Class 11 Mathematics Straight Lines MCQs Set 30

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. Slope \( = \frac{97 - 92}{1995 - 1985} = \frac{5}{10} = \frac{1}{2} \)

Teacher's Note:
1. Slope here means the increase in population per year.
2. Read the coordinates carefully from the graph: years are x-values and population is y.

Answer: (b) x − 2y = 1801

Explanation:
1. \( y - 92 = \frac{1}{2}(x - 1985) \Rightarrow 2y - 184 = x - 1985 \)
2. x − 2y = 1801.

Teacher's Note:
1. Use point-slope form with one of the given points and the slope 1/2.
2. Check with B: 1995 − 2(97) = 1801.

Answer: (a) 104.5

Explanation:
1. Put x = 2010 in x − 2y = 1801: 2y = 2010 − 1801 = 209.
2. y = 104.5 crores.

Teacher's Note:
1. Once the equation of the line is known, any year can be substituted to predict the population.
2. From 1985 to 2010 is 25 years at 1/2 crore per year: 92 + 12.5 = 104.5.

Answer: (b) 2x + y = 4087

Explanation:
1. Slope of AB = 1/2, so the perpendicular slope is −2.
2. y − 97 = −2(x − 1995) ⇒ 2x + y = 3990 + 97 = 4087.

Teacher's Note:
1. Negative reciprocal of 1/2 is −2.
2. Check: 2(1995) + 97 = 4087.

Answer: (c) 2021

Explanation:
1. Put y = 110 in x − 2y = 1801: x = 1801 + 220 = 2021.

Teacher's Note:
1. Solve the line equation for x when y is known.
2. Population rises by 1 crore every 2 years, and 110 − 92 = 18 crores needs 36 years after 1985.

Answer: (b) 3x + y = 4

Explanation:
1. Slope \( = \frac{1 + 2}{1 - 2} = -3 \)
2. y − 1 = −3(x − 1) ⇒ 3x + y = 4.

Teacher's Note:
1. Check that both points satisfy the answer: 6 − 2 = 4 and 3 + 1 = 4.
2. Point-slope form with either point gives the same line.

Answer: (c) \( \frac{-2}{3} \)

Explanation:
1. Slope \( = \frac{0 - (-2)}{-1 - 2} = \frac{2}{-3} = -\frac{2}{3} \)

Teacher's Note:
1. Subtract coordinates in the same order in the numerator and denominator.
2. A line falling from left to right has a negative slope.

Answer: (a) 5x + 4y = 2

Explanation:
1. Mid-point of Mansi and Sneha \( = \left(0, \frac{1}{2}\right) \).
2. Slope from (2, −2) to \( \left(0, \frac{1}{2}\right) \) \( = \frac{\frac{1}{2} + 2}{0 - 2} = -\frac{5}{4} \).
3. \( y - \frac{1}{2} = -\frac{5}{4}x \Rightarrow 5x + 4y = 2 \)

Teacher's Note:
1. A median joins a vertex to the mid-point of the opposite side.
2. Check: Rani (2, −2) gives 10 − 8 = 2.

Answer: (a) 3x − 2y = 1

Explanation:
1. The altitude through Mansi is perpendicular to the side Rani–Sneha, whose slope is \( -\frac{2}{3} \).
2. Its slope is \( \frac{3}{2} \): \( y - 1 = \frac{3}{2}(x - 1) \Rightarrow 3x - 2y = 1 \).

Teacher's Note:
1. An altitude is perpendicular to the opposite side, so use the negative reciprocal slope.
2. Do not confuse altitude (perpendicular) with median (through the mid-point).

Answer: (c) x − 2y = 6

Explanation:
1. Slope of Mansi–Sneha \( = \frac{0 - 1}{-1 - 1} = \frac{1}{2} \).
2. \( y + 2 = \frac{1}{2}(x - 2) \Rightarrow 2y + 4 = x - 2 \Rightarrow x - 2y = 6 \)

Teacher's Note:
1. A parallel line keeps the same slope.
2. Check with Rani (2, −2): 2 + 4 = 6.

Answer: (a) \( \sqrt{40} \) units

Explanation:
1. \( AC = \sqrt{(-1 - 1)^2 + (-2 - 4)^2} = \sqrt{4 + 36} = \sqrt{40} \) units

Teacher's Note:
1. Square the differences of coordinates; signs do not matter after squaring.
2. √40 can also be written as 2√10.

Answer: (b) \( -\frac{1}{3} \)

Explanation:
1. Slope of BC \( = \frac{-2 - (-3)}{-1 - 2} = \frac{1}{-3} = -\frac{1}{3} \)

Teacher's Note:
1. Keep the order of subtraction the same in numerator and denominator.
2. This slope is needed again for the altitude and the right bisector.

Answer: (c) 13x − y − 9 = 0

Explanation:
1. D, the mid-point of BC \( = \left(\frac{1}{2}, -\frac{5}{2}\right) \).
2. Slope of AD \( = \frac{4 + \frac{5}{2}}{1 - \frac{1}{2}} = 13 \).
3. y − 4 = 13(x − 1) ⇒ 13x − y − 9 = 0.

Teacher's Note:
1. The median through A goes to the mid-point of the opposite side BC.
2. Check with D: 13/2 + 5/2 − 9 = 0.

Answer: (a) 3x − y + 1 = 0

Explanation:
1. The altitude through A is perpendicular to BC (slope −1/3), so its slope is 3.
2. y − 4 = 3(x − 1) ⇒ 3x − y + 1 = 0.

Teacher's Note:
1. Altitude = perpendicular from a vertex to the opposite side.
2. Check with A: 3 − 4 + 1 = 0.

Answer: (c) 3x − y − 4 = 0

Explanation:
1. The right (perpendicular) bisector passes through D \( \left(\frac{1}{2}, -\frac{5}{2}\right) \) with slope 3.
2. \( y + \frac{5}{2} = 3\left(x - \frac{1}{2}\right) \Rightarrow y = 3x - 4 \Rightarrow 3x - y - 4 = 0 \)

Teacher's Note:
1. The perpendicular bisector uses the mid-point of BC and the slope perpendicular to BC.
2. It is parallel to the altitude through A but passes through D instead of A.

Answer: (c) x − 2y + 3 = 0

Explanation:
1. Slope through (1, 2) and (5, 4) \( = \frac{2}{4} = \frac{1}{2} \).
2. \( y - 2 = \frac{1}{2}(x - 1) \Rightarrow 2y - 4 = x - 1 \Rightarrow x - 2y + 3 = 0 \)

Teacher's Note:
1. This line is the diagonal BD of the rectangle.
2. Check with (5, 4): 5 − 8 + 3 = 0.

Answer: (b) x + 2y − 9 = 0

Explanation:
1. Slope through (1, 4) and (5, 2) \( = \frac{-2}{4} = -\frac{1}{2} \).
2. \( y - 4 = -\frac{1}{2}(x - 1) \Rightarrow 2y - 8 = -x + 1 \Rightarrow x + 2y - 9 = 0 \)

Teacher's Note:
1. This is the other diagonal AC of the rectangle.
2. Check with (5, 2): 5 + 4 − 9 = 0.

Answer: (c) (3, 3)

Explanation:
1. Solve x − 2y = −3 and x + 2y = 9.
2. Adding: 2x = 6 ⇒ x = 3; then 2y = 6 ⇒ y = 3. The point is (3, 3).

Teacher's Note:
1. The diagonals of a rectangle bisect each other, so the point is also the mid-point of each diagonal.
2. Mid-point of (1, 4) and (5, 2) is (3, 3), which confirms the answer.

Answer: (a) zero

Explanation:
1. Slope through (1, 4) and (5, 4) \( = \frac{4 - 4}{5 - 1} = 0 \)

Teacher's Note:
1. Points with the same y-coordinate lie on a horizontal line.
2. Horizontal lines have slope zero.

Answer: (a) Rishabh-Rajkumar and Shubham-Vikram

Explanation:
1. Rishabh–Rajkumar (y = 4) and Shubham–Vikram (y = 2) are both horizontal, slope 0.
2. Rajkumar–Vikram and Rishabh–Shubham are vertical, with undefined slope.
3. So the pair with the same slope is Rishabh–Rajkumar and Shubham–Vikram.

Teacher's Note:
1. Opposite sides of a rectangle are parallel and therefore have equal slopes.
2. Adjacent sides of a rectangle are perpendicular.

Answer: (b) \( 2\sqrt{5} \)

Explanation:
1. \( AB = \sqrt{(6 - 2)^2 + (-5 + 3)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)

Teacher's Note:
1. Simplify surds by taking out perfect squares: √20 = √4 × √5.
2. AC and CB should add up to AB, which is a good check later.

Answer: (a) x + 2y + 4 = 0

Explanation:
1. Slope \( = \frac{-5 + 3}{6 - 2} = -\frac{1}{2} \).
2. \( y + 3 = -\frac{1}{2}(x - 2) \Rightarrow 2y + 6 = -x + 2 \Rightarrow x + 2y + 4 = 0 \)

Teacher's Note:
1. Check with B(6, −5): 6 − 10 + 4 = 0.
2. Use two-point form when two points are given.

Answer: (d) \( \left(3, -\frac{7}{2}\right) \)

Explanation:
1. \( x = \frac{1 \times 6 + 3 \times 2}{4} = 3,\quad y = \frac{1 \times (-5) + 3 \times (-3)}{4} = -\frac{14}{4} = -\frac{7}{2} \)
2. C \( = \left(3, -\frac{7}{2}\right) \).

Teacher's Note:
1. In ratio 1 : 3 from A, the point C is one quarter of the way from A to B.
2. C must also satisfy x + 2y + 4 = 0: 3 − 7 + 4 = 0.

Answer: (c) \( \frac{\sqrt{5}}{2} \)

Explanation:
1. \( AC = \sqrt{(3 - 2)^2 + \left(-\frac{7}{2} + 3\right)^2} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} \)

Teacher's Note:
1. AC is one quarter of AB = 2√5, which gives √5/2 directly.
2. Using the ratio saves calculation in such questions.

Answer: (a) \( \frac{3\sqrt{5}}{2} \)

Explanation:
1. \( CB = \sqrt{(6 - 3)^2 + \left(-5 + \frac{7}{2}\right)^2} = \sqrt{9 + \frac{9}{4}} = \frac{3\sqrt{5}}{2} \)

Teacher's Note:
1. CB is three quarters of AB = 2√5.
2. Check: AC + CB = √5/2 + 3√5/2 = 2√5 = AB.

Answer: (c) A is true, but R is false.

Explanation:
1. (3, 0) lies on the positive X-axis, 3 units from the origin.
2. Its distance from the Y-axis is |x| = 3 and from the X-axis is |y| = 0, so A is true.
3. R interchanges the axes, so R is false.

Teacher's Note:
1. The distance of (x, y) from the Y-axis is |x| and from the X-axis is |y|.
2. Points on the X-axis are at zero distance from the X-axis.

Answer: (d) A is false, but R is true.

Explanation:
1. h + k + 1 = 0, so h = −1 − k; also h² + k² = 25.
2. (1 + k)² + k² = 25 ⇒ k² + k − 12 = 0 ⇒ k = 3 or k = −4.
3. k negative gives k = −4 and h = 3, not −3. So A is false; R is the correct distance formula.

Teacher's Note:
1. Use both conditions: the point lies on the line and its distance from the origin is 5.
2. Always substitute back: (3, −4) is on x + y + 1 = 0 and is 5 units from O.

Answer: (c) A is true, but R is false.

Explanation:
1. Let the ratio be k : 1. The point \( \left(\frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1}\right) \) lies on x + y = 4.
2. 5k − 1 + 7k + 1 = 4(k + 1) ⇒ 8k = 4 ⇒ k = 1/2, i.e. ratio 1 : 2. A is true.
3. The formula in R (with minus signs) is the external division formula, so R is false.

Teacher's Note:
1. Taking the ratio as k : 1 keeps the algebra to a single unknown.
2. Internal division uses m₁ + m₂ in the denominator; external division uses m₁ − m₂.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Centroid \( = \left(\frac{3 + c^2}{3}, \frac{a + b - 3}{3}\right) \).
2. Its x-coordinate (3 + c²)/3 is never 0 since c² ≥ 0, so A is true.
3. On the X-axis the y-coordinate is 0, i.e. a + b = 3, so R is true; but R concerns the X-axis, not the Y-axis, so it does not explain A.

Teacher's Note:
1. A point lies on the Y-axis when its x-coordinate is zero, and on the X-axis when its y-coordinate is zero.
2. c² ≥ 0 for every real c is the key fact behind the assertion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Area \( = \frac{1}{2}\left|4(-2 - 16) + 3(16 - 4) + (-3)(4 + 2)\right| \)
2. \( = \frac{1}{2}\left|-72 + 36 - 18\right| = 27 \) sq units, so A is true.
3. R is the area formula used to get A, so R is true and explains A.

Teacher's Note:
1. The modulus in the formula makes the area positive whatever order the vertices are taken in.
2. A zero value of this expression means the three points are collinear.

Answer: (c) A is true, but R is false.

Explanation:
1. The X-axis has inclination 0°, so slope = tan 0° = 0.
2. The Y-axis has inclination 90°, and tan 90° is not defined.
3. A is true and R (which swaps them) is false.

Teacher's Note:
1. Horizontal lines have slope 0 and vertical lines have undefined slope.
2. R simply interchanges the two facts, which makes it false.

Answer: (d) A is false, but R is true.

Explanation:
1. Slope AB \( = \frac{1}{6} \), slope BC \( = \frac{3}{-1} = -3 \), slope CD \( = \frac{-1}{-6} = \frac{1}{6} \), slope AD \( = \frac{3}{-1} = -3 \).
2. So slope AB ≠ slope BC and slope CD ≠ slope AD; A is false (in fact AB ∥ CD and BC ∥ AD).
3. Mid-point of AC \( = \left(\frac{1}{2}, 1\right) \) = mid-point of BD, so R is true.

Teacher's Note:
1. In a parallelogram opposite sides are parallel, not adjacent sides.
2. Diagonals of a parallelogram bisect each other, which is why the two mid-points coincide.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( m_1 = -\frac{1}{2} \), \( m_2 = -3 \).
2. \( \tan\theta = \pm\frac{-3 + \frac{1}{2}}{1 + \frac{3}{2}} = \pm 1 \), so θ = tan⁻¹(1) (or π − tan⁻¹(1)).
3. A is true and is obtained from the formula in R, so R explains A.

Teacher's Note:
1. The ± sign gives both the acute and the obtuse angle between the lines.
2. tan⁻¹(1) = 45°, so these lines meet at 45° and 135°.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( y = \frac{3}{4}x + \frac{5}{2} \), so slope \( = \frac{3}{4} \); A is true.
2. y = 0 gives \( x = -\frac{10}{3} \); x = 0 gives \( y = \frac{10}{4} = \frac{5}{2} \); R is true.
3. Intercepts do not give the slope directly here, so R is not the explanation of A.

Teacher's Note:
1. Slope of ax + by + c = 0 is −a/b.
2. Intercepts are found by putting y = 0 and x = 0 in turn.

Answer: (c) A is true, but R is false.

Explanation:
1. Slope of x cos θ + y sin θ = 2 is −cot θ; slope of x − y = 3 is 1.
2. Perpendicular: (−cot θ)(1) = −1 ⇒ cot θ = 1 ⇒ θ = π/4. A is true.
3. Perpendicular lines satisfy m₁m₂ = −1, not m₁ = m₂, so R is false.

Teacher's Note:
1. m₁ = m₂ is the condition for parallel lines; m₁m₂ = −1 is for perpendicular lines.
2. The slope of x cos θ + y sin θ = p is −cos θ/sin θ = −cot θ.

Answer: (d) A is false, but R is true.

Explanation:
1. x + 7y = 0 ⇒ \( y = -\frac{1}{7}x \); slope \( -\frac{1}{7} \), so A is false.
2. 6x + 3y − 5 = 0 ⇒ \( y = -2x + \frac{5}{3} \); slope −2 and y-intercept \( \frac{5}{3} \), so R is true.

Teacher's Note:
1. Always make y the subject to read the slope and y-intercept.
2. A line through the origin has y-intercept 0 but its slope can still be negative.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Write 8x + 6y = 15 as \( 4x + 3y = \frac{15}{2} \).
2. \( d = \frac{\left|11 - \frac{15}{2}\right|}{\sqrt{16 + 9}} = \frac{7/2}{5} = \frac{7}{10} \); A is true.
3. R is the correct formula used to obtain A, so R explains A.

Teacher's Note:
1. Make the coefficients of x and y the same in both equations before using the formula.
2. Parallel lines have proportional x and y coefficients, here 4 : 3 = 8 : 6.