CBSE Class 11 Mathematics Straight Lines Competitive Exam MCQs Set 01

Answer: (c) 3

Explanation:
1. Distance formula: \( \sqrt{(x - 3)^2 + (2 - 4)^2} = 2 \).
2. Squaring: (x − 3)² + 4 = 4, so (x − 3)² = 0.
3. Hence x = 3.

Teacher's Note:
1. Square both sides first to remove the root; it keeps the algebra simple.
2. Here the two points lie on the vertical line x = 3, so the distance is just the difference of y-coordinates.

Answer: (d) (4, 4)

Explanation:
1. Let the point be P(a, a). PA² = PB² gives (a − 5)² + a² = a² + (a − 3)².
2. So −10a + 25 = −6a + 9, i.e. 4a = 16.
3. Hence a = 4 and the point is (4, 4).

Teacher's Note:
1. Use PA² = PB² instead of PA = PB to avoid square roots.
2. The a² terms cancel, so the condition becomes a simple linear equation.

Answer: (c) are collinear

Explanation:
1. Slope of the join of (1, 1) and (−2, 7) = \( \frac{7 - 1}{-2 - 1} = -2 \).
2. Slope of the join of (1, 1) and (3, −3) = \( \frac{-3 - 1}{3 - 1} = -2 \).
3. Equal slopes through the common point (1, 1), so the points are collinear.

Teacher's Note:
1. Equal slopes from a common point is the quickest test for collinearity.
2. Always test collinearity first; if the points are collinear, no triangle is formed at all.

Answer: (b) an equilateral triangle

Explanation:
1. AB = |6a − 4a| = 2a.
2. AC = \( \sqrt{3a^2 + a^2} = 2a \) and BC = \( \sqrt{3a^2 + a^2} = 2a \).
3. All three sides are equal, so the triangle is equilateral.

Teacher's Note:
1. Compute all three sides before choosing; isosceles and equilateral are easy to confuse.
2. Since a > 0, |2a| = 2a, so no sign trouble arises.

Answer: (b) (3, 4)

Explanation:
1. Section formula with m : n = 1 : 2: x = \( \frac{1 \times 9 + 2 \times 0}{3} = 3 \).
2. y = \( \frac{1 \times 12 + 2 \times 0}{3} = 4 \).
3. The point is (3, 4).

Teacher's Note:
1. In the section formula, m multiplies the coordinates of the second point B.
2. A point dividing a segment internally always lies between the two end points, which rules out the negative options.

Answer: (a) 2 : 3

Explanation:
1. Let the x-axis divide the join in the ratio k : 1; the y-coordinate of the point is \( \frac{6k - 4}{k + 1} \).
2. On the x-axis this is 0, so 6k − 4 = 0 and k = 2/3.
3. The ratio is 2 : 3 (internal, as k > 0).

Teacher's Note:
1. For the x-axis set the y-coordinate to zero; for the y-axis set the x-coordinate to zero.
2. Quick check: the ratio equals |y₁| : |y₂| = 4 : 6 = 2 : 3 when the points are on opposite sides of the axis.

Answer: (d) cannot be found

Explanation:
1. External division formula: \( x = \frac{mx_2 - nx_1}{m - n} \).
2. For m : n = 1 : 1 the denominator m − n = 0.
3. So no such point exists; it cannot be found.

Teacher's Note:
1. External division in the ratio 1 : 1 would need a point equally far from both ends but outside the segment, which is impossible.
2. Always check the denominator m − n before substituting in the external formula.

Answer: (a) \( \left( \frac{1}{3}, 2 \right) \)

Explanation:
1. AC = 2BC means C divides AB internally in the ratio 2 : 1.
2. x = \( \frac{2 \times 2 + 1 \times (-3)}{3} = \frac{1}{3} \), y = \( \frac{2 \times 1 + 1 \times 4}{3} = 2 \).
3. So C = (1/3, 2).

Teacher's Note:
1. AC : CB = 2 : 1 gives the ratio directly; write it in the order A to B.
2. Since C is on the segment, its x-coordinate must lie between −3 and 2, which rules out (c) and (d).

Answer: (a) 1 : 2

Explanation:
1. Let the dividing point be \( \left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right) \) for ratio k : 1.
2. It lies on x + y = 4: (5k − 1) + (7k + 1) = 4(k + 1), so 12k = 4k + 4.
3. k = 1/2, so the ratio is 1 : 2.

Teacher's Note:
1. Substitute the section-formula point into the line and solve for k.
2. A positive k means the line cuts the segment internally.

Answer: (b) \( \frac{25}{2} \) sq unit

Explanation:
1. Area = \( \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).
2. = \( \frac{1}{2} |1(-6 - 5) + 4(5 - 2) + 3(2 + 6)| = \frac{1}{2} |-11 + 12 + 24| \).
3. = 25/2 sq units.

Teacher's Note:
1. Keep the brackets and signs carefully; one sign slip changes the answer.
2. Area is always taken as the absolute value of the expression.

Answer: (d) none of these

Explanation:
1. Split the quadrilateral ABCD with A(4, 3), B(2, −1), C(−1, 2), D(−3, 2) along the diagonal AC.
2. Area ABC = \( \frac{1}{2} |4(-1 - 2) + 2(2 - 3) + (-1)(3 + 1)| = \frac{1}{2} |-18| = 9 \).
3. Area ACD = \( \frac{1}{2} |4(2 - 2) + (-1)(2 - 3) + (-3)(3 - 2)| = \frac{1}{2} |-2| = 1 \).
4. Total area = 10 sq units, which is not among (a), (b), (c).

Teacher's Note:
1. Find the area of a quadrilateral by splitting it into two triangles along a diagonal.
2. Take the vertices in order around the figure so that the two triangles do not overlap.

Answer: (a) 1

Explanation:
1. Slope of the join of (2, 1) and (4, 5) = 4/2 = 2.
2. Slope of the join of (a, −1) and (2, 1) = \( \frac{2}{2 - a} \).
3. Collinear: \( \frac{2}{2 - a} = 2 \), so 2 − a = 1 and a = 1.

Teacher's Note:
1. Equate slopes through a common point to get a linear equation in the unknown.
2. The area-equals-zero method gives the same answer and is useful when a slope may be undefined.

Answer: (c) (8, 8)

Explanation:
1. Let the third vertex be (x, y). Then \( \frac{4 - 9 + x}{3} = 1 \) and \( \frac{-3 + 7 + y}{3} = 4 \).
2. x − 5 = 3 gives x = 8; y + 4 = 12 gives y = 8.
3. The third vertex is (8, 8).

Teacher's Note:
1. The centroid is the average of the three vertices, so sum of x-coordinates = 3 × (x of centroid).
2. Check: (4 − 9 + 8)/3 = 1 and (−3 + 7 + 8)/3 = 4.

Answer: (b) x − y = 0

Explanation:
1. Let the point be (h, k) with h = k.
2. Replacing h, k by x, y gives y = x.
3. So the locus is x − y = 0, the line through the origin with slope 1.

Teacher's Note:
1. To find a locus, write the condition on (h, k) and then replace h by x and k by y.
2. Points such as (1, 1), (−2, −2) all lie on y = x, confirming the answer.

Answer: (d) parallel to Y-axis

Explanation:
1. Slope m = tan θ, where θ is the inclination with the positive x-axis.
2. For a line parallel to the y-axis, θ = 90° and tan 90° is not defined.
3. So the slope is not defined for a line parallel to the Y-axis.

Teacher's Note:
1. A vertical line has x₁ = x₂ for any two of its points, so the slope formula has zero in the denominator.
2. A horizontal line has slope 0, which is defined; do not confuse 'zero' with 'not defined'.

Answer: (b) \( \tan^{-1}\left( \frac{2}{3} \right) \)

Explanation:
1. Slope of BA = \( \frac{3 - 1}{2 + 2} = \frac{1}{2} \); slope of BC = \( \frac{3 + 4}{2 + 2} = \frac{7}{4} \).
2. \( \tan \theta = \left| \frac{\frac{7}{4} - \frac{1}{2}}{1 + \frac{7}{8}} \right| = \frac{5/4}{15/8} = \frac{2}{3} \).
3. So θ = tan⁻¹(2/3).

Teacher's Note:
1. The angle between two lines is found from their slopes using tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
2. Here both slopes are positive and the angle at B is acute, so no supplementary angle is needed.

Answer: (b) y = −3

Explanation:
1. A line parallel to the x-axis has the form y = constant.
2. It passes through (−4, −3), so the constant is −3.
3. The equation is y = −3.

Teacher's Note:
1. Every point on a horizontal line has the same y-coordinate.
2. The x-coordinate of the given point plays no role here.

Answer: (a) 3x − 4y = 25

Explanation:
1. Slope of OP, where P(3, −4), is −4/3, so the line has slope 3/4.
2. Equation through P: \( y + 4 = \frac{3}{4}(x - 3) \).
3. 4y + 16 = 3x − 9, i.e. 3x − 4y = 25.

Teacher's Note:
1. The line is perpendicular to the segment joining the origin and the foot.
2. Check: (3, −4) satisfies 3x − 4y = 9 + 16 = 25.

Answer: (c) x + y + 1 = 0

Explanation:
1. Equal intercepts a, a: \( \frac{x}{a} + \frac{y}{a} = 1 \), i.e. x + y = a.
2. Through (1, −2): a = 1 − 2 = −1.
3. The line is x + y + 1 = 0.

Teacher's Note:
1. With equal intercepts the intercept form reduces to x + y = a.
2. Substitute the given point to find a; the point must satisfy the final answer.

Answer: (a) (3a, −2b)

Explanation:
1. The line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
2. For (3a, −2b): \( \frac{3a}{a} + \frac{-2b}{b} = 3 - 2 = 1 \). ✓
3. So (3a, −2b) lies on the line.

Teacher's Note:
1. Use the intercept form and test each option by substitution.
2. For (−3a, 2b) the sum is −1, and for (a, b) it is 2, so they are not on the line.

Answer: (b) −2

Explanation:
1. Slope = (6 − 4)/(5 − 3) = 1, so the line is y − 4 = x − 3, i.e. y = x + 1.
2. Put y = −1: −1 = x + 1.
3. x = −2.

Teacher's Note:
1. Find the equation first, then substitute the given ordinate.
2. Ordinate means y-coordinate and abscissa means x-coordinate.

Answer: (b) 4x + 3y = 24, x + y = 7

Explanation:
1. Let the intercepts be a and 14 − a: \( \frac{x}{a} + \frac{y}{14 - a} = 1 \).
2. Through (3, 4): 3(14 − a) + 4a = a(14 − a), i.e. a² − 13a + 42 = 0, so a = 6 or 7.
3. a = 6: x/6 + y/8 = 1, i.e. 4x + 3y = 24; a = 7: x/7 + y/7 = 1, i.e. x + y = 7.

Teacher's Note:
1. Use one unknown intercept and express the other using the given sum.
2. Both roots of the quadratic give valid lines, so expect two answers.

Answer: (a) \( \frac{x}{x_1} + \frac{y}{y_1} = 2 \)

Explanation:
1. Let the line meet the axes at (a, 0) and (0, b); the mid-point is (a/2, b/2) = (x₁, y₁).
2. So a = 2x₁ and b = 2y₁.
3. Line: \( \frac{x}{2x_1} + \frac{y}{2y_1} = 1 \), i.e. \( \frac{x}{x_1} + \frac{y}{y_1} = 2 \).

Teacher's Note:
1. If (x₁, y₁) bisects the intercept, the intercepts are 2x₁ and 2y₁.
2. Check: the point (x₁, y₁) itself gives 1 + 1 = 2.

Answer: (c) \( \sqrt{3}x - y - 2\sqrt{3} = 0 \)

Explanation:
1. Slope of AB = (1 − 0)/(3 − 2) = 1, so AB makes 45° with the x-axis.
2. After rotating anticlockwise by 15° the inclination is 60°, so the slope is √3.
3. Line through A(2, 0): y = √3(x − 2), i.e. \( \sqrt{3}x - y - 2\sqrt{3} = 0 \).

Teacher's Note:
1. Rotating a line about a point changes only its inclination; the point stays on the line.
2. Anticlockwise rotation adds to the angle, clockwise rotation subtracts from it.

Answer: (b) y = (5/12)x

Explanation:
1. For y = 0.2x, tan θ = 1/5.
2. \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{5}{12} \).
3. Required line: y = (5/12)x.

Teacher's Note:
1. Doubling the angle does not double the slope; use the tan 2θ identity.
2. A line through the origin with slope m is simply y = mx.

Answer: (c) 2x + y − 1 = 0

Explanation:
1. A line perpendicular to x − 2y + 11 = 0 has the form 2x + y + k = 0.
2. Through (0, 1): 0 + 1 + k = 0, so k = −1.
3. Required line: 2x + y − 1 = 0.

Teacher's Note:
1. Perpendicular to ax + by + c = 0 is bx − ay + k = 0: swap the coefficients and change one sign.
2. Then use the given point to find k.

Answer: (d) x − 2y = 0

Explanation:
1. A line parallel to x − 2y + 5 = 0 has the form x − 2y + k = 0.
2. Through the origin: k = 0.
3. Required line: x − 2y = 0.

Teacher's Note:
1. Parallel lines keep the same x and y coefficients; only the constant changes.
2. Any line through the origin has no constant term.

Answer: (b) 2x − 3y + 1 = 0

Explanation:
1. Parallel line: 2x − 3y + k = 0.
2. Through (1, 1): 2 − 3 + k = 0, so k = 1.
3. Required line: 2x − 3y + 1 = 0.

Teacher's Note:
1. Keep the coefficients 2 and −3 unchanged for a parallel line.
2. Substitute the point to fix the constant term.

Answer: (b) 4/3

Explanation:
1. Perpendicular to 3x + y = 3: x − 3y + k = 0.
2. Through (2, 2): 2 − 6 + k = 0, so k = 4, and the line is x − 3y + 4 = 0.
3. Put x = 0: y = 4/3, which is the y-intercept.

Teacher's Note:
1. The y-intercept is the value of y where the line meets the y-axis (x = 0).
2. Slope of 3x + y = 3 is −3, so the perpendicular slope is 1/3.

Answer: (c) a(x − c) + b(y − d) = 0

Explanation:
1. Parallel line: ax + by + k = 0.
2. Through (c, d): k = −ac − bd.
3. So ax + by − ac − bd = 0, i.e. a(x − c) + b(y − d) = 0.

Teacher's Note:
1. A line through (x₁, y₁) parallel to ax + by + c = 0 is a(x − x₁) + b(y − y₁) = 0.
2. Check: putting x = c, y = d makes the left side zero.

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

Explanation:
1. Altitude p = distance of (2, −1) from x + y − 2 = 0 = \( \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \).
2. For an equilateral triangle of side a, p = \( \frac{\sqrt{3}}{2}a \), so a = \( \frac{2p}{\sqrt{3}} = \frac{2}{\sqrt{6}} \).
3. a = \( \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}} \).

Teacher's Note:
1. The altitude of the triangle is the perpendicular distance of the vertex from the base line.
2. Remember altitude = (√3/2) × side for an equilateral triangle.

Answer: (b) 3x + 4y − 24 = 0

Explanation:
1. Let the line be y = −(3/4)x + c with c > 0; its x-intercept is 4c/3.
2. Area = \( \frac{1}{2} \cdot c \cdot \frac{4c}{3} = \frac{2c^2}{3} = 24 \), so c² = 36 and c = 6.
3. Line: y = −(3/4)x + 6, i.e. 3x + 4y − 24 = 0.

Teacher's Note:
1. Area of the triangle cut off by a line with the axes is ½ × |x-intercept| × |y-intercept|.
2. The condition 'positive y-intercept' picks c = 6 out of c = ±6.

Answer: (c) 4x + 3y − 10 = 0

Explanation:
1. Line through (4, −2): y + 2 = m(x − 4), i.e. mx − y − 4m − 2 = 0.
2. Distance from O: \( \frac{|4m + 2|}{\sqrt{m^2 + 1}} = 2 \) gives 12m² + 16m = 0, so m = 0 or m = −4/3.
3. Non-horizontal, so m = −4/3: 3y + 6 = −4x + 16, i.e. 4x + 3y − 10 = 0.

Teacher's Note:
1. Use the perpendicular distance formula with the point-slope form to get an equation in m.
2. m = 0 gives the horizontal line y = −2, which the question excludes.

Answer: (b) parallel

Explanation:
1. Slope of 2x − 3y = 5 is 2/3; slope of 6x − 9y − 7 = 0 is 6/9 = 2/3.
2. \( \frac{2}{6} = \frac{-3}{-9} \neq \frac{-5}{-7} \).
3. Equal slopes but different lines, so they are parallel.

Teacher's Note:
1. Compare a/a′, b/b′ and c/c′: equal first two ratios with a different third ratio means parallel.
2. If all three ratios were equal the lines would be coincident.

Answer: (b) less than 60°

Explanation:
1. \( \tan \theta = \left| \frac{2 - (-2)}{1 + 2(-2)} \right| = \left| \frac{4}{-3} \right| = \frac{4}{3} \).
2. tan 45° = 1 and tan 60° = √3 ≈ 1.732.
3. Since 1 < 4/3 < √3, the acute angle lies between 45° and 60°, so it is less than 60°.

Teacher's Note:
1. The absolute value in the formula gives the acute angle directly.
2. Compare tan θ with tan of standard angles to locate θ.

Answer: (c) 90°

Explanation:
1. A = AB ∩ DA = (−3, 3); B = AB ∩ BC = (1, 1); C = BC ∩ CD = (1, −1); D = CD ∩ DA = (−2, −2).
2. Slope of AC = \( \frac{-1 - 3}{1 + 3} = -1 \); slope of BD = \( \frac{-2 - 1}{-2 - 1} = 1 \).
3. Product of slopes = −1, so the diagonals are at 90°.

Teacher's Note:
1. Each vertex is the intersection of the two sides that meet there, e.g. A lies on AB and DA.
2. Perpendicularity is checked quickly by the product of slopes being −1.

Answer: (c) 3x + y + 7 = 0 or x − 3y − 31 = 0

Explanation:
1. The third side makes equal angles with the equal sides (slopes 7 and −1): \( \left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{m + 1}{1 - m} \right| \).
2. Solving gives m = −3 or m = 1/3 (check: for m = −3 both sides equal 1/2; for m = 1/3 both equal 2).
3. Through (1, −10): y + 10 = −3(x − 1), i.e. 3x + y + 7 = 0, or y + 10 = (1/3)(x − 1), i.e. x − 3y − 31 = 0.

Teacher's Note:
1. In an isosceles triangle the base makes equal angles with the two equal sides.
2. Both slopes satisfy the condition, so both lines are possible third sides.

Answer: (a) \( x = 7,\ x + \sqrt{3}y = 7 + 9\sqrt{3} \)

Explanation:
1. Slope of the given line is \( \frac{1}{\sqrt{3}} \), so its inclination is 30°.
2. Lines at 60° to it have inclinations 30° + 60° = 90° and 30° − 60° = −30° (slope \( -\frac{1}{\sqrt{3}} \)).
3. Through (7, 9): x = 7 and \( y - 9 = -\frac{1}{\sqrt{3}}(x - 7) \), i.e. \( x + \sqrt{3}y = 7 + 9\sqrt{3} \).

Teacher's Note:
1. Use inclinations θ ± α when the angle with a given line is known.
2. An inclination of 90° gives a vertical line, which has no slope but is still a valid answer.

Answer: (d) x + 1 = 0, y + 1 = 0

Explanation:
1. x + y = 0 has slope −1, i.e. inclination 135°.
2. Required inclinations: 135° ± 45° = 180° (horizontal) and 90° (vertical).
3. Through (−1, −1): y + 1 = 0 and x + 1 = 0.

Teacher's Note:
1. Work with inclinations: θ ± α gives the two required directions.
2. An inclination of 180° is the same direction as 0°, i.e. a horizontal line.

Answer: (c) \( \left( 0, \frac{a}{m} \right) \)

Explanation:
1. The perpendicular through (a, 0) has slope \( -\frac{1}{m} \): \( y = -\frac{1}{m}(x - a) \).
2. Intersect with \( y = mx + \frac{a}{m} \): \( mx + \frac{a}{m} = -\frac{x}{m} + \frac{a}{m} \), so (m + 1/m)x = 0 and x = 0.
3. Then y = a/m, so the foot is (0, a/m).

Teacher's Note:
1. The foot of the perpendicular is the intersection of the given line with the perpendicular through the point.
2. Check: (0, a/m) lies on the given line since m·0 + a/m = a/m.

Answer: (a) 2x + y − 5 = 0

Explanation:
1. Solving x − 3y + 1 = 0 and 2x + 5y − 9 = 0 gives the point P(2, 1).
2. OP = √5, equal to the required distance, so OP itself is the perpendicular from O and the line is ⊥ OP.
3. Slope of OP = 1/2, so the line has slope −2: y − 1 = −2(x − 2), i.e. 2x + y − 5 = 0.

Teacher's Note:
1. If the given point is exactly at the required distance from the origin, the line must be perpendicular to OP at P.
2. Check: distance of 2x + y − 5 = 0 from O is 5/√5 = √5.

Answer: (a) y = 3

Explanation:
1. From x − y = 1, x = y + 1; then 2(y + 1) − 3y + 1 = 0 gives y = 3, x = 4.
2. The point of intersection is (4, 3).
3. Line parallel to the x-axis through it: y = 3.

Teacher's Note:
1. Find the point of intersection first by solving the two equations.
2. A line parallel to the x-axis depends only on the y-coordinate of the point.

Answer: (d) 4x − 3y + 2 = 0

Explanation:
1. From x = 5 − 2y: 3(5 − 2y) + 7y = 17 gives y = 2, x = 1; point (1, 2).
2. Perpendicular to 3x + 4y = 10: 4x − 3y + k = 0.
3. Through (1, 2): 4 − 6 + k = 0, k = 2; line 4x − 3y + 2 = 0.

Teacher's Note:
1. Solve for the intersection point, then use the perpendicular form bx − ay + k = 0.
2. Check: (1, 2) gives 4 − 6 + 2 = 0.

Answer: (c) 25x + 25y − 4 = 0

Explanation:
1. Doubling the second equation: 150x + 50y + 6 = 0; subtracting the first gives 50x + 7 = 0, so x = −7/50.
2. Then 50y = 1 − 100x = 15, so y = 3/10 = 15/50.
3. Equal intercepts: x + y = a with a = −7/50 + 15/50 = 4/25, i.e. 25x + 25y − 4 = 0.

Teacher's Note:
1. Eliminate y by making its coefficients equal, then find x.
2. A line with equal intercepts has the form x + y = a; the point fixes a.