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Straight Lines and Pair of Straight Lines question

2021 · 31 Aug · Shift 2 · Q33
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  5. /2021 · 31 Aug · Shift 2 · Q33

Straight Lines and Pair of Straight Lines question

2021 · 31 Aug · Shift 2 · Q33

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let A be the set of all points (α\alphaα, β\betaβ) such that the area of triangle formed by the points (5, 6), (3, 2) and (α\alphaα, β\betaβ) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :
  1. A
    45{4 \over {\sqrt 5 }}5​4​
  2. B
    165{16 \over {\sqrt 5 }}5​16​
  3. C
    85{8 \over {\sqrt 5 }}5​8​
  4. D
    125{12 \over {\sqrt 5 }}5​12​
View written solutionFree

Correct answer: C

  1. Let the variable point be P(α,β)P(\alpha,\beta)P(α,β), and the fixed points be A(5,6),B(3,2).A(5,6),\quad B(3,2).A(5,6),B(3,2).

  2. The area of triangle formed by A,B,PA,B,PA,B,P is 12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣=12.\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=12.21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣=12. Substituting: 12∣5(2−β)+3(β−6)+α(6−2)∣=12.\frac12\left|5(2-\beta)+3(\beta-6)+\alpha(6-2)\right|=12.21​∣5(2−β)+3(β−6)+α(6−2)∣=12.

  3. Simplify the expression inside modulus: 5(2−β)+3(β−6)+4α=10−5β+3β−18+4α5(2-\beta)+3(\beta-6)+4\alpha = 10-5\beta+3\beta-18+4\alpha5(2−β)+3(β−6)+4α=10−5β+3β−18+4α =4α−2β−8.=4\alpha-2\beta-8.=4α−2β−8. So, 12∣4α−2β−8∣=12\frac12|4\alpha-2\beta-8|=1221​∣4α−2β−8∣=12 ∣4α−2β−8∣=24.|4\alpha-2\beta-8|=24.∣4α−2β−8∣=24.

  4. Divide by 222: ∣2α−β−4∣=12.|2\alpha-\beta-4|=12.∣2α−β−4∣=12. Hence the set AAA consists of points lying on the two lines: 2α−β−4=12or2α−β−4=−12.2\alpha-\beta-4=12 \quad \text{or} \quad 2\alpha-\beta-4=-12.2α−β−4=12or2α−β−4=−12. That is, 2α−β=16or2α−β=−8.2\alpha-\beta=16 \quad \text{or} \quad 2\alpha-\beta=-8.2α−β=16or2α−β=−8.

  5. We need the least possible distance from the origin (0,0)(0,0)(0,0) to any point on these lines. This is the perpendicular distance from the origin to either line.

    For line 2x−y−16=02x-y-16=02x−y−16=0: d1=∣−16∣22+(−1)2=165.d_1=\frac{| -16 |}{\sqrt{2^2+(-1)^2}}=\frac{16}{\sqrt5}.d1​=22+(−1)2​∣−16∣​=5​16​.

    For line 2x−y+8=02x-y+8=02x−y+8=0: d2=∣8∣22+(−1)2=85.d_2=\frac{|8|}{\sqrt{2^2+(-1)^2}}=\frac{8}{\sqrt5}.d2​=22+(−1)2​∣8∣​=5​8​.

  6. Therefore, the least possible length is min⁡(165,85)=85.\min\left(\frac{16}{\sqrt5},\frac{8}{\sqrt5}\right)=\frac{8}{\sqrt5}.min(5​16​,5​8​)=5​8​.

  7. Matching with options, this is Option C.

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