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Straight Lines and Pair of Straight Lines question

2022 · 29 Jul · Shift 2 · Q32
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  5. /2022 · 29 Jul · Shift 2 · Q32

Straight Lines and Pair of Straight Lines question

2022 · 29 Jul · Shift 2 · Q32

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let A(α,−2),B(α,6)\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)A(α,−2),B(α,6) and C(α4,−2)\mathrm{C}\left(\frac{\alpha}{4},-2\right)C(4α​,−2) be vertices of a △ABC\triangle \mathrm{ABC}△ABC. If (5,α4)\left(5, \frac{\alpha}{4}\right)(5,4α​) is the circumcentre of △ABC\triangle \mathrm{ABC}△ABC, then which of the following is NOT correct about △ABC\triangle \mathrm{ABC}△ABC?
  1. A
    area is 24
  2. B
    perimeter is 25
  3. C
    circumradius is 5
  4. D
    inradius is 2
View written solutionFree

Correct answer: B

  1. List the points and use the circumcentre condition

Given: A(α,−2),B(α,6),C(α4,−2)A(\alpha,-2),\quad B(\alpha,6),\quad C\left(\frac{\alpha}{4},-2\right)A(α,−2),B(α,6),C(4α​,−2)

Circumcentre is: O(5,α4)O\left(5,\frac{\alpha}{4}\right)O(5,4α​)

Since OOO is the circumcentre, it is equidistant from all three vertices: OA=OB=OCOA=OB=OCOA=OB=OC

  1. First observe the shape of the triangle
  • A(α,−2)A(\alpha,-2)A(α,−2) and B(α,6)B(\alpha,6)B(α,6) have same xxx-coordinate, so ABABAB is vertical.
  • A(α,−2)A(\alpha,-2)A(α,−2) and C(α4,−2)C\left(\frac{\alpha}{4},-2\right)C(4α​,−2) have same yyy-coordinate, so ACACAC is horizontal.

Hence, AB⊥ACAB \perp ACAB⊥AC so △ABC\triangle ABC△ABC is right-angled at AAA.

  1. Use property of circumcentre of a right triangle

In a right triangle, the circumcentre is the midpoint of the hypotenuse.

Here hypotenuse is BCBCBC.

Midpoint of BCBCBC is:

=\left(\frac{5\alpha}{8},2\right)$$ This must equal the given circumcentre $\left(5,\frac{\alpha}{4}\right)$. So, $$\frac{5\alpha}{8}=5 \implies \alpha=8$$ Also, $$2=\frac{\alpha}{4} \implies \alpha=8$$ Consistent. Thus the vertices are: $$A(8,-2),\quad B(8,6),\quad C(2,-2)$$ 4. **Find the side lengths** $$AB=|6-(-2)|=8$$ $$AC=|8-2|=6$$ Now, $$BC=\sqrt{(8-2)^2+(6-(-2))^2} =\sqrt{6^2+8^2} =\sqrt{100}=10$$ So the triangle is a $6$-$8$-$10$ right triangle. 5. **Check each option** ### Option A: area is 24 For a right triangle, $$\text{Area}=\frac12\cdot AB\cdot AC=\frac12\cdot 8\cdot 6=24$$ So **A is correct**. ### Option B: perimeter is 25 $$\text{Perimeter}=AB+AC+BC=8+6+10=24$$ So perimeter is **not** $25$. Hence **B is NOT correct**. ### Option C: circumradius is 5 For a right triangle, circumradius is half the hypotenuse: $$R=\frac{BC}{2}=\frac{10}{2}=5$$ So **C is correct**. ### Option D: inradius is 2 For a right triangle with legs $6,8$ and hypotenuse $10$, $$r=\frac{a+b-c}{2}=\frac{6+8-10}{2}=2$$ So **D is correct**. 6. **Conclusion** The only statement which is NOT correct is: $$\boxed{\text{B}}$$
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