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Chapter 10 of 18
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Straight Lines

ICSE · Class 11 · Mathematics

Flashcards for Straight Lines — ICSE Class 11 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

59 questions28 flashcards5 concepts

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A 3D diagram showing three points A, B, and C that lie on the same straight line, illustrating the concept of collinearity.
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28 Flashcards
Card 1Distance formula

Find 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 =

Card 2Distance formula

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 =

Card 3Section formula

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

Card 4Section formula

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

Card 5Centroid of triangle

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

Card 6Incentre of triangle

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

Card 7Area of triangle

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

Card 8Shifting of origin

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

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Frequently Asked Questions

What are the important topics in Straight Lines for ICSE Class 11 Mathematics?
Straight Lines covers several key topics that are frequently asked in ICSE Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Straight Lines — ICSE Class 11 Mathematics?
Understand the core concepts first, then work through the 59 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many flashcards are available for Straight Lines?
There are 28 flashcards for Straight Lines covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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