Straight Lines
ICSE · Class 11 · Mathematics
Flashcards for Straight Lines — ICSE Class 11 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
Interactive on Super Tutor
Studying Straight Lines? Get the full interactive chapter.
Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for flashcards and more.
1,000+ Class 11 students started this chapter today

Super Tutor has 12+ illustrations like this for Straight Lines alone — flashcards, concept maps, and step-by-step visuals.
See them allFind the distance between A(2,1) and B(-3,0).
Answer
Step 1: Use the distance formula: AB = √((x2 - x1)^2 + (y2 - y1)^2) Step 2: Substitute A(2,1) and B(-3,0): AB = √(( -3 - 2 )^2 + (0 - 1)^2) Step 3: AB = √(( -5 )^2 + ( -1 )^2) = √(25 + 1) Step 4: AB =…
A point P(x,0) on the x-axis is equidistant from (7,6) and (3,4). Find P.
Answer
Step 1: Since P is equidistant from both points, set the distances equal. Step 2: Use distance formula: √((x - 7)^2 + (0 - 6)^2) = √((x - 3)^2 + (0 - 4)^2) Step 3: Square both sides: (x - 7)^2 + 36 = …
Use the section formula to find the point dividing A(4,-1) and B(-2,3) internally in the ratio 1:3.
Answer
Step 1: Use the section formula: R = ((m x2 + n x1)/(m+n), (m y2 + n y1)/(m+n)) Step 2: Here, m = 1 and n = 3, A(x1,y1) = (4,-1), B(x2,y2) = (-2,3) Step 3: Substitute: R = ((1·(-2) + 3·4)/(1+3), (1·3 …
Find the midpoint of the line segment joining A(-1,4) and B(3,-2).
Answer
Step 1: Midpoint is the special case of section formula when m = n. Step 2: Use midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2) Step 3: Substitute A(-1,4) and B(3,-2): M = ((-1 + 3)/2, (4 - 2)/2) Ste…
Find the centroid of the triangle with vertices (5,6), (1,2), and (3,0).
Answer
Step 1: Use the centroid formula: G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3) Step 2: Substitute the three vertices: G = ((5 + 1 + 3)/3, (6 + 2 + 0)/3) Step 3: Simplify: G = (9/3, 8/3) Answer: (3, 8/3) Q…
Find the incentre of the triangle with vertices A(20,7), B(-36,7), and C(0,-8).
Answer
Step 1: Find side lengths opposite the vertices. BC = √((0 + 36)^2 + (-8 - 7)^2) = √(1296 + 225) = 39 CA = √((20 - 0)^2 + (7 + 8)^2) = √(400 + 225) = 25 AB = √((-36 - 20)^2 + (7 - 7)^2) = √(3136) = 56…
Find the area of the triangle with vertices A(-3,16), B(4,4), and C(3,-2).
Answer
Step 1: Use the area formula: Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| Step 2: Substitute: Area = (1/2) |(-3)(4 - (-2)) + 4((-2) - 16) + 3(16 - 4)| Step 3: Simplify inside the modulus: A…
Why is the area of a triangle unchanged when the origin is shifted?
Answer
Step 1: Shifting the origin changes coordinates, not the shape or size of the triangle. Step 2: If the origin shifts to (h,k), then x' = x - h and y' = y - k. Step 3: In the area formula, the extra te…
+20 more flashcards available
Practice AllFrequently Asked Questions
What are the important topics in Straight Lines for ICSE Class 11 Mathematics?
How to score full marks in Straight Lines — ICSE Class 11 Mathematics?
How many flashcards are available for Straight Lines?
Sources & Official References
Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.
More resources for Straight Lines
Practice Quiz
Test yourself with a quick quiz
Important Questions
Practice with board exam-style questions
Revision Notes
Key points for last-minute revision
Formula Sheet
All formulas in one place
Chapter Summary
Understand the chapter at a glance
Concept Maps
See how topics connect visually
Study Plan
Step-by-step plan to ace this chapter
Syllabus
What topics to cover
NCERT Solutions
Every textbook question solved step by step
For serious students
Get the full Straight Lines chapter — for free.
Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for ICSE Class 11 Mathematics.