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Straight Lines and Pair of Straight Lines question

2023 · 13 Apr · Shift 2 · Q29
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  5. /2023 · 13 Apr · Shift 2 · Q29

Straight Lines and Pair of Straight Lines question

2023 · 13 Apr · Shift 2 · Q29

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let (α,β)(\alpha, \beta)(α,β) be the centroid of the triangle formed by the lines 15x−y=82,6x−5y=−415 x-y=82,6 x-5 y=-415x−y=82,6x−5y=−4 and 9x+4y=179 x+4 y=179x+4y=17. Then α+2β\alpha+2 \betaα+2β and 2α−β2 \alpha-\beta2α−β are the roots of the equation :
  1. A
    x2−7x+12=0x^{2}-7 x+12=0x2−7x+12=0
  2. B
    x2−13x+42=0x^{2}-13 x+42=0x2−13x+42=0
  3. C
    x2−14x+48=0x^{2}-14 x+48=0x2−14x+48=0
  4. D
    x2−10x+25=0x^{2}-10 x+25=0x2−10x+25=0
View written solutionFree

Correct answer: B

  1. Write the three lines

Let L1:15x−y=82,L_1: 15x-y=82,L1​:15x−y=82, L2:6x−5y=−4,L_2: 6x-5y=-4,L2​:6x−5y=−4, L3:9x+4y=17.L_3: 9x+4y=17.L3​:9x+4y=17.

The triangle is formed by the pairwise intersections of these three lines.

  1. Find the three vertices

Vertex A=L1∩L2A = L_1 \cap L_2A=L1​∩L2​

From L1L_1L1​: y=15x−82.y=15x-82.y=15x−82. Substitute into L2L_2L2​: 6x−5(15x−82)=−46x-5(15x-82)=-46x−5(15x−82)=−4 6x−75x+410=−46x-75x+410=-46x−75x+410=−4 −69x=−414-69x=-414−69x=−414 x=6.x=6.x=6. Then y=15(6)−82=8.y=15(6)-82=8.y=15(6)−82=8. So, A=(6,8).A=(6,8).A=(6,8).

Vertex B=L2∩L3B = L_2 \cap L_3B=L2​∩L3​

Solve 6x−5y=−4,6x-5y=-4,6x−5y=−4, 9x+4y=17.9x+4y=17.9x+4y=17. Multiply first by 444 and second by 555: 24x−20y=−16,24x-20y=-16,24x−20y=−16, 45x+20y=85.45x+20y=85.45x+20y=85. Add: 69x=69⇒x=1.69x=69 \Rightarrow x=1.69x=69⇒x=1. Then 6(1)−5y=−4⇒−5y=−10⇒y=2.6(1)-5y=-4 \Rightarrow -5y=-10 \Rightarrow y=2.6(1)−5y=−4⇒−5y=−10⇒y=2. So, B=(1,2).B=(1,2).B=(1,2).

Vertex C=L3∩L1C = L_3 \cap L_1C=L3​∩L1​

Solve 9x+4y=17,9x+4y=17,9x+4y=17, 15x−y=82.15x-y=82.15x−y=82. From L1L_1L1​: y=15x−82.y=15x-82.y=15x−82. Substitute into L3L_3L3​: 9x+4(15x−82)=179x+4(15x-82)=179x+4(15x−82)=17 9x+60x−328=179x+60x-328=179x+60x−328=17 69x=34569x=34569x=345 x=5.x=5.x=5. Then y=15(5)−82=−7.y=15(5)-82=-7.y=15(5)−82=−7. So, C=(5,−7).C=(5,-7).C=(5,−7).

  1. Find the centroid

If the centroid is (α,β)(\alpha,\beta)(α,β), then α=6+1+53=123=4,\alpha=\frac{6+1+5}{3}=\frac{12}{3}=4,α=36+1+5​=312​=4, β=8+2+(−7)3=33=1.\beta=\frac{8+2+(-7)}{3}=\frac{3}{3}=1.β=38+2+(−7)​=33​=1. Thus, (α,β)=(4,1).(\alpha,\beta)=(4,1).(α,β)=(4,1).

  1. Compute the required numbers

α+2β=4+2(1)=6,\alpha+2\beta=4+2(1)=6,α+2β=4+2(1)=6, 2α−β=2(4)−1=7.2\alpha-\beta=2(4)-1=7.2α−β=2(4)−1=7.

So the required roots are 666 and 777.

  1. Form the quadratic equation

A quadratic with roots 666 and 777 is x2−(6+7)x+(6)(7)=0x^2-(6+7)x+(6)(7)=0x2−(6+7)x+(6)(7)=0 x2−13x+42=0.x^2-13x+42=0.x2−13x+42=0.

  1. Match with the options

This is Option B: x2−13x+42=0.x^2-13x+42=0.x2−13x+42=0.

  1. Compare with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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