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Straight Lines and Pair of Straight Lines question

2002 · Shift 0 · Q72
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Straight Lines and Pair of Straight Lines question

2002 · Shift 0 · Q72

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
A triangle with vertices (4,0),(−1,−1),(3,5)\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right)(4,0),(−1,−1),(3,5) is :
  1. A
    isosceles and right angled
  2. B
    isosceles but not right angled
  3. C
    right angled but not isosceles
  4. D
    neither right angled nor isosceles
View written solutionFree

Correct answer: A

  1. Let the vertices be A(4,0),B(−1,−1),C(3,5).A(4,0),\quad B(-1,-1),\quad C(3,5).A(4,0),B(−1,−1),C(3,5).

  2. Compute the side lengths using the distance formula.

    AB=(4−(−1))2+(0−(−1))2=52+12=26AB=\sqrt{(4-(-1))^2+(0-(-1))^2}=\sqrt{5^2+1^2}=\sqrt{26}AB=(4−(−1))2+(0−(−1))2​=52+12​=26​

    AC=(4−3)2+(0−5)2=12+(−5)2=26AC=\sqrt{(4-3)^2+(0-5)^2}=\sqrt{1^2+(-5)^2}=\sqrt{26}AC=(4−3)2+(0−5)2​=12+(−5)2​=26​

    BC=(3−(−1))2+(5−(−1))2=42+62=52=213BC=\sqrt{(3-(-1))^2+(5-(-1))^2}=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}BC=(3−(−1))2+(5−(−1))2​=42+62​=52​=213​

  3. Check if the triangle is isosceles.

    Since AB=AC=26,AB=AC=\sqrt{26},AB=AC=26​, the triangle is isosceles.

  4. Check if the triangle is right angled using Pythagoras theorem.

    AB2=26,AC2=26,BC2=52AB^2=26,\quad AC^2=26,\quad BC^2=52AB2=26,AC2=26,BC2=52

    Now, AB2+AC2=26+26=52=BC2AB^2+AC^2=26+26=52=BC^2AB2+AC2=26+26=52=BC2

    Hence, the triangle is right angled.

  5. Therefore, the triangle is isosceles and right angled.

So the correct option is A.

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