Class 11 Mathematics Straight Lines MCQs Set 29

Answer: (b) quadrant II

Explanation:
1. The x-coordinate −3 is negative and the y-coordinate 2 is positive.
2. Points of the form (−, +) lie in the second quadrant.
3. Hence (−3, 2) lies in quadrant II.

Teacher's Note:
1. Remember the sign pattern (+, +), (−, +), (−, −), (+, −) for quadrants I to IV.
2. Plotting the point roughly on a rough axis is a quick check.

Answer: (a) 3

Explanation:
1. \( PQ^2 = (10 - 2)^2 + (y + 3)^2 = 100 \)
2. \( 64 + (y + 3)^2 = 100 \Rightarrow (y + 3)^2 = 36 \Rightarrow y + 3 = \pm 6 \)
3. So y = 3 or y = −9; of these only y = 3 is among the options.

Teacher's Note:
1. Square both sides of the distance formula first; it removes the root and avoids errors.
2. A distance condition usually gives two possible points, so check which value the options contain.

Answer: (c) (−1, 0)

Explanation:
1. Let the point be P(x, 0). Then PA² = PB².
2. \( (3 - x)^2 + 4 = (x + 5)^2 + 4 \Rightarrow 9 - 6x = 10x + 25 \)
3. 16x = −16, so x = −1 and the point is (−1, 0).

Teacher's Note:
1. A point on the X-axis always has y-coordinate 0, so only one unknown remains.
2. Equate the squares of the distances; the x² terms cancel and a linear equation is left.

Answer: (b) (0, 0)

Explanation:
1. Section formula: \( x = \frac{m x_2 + n x_1}{m + n},\ y = \frac{m y_2 + n y_1}{m + n} \) with m : n = 1 : 3.
2. \( x = \frac{1(-3) + 3(1)}{4} = 0,\quad y = \frac{1(9) + 3(-3)}{4} = 0 \)
3. The required point is (0, 0).

Teacher's Note:
1. In the section formula the ratio number m goes with the far end point B and n with A.
2. Check that the answer lies between A and B; (0, 0) is a quarter of the way from A towards B.

Answer: (d) (2, 1)

Explanation:
1. Centroid \( = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)
2. \( = \left(\frac{1 + 2 + 3}{3}, \frac{3 + 5 - 5}{3}\right) = (2, 1) \)

Teacher's Note:
1. The centroid is simply the average of the three x-coordinates and the three y-coordinates.
2. Watch the signs of negative coordinates while adding.

Answer: (a) Yes

Explanation:
1. Let A(1, −1), B(5, 2), C(9, 5).
2. Slope of AB \( = \frac{2 + 1}{5 - 1} = \frac{3}{4} \), slope of BC \( = \frac{5 - 2}{9 - 5} = \frac{3}{4} \)
3. The slopes are equal and B is common, so A, B, C are collinear (also AB + BC = 5 + 5 = 10 = AC).

Teacher's Note:
1. Equal slopes of AB and BC with a common point B is the quickest test for collinearity.
2. Equivalent checks are AB + BC = AC or zero area of triangle ABC.

Answer: (b) 27

Explanation:
1. Area \( = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right| \)
2. \( = \frac{1}{2}\left|4(-2 - 16) + 3(16 - 4) + (-3)(4 + 2)\right| = \frac{1}{2}\left|-72 + 36 - 18\right| \)
3. \( = \frac{54}{2} = 27 \) square units.

Teacher's Note:
1. Always take the modulus at the end because area cannot be negative.
2. Forgetting the factor one-half gives 54, which is a common wrong choice.

Answer: (d) not defined

Explanation:
1. Slope m = tan θ, where θ is the inclination.
2. tan 90° is not defined, so the slope of a vertical line is not defined.

Teacher's Note:
1. A line with inclination 90° is parallel to the Y-axis.
2. Do not confuse 'not defined' with zero; zero slope belongs to horizontal lines.

Answer: (c) \( \sqrt{3} \)

Explanation:
1. m = tan 60° \( = \sqrt{3} \)

Teacher's Note:
1. Learn tan 30° = \( \frac{1}{\sqrt{3}} \), tan 45° = 1 and tan 60° = \( \sqrt{3} \) by heart.
2. A steeper line has a larger inclination and hence a larger positive slope.

Answer: (d) \( \frac{2}{t_2 + t_1} \)

Explanation:
1. \( m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2a(t_2 - t_1)}{a(t_2 - t_1)(t_2 + t_1)} \)
2. \( = \frac{2}{t_2 + t_1} \)

Teacher's Note:
1. Factorise the denominator as a difference of squares before cancelling.
2. Cancelling (t₂ − t₁) is valid because the two points are distinct.

Answer: (d) 120°

Explanation:
1. Slopes: \( m_1 = 2 - \sqrt{3} \), \( m_2 = 2 + \sqrt{3} \); \( m_1 m_2 = 4 - 3 = 1 \).
2. \( \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \frac{2\sqrt{3}}{2} = \sqrt{3} \), so the acute angle is 60°.
3. The lines therefore make angles of 60° and 120°; of these, 120° is the one in the options.

Teacher's Note:
1. Two intersecting lines form a pair of angles θ and 180° − θ, so either may appear in the options.
2. 2 − √3 = tan 15° and 2 + √3 = tan 75°, which confirms the angle 75° − 15° = 60°.

Answer: (b) 135°

Explanation:
1. Slope \( = \frac{-2 - (-1)}{4 - 3} = -1 \)
2. tan θ = −1 with 0° ≤ θ < 180° gives θ = 135°.

Teacher's Note:
1. Inclination is measured anticlockwise from the positive X-axis, so a negative slope means an obtuse angle.
2. tan 135° = −1, not tan 45°.

Answer: (c) \( \frac{b^2 - a^2}{2ab} \)

Explanation:
1. First line \( \frac{x}{a} - \frac{y}{b} = 1 \) has slope \( m_1 = \frac{b}{a} \); second line \( \frac{x}{b} - \frac{y}{a} = 1 \) has slope \( m_2 = \frac{a}{b} \).
2. \( \tan\theta = \frac{m_1 - m_2}{1 + m_1 m_2} = \frac{\frac{b}{a} - \frac{a}{b}}{1 + 1} = \frac{b^2 - a^2}{2ab} \)

Teacher's Note:
1. For the intercept form x/a + y/b = 1 the slope is −b/a.
2. The product of the two slopes here is 1, which makes the denominator simply 2.

Answer: (c) 9

Explanation:
1. Slope of the first line \( = \frac{7 - y}{2 - 3} = y - 7 \)
2. Slope of the second line \( = \frac{6 - 4}{0 + 1} = 2 \)
3. Parallel lines have equal slopes: y − 7 = 2, so y = 9.

Teacher's Note:
1. Parallel lines have equal slopes; perpendicular lines have slopes whose product is −1.
2. Be careful with the negative denominator 2 − 3 = −1.

Answer: (a) Yes

Explanation:
1. Slope of AB \( = \frac{5 - 4}{3 - 4} = -1 \), slope of BC \( = \frac{1 - 5}{-1 - 3} = 1 \)
2. Product of slopes = (−1)(1) = −1, so AB ⊥ BC.
3. Hence the triangle is right angled at B.

Teacher's Note:
1. Find all three slopes if needed; any pair with product −1 gives the right angle.
2. The common vertex of the two perpendicular sides is where the right angle lies.

Answer: (a) 1

Explanation:
1. Slope of BC \( = \frac{-5 + 2}{4 - 3} = -3 \)
2. Slope of AB \( = \frac{-2 - 4}{3 - x} = \frac{-6}{3 - x} \)
3. Collinear: \( \frac{-6}{3 - x} = -3 \Rightarrow 3 - x = 2 \Rightarrow x = 1 \)

Teacher's Note:
1. For collinearity equate the slopes of two segments sharing a common point.
2. Compute the slope that contains no unknown first; it makes the equation simple.

Answer: (d) y = 5

Explanation:
1. A line parallel to the X-axis has the form y = k.
2. It passes through (−3, 5), so k = 5 and the line is y = 5.

Teacher's Note:
1. Lines parallel to the X-axis fix the y-coordinate; lines parallel to the Y-axis fix x.
2. x = −3 is the line through the same point but parallel to the Y-axis.

Answer: (b) 4x + y + 5 = 0

Explanation:
1. Point-slope form: y − 3 = −4(x + 2)
2. y − 3 = −4x − 8, so 4x + y + 5 = 0.

Teacher's Note:
1. Point-slope form is \( y - y_1 = m(x - x_1) \).
2. Substitute the given point into your answer to check it: 4(−2) + 3 + 5 = 0.

Answer: (b) y − x − 1 = 0

Explanation:
1. Slope of x + y + 1 = 0 is −1, so the perpendicular slope is 1.
2. y − 2 = 1(x − 1), so y − x − 1 = 0.

Teacher's Note:
1. For ax + by + c = 0 the slope is −a/b; the perpendicular slope is its negative reciprocal.
2. Check: (1, 2) satisfies 2 − 1 − 1 = 0.

Answer: (d) y = −2x

Explanation:
1. Slope of x − 2y + 3 = 0 is \( \frac{1}{2} \), so the perpendicular slope is −2.
2. y + 2 = −2(x − 1), so y = −2x.

Teacher's Note:
1. The product of slopes of perpendicular lines is −1.
2. The line y = −2x passes through the origin, and (1, −2) satisfies it.

Answer: (d) 5x + 3y + 2 = 0

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

Teacher's Note:
1. Two-point form is point-slope form with the slope found from the two points.
2. Verify with the second point: 5(2) + 3(−4) + 2 = 0.

Answer: (d) 0

Explanation:
1. Slope \( = \frac{7 - 5}{-5 + 4} = -2 \)
2. y − 5 = −2(x + 4), i.e. 2x + y + 3 = 0.
3. (l, m) lies on it, so 2l + m + 3 = 0.

Teacher's Note:
1. A point lies on a line exactly when its coordinates satisfy the equation of the line.
2. Writing the line in the form asked for (2x + y + 3) gives the answer at once.

Answer: (a) 5y − 3x + 15 = 0

Explanation:
1. Slope \( m = \frac{3}{5} \), y-intercept c = −3.
2. \( y = \frac{3}{5}x - 3 \Rightarrow 5y = 3x - 15 \Rightarrow 5y - 3x + 15 = 0 \)

Teacher's Note:
1. Slope-intercept form y = mx + c needs only the slope and the y-intercept.
2. The tangent of the inclination is the slope.

Answer: (a) x − 2y − 4 = 0

Explanation:
1. The line passes through (4, 0) with slope 1/2.
2. \( y - 0 = \frac{1}{2}(x - 4) \Rightarrow 2y = x - 4 \Rightarrow x - 2y - 4 = 0 \)

Teacher's Note:
1. An x-intercept of 4 means the point (4, 0) is on the line.
2. Then the point-slope form finishes the job.

Answer: (a) −1

Explanation:
1. Equal intercepts a and a: \( \frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a \)
2. Slope = −1.

Teacher's Note:
1. Equal intercepts (same sign) always give slope −1, whatever the value of a.
2. If the intercepts were equal in length but opposite in sign, the slope would be 1.

Answer: (d) (−1, −1)

Explanation:
1. Put \( u = \frac{1}{a},\ v = \frac{1}{b} \): 2u − 3v = 1 and 4u − 5v = 1.
2. Doubling the first: 4u − 6v = 2; subtracting from the second gives v = −1.
3. Then 2u + 3 = 1, so u = −1. Hence a = −1, b = −1.

Teacher's Note:
1. Substituting 1/a and 1/b as new variables turns the equations into linear ones.
2. Check: −x − y = 1 is satisfied by both (2, −3) and (4, −5).

Answer: (a) 2x + 3y = 12

Explanation:
1. Let the line meet the axes at (a, 0) and (0, b). Midpoint (a/2, b/2) = (3, 2), so a = 6, b = 4.
2. \( \frac{x}{6} + \frac{y}{4} = 1 \Rightarrow 2x + 3y = 12 \)

Teacher's Note:
1. The intercepts are simply double the coordinates of the midpoint.
2. Then use the intercept form x/a + y/b = 1.

Answer: (c) \( -\frac{3}{2}, \frac{7}{2} \)

Explanation:
1. \( 2y = -3x + 7 \Rightarrow y = -\frac{3}{2}x + \frac{7}{2} \)
2. Slope \( = -\frac{3}{2} \), y-intercept \( = \frac{7}{2} \).

Teacher's Note:
1. Make y the subject; the coefficient of x is the slope and the constant term is the y-intercept.
2. Keep the sign of the constant when moving it across.

Answer: (c) 45°

Explanation:
1. Slopes: \( m_1 = \sqrt{3} \) (inclination 60°) and \( m_2 = \frac{1}{\sqrt{3}} \) (inclination 30°).
2. The lines meet at 60° − 30° = 30° and hence also at 180° − 30° = 150°.
3. So the angle cannot be 45°.

Teacher's Note:
1. When slopes are standard values, comparing inclinations is faster than the tan formula.
2. Remember that both θ and 180° − θ are angles between two lines.

Answer: (c) \( \pi - \tan^{-1}(7) \)

Explanation:
1. Slopes: \( m_1 = \frac{1}{2} \), \( m_2 = -3 \).
2. \( \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{-3 - \frac{1}{2}}{1 - \frac{3}{2}}\right| = 7 \)
3. The angles are \( \tan^{-1}7 \) and \( \pi - \tan^{-1}7 \); the second is in the options.

Teacher's Note:
1. An angle between lines lies between 0 and π, so negative angles or angles beyond π are ruled out.
2. Both θ and π − θ are angles between the same pair of lines.

Answer: (a) 2x − 3y + 1 = 0

Explanation:
1. A parallel line has the form 2x − 3y = k.
2. Through (4, 3): k = 8 − 9 = −1, so 2x − 3y + 1 = 0.

Teacher's Note:
1. Lines parallel to ax + by + c = 0 keep the same a and b; only the constant changes.
2. Find the constant by substituting the given point.

Answer: (d) 4

Explanation:
1. \( m_1 = \frac{8 - 6}{4 + 2} = \frac{1}{3} \), \( m_2 = \frac{24 - 12}{x - 8} = \frac{12}{x - 8} \)
2. \( m_1 m_2 = -1 \Rightarrow \frac{4}{x - 8} = -1 \Rightarrow x - 8 = -4 \)
3. x = 4.

Teacher's Note:
1. For perpendicular lines use m₁m₂ = −1, provided neither line is vertical.
2. Simplify the slope products before cross-multiplying.

Answer: (d) \( \frac{3}{5} \)

Explanation:
1. \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}} \)
2. \( = \frac{3}{5} \)

Teacher's Note:
1. Substitute the point into the left side of the line and divide by √(A² + B²).
2. Take the modulus since a distance is never negative.

Answer: (d) 1 unit

Explanation:
1. \( d = \frac{|0 + 0 - 13|}{\sqrt{25 + 144}} = \frac{13}{13} = 1 \) unit

Teacher's Note:
1. From the origin the distance is simply |C|/√(A² + B²).
2. 5, 12, 13 is a Pythagorean triple, which makes √(25 + 144) = 13.

Answer: (a) \( \frac{130}{17\sqrt{29}} \)

Explanation:
1. From 3x + 4y = 0, \( x = -\frac{4y}{3} \); then \( -\frac{8y}{3} - 3y + 5 = 0 \Rightarrow y = \frac{15}{17} \), \( x = -\frac{20}{17} \).
2. Distance from 5x − 2y = 0: \( \frac{\left|5\left(-\frac{20}{17}\right) - 2\left(\frac{15}{17}\right)\right|}{\sqrt{25 + 4}} = \frac{130}{17\sqrt{29}} \)

Teacher's Note:
1. First find the point of intersection by solving the two equations together.
2. Keep fractions exact; the denominator 17 carries through to the answer.

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

Explanation:
1. \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|7 - 5|}{\sqrt{9 + 16}} = \frac{2}{5} \)

Teacher's Note:
1. The formula needs the coefficients of x and y to be identical in both equations.
2. Only the constants differ for parallel lines written in the same form.

Answer: (a) \( \frac{3}{10} \)

Explanation:
1. Write the second line as \( 3x + 4y = \frac{15}{2} \).
2. \( d = \frac{\left|9 - \frac{15}{2}\right|}{\sqrt{9 + 16}} = \frac{3/2}{5} = \frac{3}{10} \)

Teacher's Note:
1. Divide the second equation by 2 so that both lines have the same x and y coefficients.
2. Using 9 and 15 directly gives a wrong answer.