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Straight Lines and Pair of Straight Lines question

2022 · 27 Jul · Shift 1 · Q36
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  5. /2022 · 27 Jul · Shift 1 · Q36

Straight Lines and Pair of Straight Lines question

2022 · 27 Jul · Shift 1 · Q36

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let A(1,1),B(−4,3),C(−2,−5)A(1,1), B(-4,3), C(-2,-5)A(1,1),B(−4,3),C(−2,−5) be vertices of a triangle ABC,PA B C, PABC,P be a point on side BCB CBC, and Δ1\Delta_{1}Δ1​ and Δ2\Delta_{2}Δ2​ be the areas of triangles APBA P BAPB and ABCA B CABC, respectively. If Δ1:Δ2=4:7\Delta_{1}: \Delta_{2}=4: 7Δ1​:Δ2​=4:7, then the area enclosed by the lines AP,ACA P, A CAP,AC and the xxx-axis is :
  1. A
    14\frac{1}{4}41​
  2. B
    34\frac{3}{4}43​
  3. C
    12\frac{1}{2}21​
  4. D
    1
View written solutionFree

Correct answer: C

  1. Use the area ratio to locate point PPP on BCBCBC

Since PPP lies on side BCBCBC, triangles APBAPBAPB and ABCABCABC have:

  • the same altitude from AAA to line BCBCBC
  • bases PBPBPB and CBCBCB respectively

Hence,

Δ1Δ2=PBCB=47.\frac{\Delta_1}{\Delta_2} = \frac{PB}{CB} = \frac{4}{7}.Δ2​Δ1​​=CBPB​=74​.

So,

PB:PC=4:3.PB:PC = 4:3.PB:PC=4:3.

Thus PPP divides BCBCBC internally in the ratio 4:34:34:3.

  1. Find the coordinates of PPP

Given

B(−4,3),C(−2,−5).B(-4,3), \quad C(-2,-5).B(−4,3),C(−2,−5).

If PPP divides BCBCBC in the ratio BP:PC=4:3BP:PC = 4:3BP:PC=4:3, then by section formula,

P=(4xC+3xB4+3, 4yC+3yB4+3).P = \left(\frac{4x_C+3x_B}{4+3},\, \frac{4y_C+3y_B}{4+3}\right).P=(4+34xC​+3xB​​,4+34yC​+3yB​​).

So,

P=(4(−2)+3(−4)7, 4(−5)+3(3)7)=(−8−127, −20+97)=(−207,−117).P = \left(\frac{4(-2)+3(-4)}{7},\, \frac{4(-5)+3(3)}{7}\right) = \left(\frac{-8-12}{7},\, \frac{-20+9}{7}\right) = \left(-\frac{20}{7}, -\frac{11}{7}\right).P=(74(−2)+3(−4)​,74(−5)+3(3)​)=(7−8−12​,7−20+9​)=(−720​,−711​).
  1. Find equations of lines ACACAC and APAPAP

Line ACACAC

Points A(1,1)A(1,1)A(1,1) and C(−2,−5)C(-2,-5)C(−2,−5). Slope:

mAC=−5−1−2−1=−6−3=2.m_{AC} = \frac{-5-1}{-2-1} = \frac{-6}{-3}=2.mAC​=−2−1−5−1​=−3−6​=2.

Equation through (1,1)(1,1)(1,1):

y−1=2(x−1)  ⟹  y=2x−1.y-1=2(x-1) \implies y=2x-1.y−1=2(x−1)⟹y=2x−1.

Its xxx-intercept is obtained by putting y=0y=0y=0:

0=2x−1  ⟹  x=12.0=2x-1 \implies x=\frac12.0=2x−1⟹x=21​.

So ACACAC meets the xxx-axis at

(12,0).\left(\frac12,0\right).(21​,0).

Line APAPAP

Points A(1,1)A(1,1)A(1,1) and P(−207,−117)P\left(-\frac{20}{7},-\frac{11}{7}\right)P(−720​,−711​). Slope:

mAP=−117−1−207−1=−187−277=23.m_{AP} = \frac{-\frac{11}{7}-1}{-\frac{20}{7}-1} = \frac{-\frac{18}{7}}{-\frac{27}{7}} = \frac{2}{3}.mAP​=−720​−1−711​−1​=−727​−718​​=32​.

Equation through (1,1)(1,1)(1,1):

y−1=23(x−1).y-1=\frac23(x-1).y−1=32​(x−1).

Thus,

y=23x+13.y=\frac23x+\frac13.y=32​x+31​.

Its xxx-intercept is:

0=23x+13  ⟹  2x+1=0  ⟹  x=−12.0=\frac23x+\frac13 \implies 2x+1=0 \implies x=-\frac12.0=32​x+31​⟹2x+1=0⟹x=−21​.

So APAPAP meets the xxx-axis at

(−12,0).\left(-\frac12,0\right).(−21​,0).
  1. Find the enclosed region

The lines APAPAP, ACACAC, and the xxx-axis form a triangle with vertices:

A(1,1),(−12,0),(12,0).A(1,1), \quad \left(-\frac12,0\right), \quad \left(\frac12,0\right).A(1,1),(−21​,0),(21​,0).

Base on the xxx-axis:

12−(−12)=1.\frac12 - \left(-\frac12\right)=1.21​−(−21​)=1.

Height from AAA to the xxx-axis:

1.1.1.

Therefore area is

12×1×1=12.\frac12 \times 1 \times 1 = \frac12.21​×1×1=21​.
  1. Match with options

The required area is

12.\boxed{\frac12}.21​​.

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Hence, the derived answer agrees with the stored answer.

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