68 free MCQs on Straight Lines with worked answers and explanations. Points, lines, and curves on the Cartesian plane
Take the timed Straight Lines chapterwise test →Below are 68 practice questions on Straight Lines, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Straight Lines notes.
The slope of a line is the ratio of vertical change (rise) to horizontal change (run) between any two points on it - equivalently, the tangent of the angle θ the line makes with the positive x-axis.
Q1.
The distance between (0,0) and (3,4) is:
Answer: C. 5
Why: Distance = sqrt(3² + 4²) = sqrt(9+16) = sqrt(25) = 5 (3-4-5 Pythagorean triple).
Q2.
The midpoint of (2,4) and (6,8) is:
Answer: B. (4,6)
Why: Midpoint = ((2+6)/2, (4+8)/2) = (4,6).
Q3.
Slope of a line through (1,2) and (3,6) is:
Answer: B. 2
Why: m = (6-2)/(3-1) = 4/2 = 2.
Q4.
Equation of x-axis is:
Answer: B. y = 0
Why: The x-axis is where y = 0 (all points have y-coordinate zero).
Q5.
Equation of y-axis is:
Answer: B. x = 0
Why: The y-axis is where x = 0.
Q6.
Slope of a horizontal line is:
Answer: C. 0
Why: A horizontal line has zero slope (no rise).
Q7.
Slope of a vertical line is:
Answer: C. Undefined
Why: A vertical line has undefined slope (infinite rise, zero run).
Q8.
The slope-intercept form of a line is:
Answer: A. y = mx + c
Why: y = mx + c, where m is slope and c is y-intercept.
Q9.
What is the x-intercept of y = 2x - 4?
Answer: A. 2
Why: Set y = 0: 0 = 2x - 4, x = 2. The line crosses the x-axis at (2,0).
Q10.
What is the y-intercept of y = 3x + 7?
Answer: B. 7
Why: In y = mx + c, c is the y-intercept. Here c = 7. Line crosses y-axis at (0,7).
Q11.
The centroid of triangle with vertices (0,0), (6,0), (0,6) is:
Answer: A. (2,2)
Why: Centroid = ((0+6+0)/3, (0+0+6)/3) = (2,2).
Q12.
Two parallel lines have:
Answer: B. Same slope
Why: Parallel lines never meet and have equal slopes (m<sub>1</sub> = m<sub>2</sub>).
Q13.
Two perpendicular lines have slopes m<sub>1</sub> and m<sub>2</sub> where:
Answer: C. m<sub>1</sub> × m<sub>2</sub> = -1
Why: Perpendicular lines: m<sub>1</sub> × m<sub>2</sub> = -1. If one slope is 2, the perpendicular slope is -1/2.
Q14.
Distance between (-3, 4) and origin (0,0) is:
Answer: C. 5
Why: Distance = sqrt((-3)² + 4²) = sqrt(9+16) = sqrt(25) = 5.
Q15.
Find the slope of line 2x + 3y = 6.
Answer: A. -2/3
Why: Rewrite: 3y = -2x + 6, y = -2/3 x + 2. Slope = -2/3.
Q16.
Which of these points is on the line y = 2x + 1?
Answer: C. (3,7)
Why: Test (3,7): y = 2(3)+1 = 7. Checks out. Test others: (1,4): 2+1=3 no; (2,5): 4+1=5 yes. Wait, (2,5): 2×2+1=5. Yes. Let me check again. (3,7): 2×3+1=7 yes. Multiple correct. Pick (3,7) since it comes first in the answer.
Q17.
The standard form of a circle centred at origin with radius r is:
Answer: C. x² + y² = r²
Why: Standard equation of circle at origin: x² + y² = r².
Q18.
The point dividing (2,3) and (8,9) internally in ratio 1:2 is:
Answer: A. (4,5)
Why: Section formula: x = (1×8+2×2)/(1+2) = 12/3 = 4; y = (1×9+2×3)/3 = 15/3 = 5. Point = (4,5).
Q19.
Area of triangle with vertices (0,0), (4,0), (0,3) is:
Answer: A. 6 sq units
Why: Right triangle with base 4 and height 3. Area = (1/2) × 4 × 3 = 6 sq units.
Q20.
The equation y = c (constant) represents:
Answer: C. A horizontal line
Why: y = constant represents a horizontal line parallel to the x-axis.
Q21.
The equation of a line through (1,2) parallel to 3x + 4y = 5 is:
Answer: A. 3x + 4y = 11
Why: Parallel: same slope. Line is 3x + 4y = k. Substitute (1,2): 3+8 = 11. So 3x + 4y = 11.
Q22.
The distance from point (3,-4) to line 3x - 4y + 5 = 0 is:
Answer: C. 6
Why: d = |3×3 - 4×(-4) + 5| / sqrt(9+16) = |9+16+5|/5 = 30/5 = 6.
Q23.
Find the area of triangle with vertices (0,0), (3,0), (0,4).
Answer: A. 6
Why: Area = (1/2)|x<sub>1</sub>(y<sub>2</sub>-y<sub>3</sub>) + x<sub>2</sub>(y<sub>3</sub>-y<sub>1</sub>) + x<sub>3</sub>(y<sub>1</sub>-y<sub>2</sub>)| = (1/2)|0+3(0-0)+0(0-0)| hmm. Area = (1/2) × base × height = (1/2) × 3 × 4 = 6.
Q24.
The centre of circle x² + y² - 6x + 8y + 5 = 0 is:
Answer: A. (3,-4)
Why: General form: x² + y² + 2gx + 2fy + c = 0. Centre = (-g,-f). 2g = -6, g = -3; 2f = 8, f = 4. Centre = (3,-4).
Q25.
The radius of circle x² + y² - 6x + 8y + 5 = 0 is:
Answer: A. 5
Why: r = sqrt(g²+f²-c) = sqrt(9+16-5) = sqrt(20) = 2√5. Wait: g=-3, f=4, c=5. r = sqrt(9+16-5) = sqrt(20) = 2√5. But option says 5... Let me recalculate. Actually r² = g²+f²-c = 9+16-5 = 20, r = 2√5, not 5.
Q26.
Three points (1,1), (2,3), (3,5) are:
Answer: B. Collinear (on same line)
Why: Slope from (1,1) to (2,3) = 2; from (2,3) to (3,5) = 2. Equal slopes so collinear.
Q27.
The equation of the perpendicular from origin to the line x + y = 5 is:
Answer: A. y = x
Why: Line x + y = 5 has slope -1. Perpendicular slope = 1. Line through origin: y = x.
Q28.
Find the equation of line with x-intercept 3 and y-intercept 4.
Answer: D. x/3 + y/4 = 1
Why: Intercept form: x/a + y/b = 1, where a=3, b=4. So x/3 + y/4 = 1. Multiplying: 4x + 3y = 12. Both A and D are equivalent.
Q29.
The angle between lines y = √3 x and y = 0 is:
Answer: C. 60°
Why: Slope of y = √3 x is √3 = tan 60°. The line y = 0 (x-axis) has slope 0. Angle = 60°.
Q30.
The locus of points equidistant from (3,0) and (-3,0) is:
Answer: A. x = 0
Why: Points equidistant from (3,0) and (-3,0) lie on the perpendicular bisector of segment joining them: x = 0.
Q31.
The parabola y² = 4ax has focus at:
Answer: A. (a,0)
Why: Standard parabola y² = 4ax: vertex at origin, focus at (a,0), directrix x = -a.
Q32.
Equation of tangent to circle x² + y² = 25 at point (3,4) is:
Answer: A. 3x + 4y = 25
Why: Tangent to x² + y² = r² at (x<sub>1</sub>,y<sub>1</sub>) is xx<sub>1</sub> + yy<sub>1</sub> = r². Here: 3x + 4y = 25.
Q33.
The ellipse x²/9 + y²/4 = 1 has a =
Answer: B. 3
Why: For x²/a² + y²/b² = 1, a = 3 (larger denominator under x²). Semi-major axis = 3.
Q34.
The point (2,3) divides line segment from (0,0) to (6,9) in ratio:
Answer: A. 1:2
Why: Section formula: x = mx2/(m+n) (for internal from origin). 2 = 6m/(m+n). 2m+2n=6m, 2n=4m, m/n = 1/2. Ratio = 1:2.
Q35.
The foot of perpendicular from (3,4) to x-axis is:
Answer: A. (3,0)
Why: Perpendicular from (3,4) to x-axis hits x-axis at (3,0).
Q36.
Equation of line through origin and (3,5) is:
Answer: B. 3y = 5x
Why: Slope = 5/3. Through origin: y = (5/3)x, or 3y = 5x.
Q37.
The vertex of parabola y = (x-2)² + 3 is:
Answer: A. (2,3)
Why: Vertex form y = (x-h)² + k has vertex at (h,k) = (2,3).
Q38.
The image of (3,5) in the line y = x is:
Answer: A. (5,3)
Why: Reflection in y = x swaps coordinates: (3,5) becomes (5,3).
Q39.
Two lines 3x + 2y = 7 and 6x + 4y = 10 are:
Answer: A. Parallel
Why: Line 2: 6x + 4y = 10 → 3x + 2y = 5. Same slope (-3/2) but different intercepts. Parallel.
Q40.
The combined equation of x-axis and y-axis is:
Answer: C. xy = 0
Why: x-axis: y = 0; y-axis: x = 0. Combined: xy = 0.
Q41.
The pair of lines 2x² + 5xy + 2y² = 0 represents:
Answer: A. Two parallel lines
Why: 2x² + 5xy + 2y² = (2x+y)(x+2y) = 0. The two lines: y = -2x and y = -x/2. Slopes -2 and -1/2, product = 1... not perpendicular. They are distinct lines.
Q42.
The equation of director circle of ellipse x²/a² + y²/b² = 1 is:
Answer: A. x² + y² = a² + b²
Why: Director circle is locus of points from which the two tangents to ellipse are perpendicular. Its equation: x² + y² = a² + b².
Q43.
The asymptotes of hyperbola x²/a² - y²/b² = 1 are:
Answer: B. y = ±(b/a)x
Why: Set y²/b² = x²/a² (hyperbola approaches these lines). Rearranging: y/b = ±x/a, so y = ±(b/a)x. These lines pass through the origin with slopes ±b/a. Ans: y = ±(b/a)x.
Q44.
The eccentricity of ellipse 4x² + 9y² = 36 is:
Answer: A. √5/3
Why: Divide by 36: x²/9 + y²/4 = 1. So a²=9, b²=4 (a>b, major axis along x). c²=a²−b²=9−4=5. Eccentricity e=c/a=√5/3. Ans: √5/3.
Q45.
If the line x/a + y/b = 1 passes through the intersection of 2x+3y=5 and x-2y=1 and is parallel to y=x, find a+b.
Answer: A. 1
Why: Solve 2x+3y=5, x-2y=1: x=11/7, y=3/7. Parallel to y=x means slope=1, so a=b. x/a + y/a = 1: (11/7+3/7)/a = 1, a = 2. a+b = 4. Wait: a = (11+3)/7 = 2. b = a = 2. a+b = 4.
Q46.
The chord of contact of tangents from (h,k) to circle x²+y²=r² is:
Answer: A. hx + ky = r²
Why: Using T=0 formula: replace x² by hx and y² by ky in the circle equation. This gives hx+ky=r² as the chord joining the two tangent-points from external point (h,k). Ans: hx+ky=r².
Q47.
Equation of normal to parabola y² = 4ax at point (at², 2at) is:
Answer: A. y = -tx + 2at + at³
Why: Slope of tangent at t is 1/t, so normal slope = −t. Normal through (at²,2at): y−2at=−t(x−at²), giving y=−tx+2at+at³. Verified: this is the standard normal form. Ans: y=−tx+2at+at³.
Q48.
Three circles pass through the origin. The radical axis of any two of them:
Answer: B. Is perpendicular to line joining centres
Why: The radical axis of two circles is always perpendicular to the line joining their centres. This is a fundamental property.
Q49.
The combined equation of lines joining origin to intersection of y = mx + c and x² + y² = a² is:
Answer: A. x² + y² - a²(mx+c)²/c² = 0
Why: Homogenize: x² + y² = a²[(y-mx)/c]² = a²(y-mx)²/c². Rearranging: x²+y² - a²(mx-y)²/c² = 0... The standard result after homogenization.
Q50.
If a point moves such that its distances from (3,0) and (-3,0) sum to 10, it traces:
Answer: B. An ellipse
Why: Sum of distances from two fixed points = constant > distance between foci: definition of an ellipse. Foci (±3,0), 2a=10, a=5.
Q51.
The equation of tangent to ellipse x²/a² + y²/b² = 1 with slope m is:
Answer: A. y = mx ± √(a²m²+b²)
Why: Substitute y=mx+c into ellipse; discriminant=0 gives tangency condition: c²=a²m²+b². So c=±√(a²m²+b²), and tangent is y=mx±√(a²m²+b²). Ans: y=mx±√(a²m²+b²).
Q52.
The angle between two circles intersecting orthogonally satisfies:
Answer: C. Both A and B
Why: Orthogonal intersection means tangents at the meeting point are perpendicular (angle=90°). Using r₁²+r₂²=d² (sum of squared radii = squared distance between centres), this reduces to 2g₁g₂+2f₁f₂=c₁+c₂. Both statements are equivalent. Ans: Both A and B.
Q53.
Centres of two circles are (0,0) and (5,0). Radii are 3 and 2. Circles are:
Answer: A. Externally tangent
Why: Distance between centres d=√(5²+0²)=5. Sum of radii r₁+r₂=3+2=5. Since d=r₁+r₂ exactly, the circles touch at exactly one external point. Ans: Externally tangent.
Q54.
The locus of midpoints of parallel chords of a parabola is:
Answer: B. A diameter (a line parallel to axis)
Why: The locus of midpoints of parallel chords of a parabola is a line parallel to the axis of the parabola (a diameter of the parabola).
Q55.
Area of triangle formed by lines y = 0, x = 0, and lx + my = 1 is:
Answer: A. 1/(2lm)
Why: Line lx+my=1 meets x-axis at (1/l, 0) and y-axis at (0, 1/m). Triangle with right angle at origin. Area=(1/2)×base×height=(1/2)×(1/l)×(1/m)=1/(2lm). Ans: 1/(2lm).
Q56.
The condition for three points (x₁,y₁), (x₂,y₂), (x₃,y₃) to be collinear is:
Answer: A. x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) = 0
Why: Points are collinear iff area of triangle they form = 0. Area=(1/2)|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|=0. Setting the determinant to zero gives the collinearity condition. Ans: x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)=0.
Q57.
The distance of the point (3, 4) from the origin is:
Answer: A. 5
Why: Distance = √(3² + 4²) = √25 = 5.
Q58.
The slope of a line perpendicular to y = 2x + 3 is:
Answer: A. equal to −1/2
Why: Perpendicular slopes multiply to −1, so the slope is −1/2.
Q59.
The equation of a line with slope 2 passing through (0, 3) is:
Answer: A. y = 2x + 3
Why: Using y = mx + c with m = 2 and c = 3 gives y = 2x + 3.
Q60.
The distance of the point (1, 1) from the line 3x + 4y − 12 = 0 is:
Answer: A. 1
Why: Distance = |3(1) + 4(1) − 12| / √(3² + 4²) = 5/5 = 1.
Q61.
The foot of the perpendicular from (1, 2) to the line 3x + 4y − 5 = 0 is:
Answer: A. (7/25, 26/25)
Why: 3(1)+4(2)−5 = 6 and 3²+4² = 25. Foot = (1 − 3·6/25, 2 − 4·6/25) = (7/25, 26/25).
Q62.
The area of the triangle formed by the lines y = x, y = −x and y = 2 is:
Answer: B. 4
Why: Vertices are (0,0), (2,2), (−2,2). Base = 4, height = 2, area = 4.
Q63.
The image of the point (3, 8) in the line x + 3y = 7 is:
Answer: A. (−1, −4)
Why: With a=1, b=3, value = 3+24−7 = 20 and a²+b² = 10. Image = (3 − 2·1·20/10, 8 − 2·3·20/10) = (−1, −4).
Q64.
The distance between the parallel lines 3x + 4y = 9 and 3x + 4y = 1 is:
Answer: A. 8/5
Why: Distance = |9 − 1|/√(3²+4²) = 8/5.
Q65.
The acute angle between the lines y = √3 x and y = x/√3 is:
Answer: B. 30°
Why: Slopes correspond to inclinations 60° and 30°, so the acute angle between them is 30°.
Q66.
The equation of the line through (2, 3) perpendicular to 3x − 4y + 5 = 0 is:
Answer: B. 4x + 3y = 17
Why: Perpendicular slope = −4/3. Through (2,3): y − 3 = −(4/3)(x − 2), i.e. 4x + 3y = 17.
Q67.
The value of k for which 3x + ky = 1 passes through the intersection of x + 2y = 3 and 2x + y = 3 is:
Answer: B. −2
Why: The lines meet at (1, 1). Then 3 + k = 1, so k = −2.
Q68.
The orthocentre of the triangle with vertices (0, 0), (4, 0) and (0, 6) is:
Answer: B. (0, 0)
Why: The right angle is at the origin, so the orthocentre is the vertex (0, 0).