = 0 \) to slope-intercept form \( y = mx + c \), rearrange the equation to isolate \( y \). First, move \( 2x \) and \( -4 \) to the right side: \( \sqrt{3}y = -2x + 4 \). Then, divide by \( \sqrt{3} \): \( y = -\frac{2}{\sqrt{3}}x + \frac{4}{\sqrt{3}} \). Now, compare this with \( y = mx + c \), so \( m = -\frac{2}{\sqrt{3}} \) and \( c = \frac{4}{\sqrt{3}} \).

🎯 Exam Tip: The slope-intercept form \( y = mx + c \) directly gives the slope \( m \) and the y-intercept \( c \). Always ensure the \( y \) term is isolated with a coefficient of 1.

Question 13. उस सरल रेखा का समीकरण ज्ञात कीजिए जो बिन्दु (2, 3) से होकर गुजरती है तथा x-अक्ष से 45° का कोण बनाती है।
Answer: रेखा द्वारा x-अक्ष से बनाया गया कोण \( = 45° \)
अतः रेखा का झुकाव \( m = \tan 45° = 1 \)
चूंकि रेखा (2, 3) से होकर जाती है अतः बिन्दु झुकाव रूप में रेखा का समीकरण।
\( y - y_1 = m(x - x_1) \)
\( \implies y - 3 = 1(x - 2) \)
\( \implies y = x - 2 + 3 \)
\( \implies y = x + 1 \)
\( \implies x - y + 1 = 0 \)
In simple words: First, find the slope (\( m \)) using the angle given. Since the angle with the x-axis is 45°, the slope \( m = \tan 45° = 1 \). Next, use the point-slope form of a line \( y - y_1 = m(x - x_1) \) with the given point (2, 3) and the slope \( m=1 \). Simplify the equation to get the final line equation.

🎯 Exam Tip: The point-slope form \( y - y_1 = m(x - x_1) \) is very useful when you have a point on the line and its slope. Remember that \( m = \tan \theta \).

Question 14. उस सरल रेखा का समीकरण ज्ञात कीजिए जो बिन्दु (-3, 2) से गुजरती है तथा अक्षों से बराबर तथा विपरीत चिह्नों वाले अन्तःखण्ड काटती है।
Answer: माना सरल रेखा का समीकरण (अन्त:खण्ड रूप) है-
\( \frac{x}{a} + \frac{y}{b} = 1 \) ....(1)
प्रश्नानुसार यह रेखा अक्षों से बराबर तथा विपरीत चिह्न वाले अन्तःखण्ड काटती है तब
\( a = -b \) ....(2)
समीकरण (1) व (2) से।
\( \frac{x}{a} + \frac{y}{-a} = 1 \)
\( \implies \frac{x - y}{a} = 1 \)
\( \implies x - y = a \)
चूंकि यह रेखा बिन्दु (-3, 2) से होकर गुजरती है।
\( \implies -3 - 2 = a \)
\( \implies a = -5 \)
अतः रेखा का समीकरण है:
\( x - y = -5 \)
\( \implies x - y + 5 = 0 \)
In simple words: Start with the intercept form of a line. If the intercepts are equal and opposite in sign, then one is \( a \) and the other is \( -a \). Substitute these into the intercept form and simplify. Since the line passes through (-3, 2), plug these coordinates into the simplified equation to find the value of \( a \). Then, write the final equation of the line.

🎯 Exam Tip: When intercepts are equal and opposite, it implies \( b = -a \). Always use the given point to find the unknown constant in the equation.

Question 15. यदि मूल बिन्दु से सरल रेखा \( 4x + 3y + a = 0 \) पर डाले गये लम्ब की लम्बाई 2 हो तो a का मान ज्ञात कीजिए।
Answer: दी गई रेखा है \( 4x + 3y + a = 0 \)
मूल बिन्दु (0,0) से इस रेखा पर डाले गये लम्ब की लम्बाई का सूत्र है \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
यहां \( (x_1, y_1) = (0,0) \), \( A = 4 \), \( B = 3 \), \( C = a \).
लम्ब की लम्बाई \( p = \frac{|4(0) + 3(0) + a|}{\sqrt{4^2 + 3^2}} \)
\( \implies p = \frac{|a|}{\sqrt{16 + 9}} \)
\( \implies p = \frac{|a|}{\sqrt{25}} \)
\( \implies p = \frac{|a|}{5} \)
परन्तु दिया है, लम्ब की लम्बाई \( p = 2 \)
\( \implies \frac{|a|}{5} = 2 \)
\( \implies |a| = 10 \)
\( \implies a = \pm 10 \). हमें \( a \) का मान 10 लेना है.
In simple words: To find the value of \( a \), use the formula for the perpendicular distance from a point to a line. Here, the point is the origin (0,0) and the distance is 2. Plug in the coefficients from the given line equation and solve for \( a \). Remember that the absolute value means \( a \) can be positive or negative.

🎯 Exam Tip: When dealing with distance, remember the absolute value, as distance is always non-negative. \( |a| = k \) means \( a = k \) or \( a = -k \).

Question 16. यदि किसी रेखा का अक्षों के मध्य का अन्तःखण्ड बिन्दु (5, 2) पर समद्विभाजित होता है, तो रेखा का समीकरण ज्ञात कीजिए।
Answer: माना अन्त:खण्ड रूप में रेखा का समीकरण है \( \frac{x}{a} + \frac{y}{b} = 1 \).
यह रेखा x-अक्ष को A पर मिलती है जिसके निर्देशांक \( (a, 0) \) हैं, और y-अक्ष को B पर मिलती है जिसके निर्देशांक \( (0, b) \) हैं।
प्रश्नानुसार अक्षों के मध्य का अन्तःखण्ड AB, बिन्दु (5, 2) पर समद्विभाजित होता है।
तो, मध्यबिन्दु सूत्र का उपयोग करके:
\( (5, 2) = \left( \frac{a+0}{2}, \frac{0+b}{2} \right) \)
\( (5, 2) = \left( \frac{a}{2}, \frac{b}{2} \right) \)
दोनों ओर तुलना करने पर:
\( \frac{a}{2} = 5 \implies a = 10 \)
\( \frac{b}{2} = 2 \implies b = 4 \)
अतः, \( a = 10 \) और \( b = 4 \).
रेखा का अभीष्ट समीकरण है:
\( \frac{x}{10} + \frac{y}{4} = 1 \)
लघुत्तम समापवर्त्य (LCM) 20 से गुणा करने पर:
\( \implies 2x + 5y = 20 \)
In simple words: Use the intercept form of a line: \( \frac{x}{a} + \frac{y}{b} = 1 \). The points where the line cuts the axes are \( (a, 0) \) and \( (0, b) \). Since (5, 2) is the midpoint of these two points, use the midpoint formula to find \( a \) and \( b \). Once you have \( a \) and \( b \), substitute them back into the intercept form and simplify to get the line's equation.

🎯 Exam Tip: The midpoint formula for two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \). Remember to simplify the final equation to a standard form.

Question 17. उस रेखा का समीकरण ज्ञात कीजिए जो बिन्दु (0, 1) से होकर जाती है तथा रेखा द्वारा x-अक्ष पर काटा गया अन्त:खण्ड y-अक्ष पर काटे गये अन्तःखण्ड का तिगुना हो ।
Answer: अन्त:खण्ड रूप में रेखा का समीकरण \( \frac{x}{a} + \frac{y}{b} = 1 \) ....(1)
प्रश्नानुसार, x-अक्ष पर काटा गया अन्त:खण्ड y-अक्ष पर काटे गये अन्तःखण्ड का तिगुना है।
\( a = 3b \) ....(2)
समीकरण (1) में \( a = 3b \) रखने पर:
\( \frac{x}{3b} + \frac{y}{b} = 1 \)
\( \implies \frac{x + 3y}{3b} = 1 \)
\( \implies x + 3y = 3b \)
चूंकि रेखा बिन्दु (0, 1) से होकर जाती है, हम \( x = 0 \) और \( y = 1 \) रखते हैं:
\( 0 + 3(1) = 3b \)
\( \implies 3 = 3b \)
\( \implies b = 1 \)
अतः \( a = 3b = 3(1) = 3 \).
रेखा का समीकरण \( x + 3y = 3(1) \)
या \( x + 3y = 3 \)
In simple words: Start with the intercept form of a line. The problem states that the x-intercept \( a \) is three times the y-intercept \( b \), so \( a = 3b \). Substitute this into the intercept form and simplify. Since the line passes through (0, 1), use this point to find the value of \( b \). Then find \( a \) and write the final equation of the line.

🎯 Exam Tip: Always use the relationships given (like \( a = 3b \)) to reduce the number of unknowns in the equation. A point on the line helps find the remaining unknown constant.

Question 18. सरल रेखाएँ \( y = 2mx + c \) एवं \( 2x - y + 5 = 0 \) परस्पर समान्तर एवं लम्बवत् हों तो m के मान ज्ञात कीजिए।
Answer: दी गई रेखाएँ हैं:
रेखा (1): \( y = 2mx + c \)
रेखा (2): \( 2x - y + 5 = 0 \implies y = 2x + 5 \)
रेखा (1) का झुकाव \( m_1 = 2m \)
रेखा (2) का झुकाव \( m_2 = 2 \)
**केस 1: यदि रेखाएँ समान्तर हैं**
समान्तर रेखाओं के लिए झुकाव बराबर होते हैं: \( m_1 = m_2 \)
\( \implies 2m = 2 \)
\( \implies m = 1 \)
**केस 2: यदि रेखाएँ लम्बवत् हैं**
लम्बवत् रेखाओं के लिए झुकावों का गुणनफल -1 होता है: \( m_1 m_2 = -1 \)
\( \implies (2m)(2) = -1 \)
\( \implies 4m = -1 \)
\( \implies m = -\frac{1}{4} \)
इस प्रकार, \( m \) के दो संभावित मान हैं: 1 (जब रेखाएँ समान्तर हों) और \( -\frac{1}{4} \) (जब रेखाएँ लम्बवत् हों).
In simple words: First, find the slopes of both lines. For parallel lines, the slopes are equal, so set them equal and solve for \( m \). For perpendicular lines, the product of their slopes is -1, so use that to find the other value of \( m \). This shows how the value of \( m \) depends on the relationship between the two lines.

🎯 Exam Tip: Remember the conditions for parallel and perpendicular lines based on their slopes: \( m_1 = m_2 \) for parallel lines and \( m_1 m_2 = -1 \) for perpendicular lines.

Question 19. मूल बिन्दु से सरल रेखा \( 4x + 3y + a = 0 \) पर डाले गये लम्ब की लम्बाई 2 हो तो a का मान ज्ञात कीजिए।
Answer: दी गई रेखा है \( 4x + 3y + a = 0 \)
मूल बिन्दु (0,0) से इस रेखा पर डाले गये लम्ब की लम्बाई का सूत्र है \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
यहां \( (x_1, y_1) = (0,0) \), \( A = 4 \), \( B = 3 \), \( C = a \).
लम्ब की लम्बाई \( p = \frac{|4(0) + 3(0) + a|}{\sqrt{4^2 + 3^2}} \)
\( \implies p = \frac{|a|}{\sqrt{16 + 9}} \)
\( \implies p = \frac{|a|}{\sqrt{25}} \)
\( \implies p = \frac{|a|}{5} \)
परन्तु दिया है, लम्ब की लम्बाई \( p = 2 \)
\( \implies \frac{|a|}{5} = 2 \)
\( \implies |a| = 10 \)
\( \implies a = \pm 10 \). हमें \( a \) का मान 10 लेना है.
In simple words: Use the formula for the perpendicular distance from a point to a line. The given point is the origin (0,0), and the distance is 2. Substitute the coefficients from the line equation and solve for \( a \). The absolute value means \( a \) can be either 10 or -10.

🎯 Exam Tip: This question is identical to Question 15. Ensure you apply the distance formula correctly, especially with the absolute value for the constant term.

Derivation: Perpendicular Form of a Line
Answer: दी गई रेखा है \( \frac{x}{a} + \frac{y}{b} = 1 \)
इस समीकरण को मानक रूप में बदलने पर \( bx + ay = ab \) ....(1)
रेखा (1) को अभिलम्ब रूप में बदलने के लिए, \( \sqrt{b^2 + a^2} \) से दोनों पक्षों में भाग देते हैं।
\( \frac{bx}{\sqrt{a^2 + b^2}} + \frac{ay}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \) ....(2)
यह समीकरण (2) रेखा का अभिलम्ब रूप \( x \cos \alpha + y \sin \alpha = p \) है। यहां, \( p \) मूल बिन्दु से रेखा पर डाले गए लम्ब की लम्बाई है।
इसलिए, \( p = \frac{ab}{\sqrt{a^2 + b^2}} \)
दोनों पक्षों का वर्ग करने पर:
\( p^2 = \frac{a^2 b^2}{a^2 + b^2} \)
दोनों पक्षों को \( \frac{1}{p^2} \) से गुणा करने पर:
\( 1 = \frac{a^2 b^2}{p^2 (a^2 + b^2)} \)
या \( \frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} \)
\( \implies \frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} \)
\( \implies \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \)
अर्थात् \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} \) इतिसिद्धम् (Hence Proved).
In simple words: This derivation shows a relationship between the intercepts of a line (\( a \) and \( b \)) and the length of the perpendicular (\( p \)) from the origin to that line. We start with the intercept form, convert it to the normal form, and then square the perpendicular distance to find the relationship. This formula is useful for connecting different properties of a straight line.

🎯 Exam Tip: This proof is a fundamental concept in coordinate geometry. Understanding how to derive it helps in solving problems related to intercepts and perpendicular distances. Remember the standard normal form of a line.