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JEE Main 2018
Straight Lines
Straight Lines and Pair of Straight Lines
Easy

Question

A triangle with vertices (4,0),(1,1),(3,5)\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right) is :

Options

Solution

Key Concepts and Formulas

  • Distance Formula: The distance between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) is given by: PQ=(x2x1)2+(y2y1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • Isosceles Triangle: A triangle is isosceles if at least two of its sides are equal in length.
  • Right-Angled Triangle: A triangle is right-angled if the square of the length of its longest side (hypotenuse) is equal to the sum of the squares of the lengths of the other two sides (Pythagorean Theorem: a2+b2=c2a^2 + b^2 = c^2, where cc is the longest side).

Step-by-Step Solution

Let the vertices of the triangle be A(4,0)A(4,0), B(1,1)B(-1,-1), and C(3,5)C(3,5).

Step 1: Calculate the Lengths of the Sides

We will use the distance formula to calculate the lengths of the sides ABAB, BCBC, and CACA. This will allow us to determine if the triangle is isosceles and to apply the Pythagorean theorem if necessary.

  • Length of side AB: AB=(14)2+(10)2=(5)2+(1)2=25+1=26AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}

  • Length of side BC: BC=(3(1))2+(5(1))2=(4)2+(6)2=16+36=52BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52}

  • Length of side CA: CA=(43)2+(05)2=(1)2+(5)2=1+25=26CA = \sqrt{(4 - 3)^2 + (0 - 5)^2} = \sqrt{(1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}

Step 2: Check if the Triangle is Isosceles

A triangle is isosceles if at least two of its sides are equal in length. Comparing the side lengths we found: AB=26AB = \sqrt{26}, BC=52BC = \sqrt{52}, and CA=26CA = \sqrt{26}. Since AB=CAAB = CA, the triangle is isosceles.

Step 3: Check if the Triangle is Right-Angled

A triangle is right-angled if it satisfies the Pythagorean theorem. We need to determine the longest side and check if its square equals the sum of the squares of the other two sides.

The longest side is BC=52BC = \sqrt{52}. Therefore, we need to check if BC2=AB2+CA2BC^2 = AB^2 + CA^2.

AB2=(26)2=26AB^2 = (\sqrt{26})^2 = 26 BC2=(52)2=52BC^2 = (\sqrt{52})^2 = 52 CA2=(26)2=26CA^2 = (\sqrt{26})^2 = 26

Now, check the Pythagorean theorem: AB2+CA2=26+26=52AB^2 + CA^2 = 26 + 26 = 52 Since BC2=AB2+CA2=52BC^2 = AB^2 + CA^2 = 52, the triangle is right-angled.

Step 4: Conclusion

The triangle is both isosceles (because AB=CAAB = CA) and right-angled (because BC2=AB2+CA2BC^2 = AB^2 + CA^2).

Common Mistakes & Tips

  • Distance Formula Errors: Be careful with signs when applying the distance formula. A common mistake is to incorrectly handle negative coordinates.
  • Pythagorean Theorem: Ensure you correctly identify the longest side as the potential hypotenuse before applying the Pythagorean theorem.
  • Simplifying Radicals: While simplifying radicals isn't strictly necessary here, it can sometimes help in identifying relationships between side lengths.

Summary

We calculated the lengths of the sides of the triangle using the distance formula. We found that two sides were equal in length, indicating that the triangle is isosceles. We then verified that the Pythagorean theorem holds true, indicating that the triangle is also right-angled. Therefore, the triangle is both isosceles and right-angled.

The final answer is isosceles and right angled\boxed{\text{isosceles and right angled}}, which corresponds to option (A).

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