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Straight Lines and Pair of Straight Lines question

2020 · 9 Jan · Shift 1 · Q37
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  5. /2020 · 9 Jan · Shift 1 · Q37

Straight Lines and Pair of Straight Lines question

2020 · 9 Jan · Shift 1 · Q37

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let C be the centroid of the triangle with vertices (3, –1), (1, 3) and (2, 4). Let P be the point of intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0. Then the line passing through the points C and P also passes through the point :
  1. A
    (–9, –7)
  2. B
    (9, 7)
  3. C
    (7, 6)
  4. D
    (–9, –6)
View written solutionFree

Correct answer: D

  1. Find the centroid CCC of the triangle

The vertices are (3,−1)(3,-1)(3,−1), (1,3)(1,3)(1,3) and (2,4)(2,4)(2,4).

The centroid of a triangle is

C(x1+x2+x33,y1+y2+y33).C\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).C(3x1​+x2​+x3​​,3y1​+y2​+y3​​).

So,

C=(3+1+23,−1+3+43)=(63,63)=(2,2).C=\left(\frac{3+1+2}{3},\frac{-1+3+4}{3}\right)=\left(\frac{6}{3},\frac{6}{3}\right)=(2,2).C=(33+1+2​,3−1+3+4​)=(36​,36​)=(2,2).
  1. Find the intersection point PPP of the given lines

The lines are:

x+3y−1=0⇒x+3y=1(1)x+3y-1=0 \quad \Rightarrow \quad x+3y=1 \tag{1}x+3y−1=0⇒x+3y=1(1) 3x−y+1=0⇒3x−y=−1(2)3x-y+1=0 \quad \Rightarrow \quad 3x-y=-1 \tag{2}3x−y+1=0⇒3x−y=−1(2)

From (2),

y=3x+1.y=3x+1.y=3x+1.

Substitute into (1):

x+3(3x+1)=1x+3(3x+1)=1x+3(3x+1)=1 x+9x+3=1x+9x+3=1x+9x+3=1 10x=−210x=-210x=−2 x=−15.x=-\frac{1}{5}.x=−51​.

Then,

y=3(−15)+1=−35+1=25.y=3\left(-\frac{1}{5}\right)+1=-\frac{3}{5}+1=\frac{2}{5}.y=3(−51​)+1=−53​+1=52​.

Hence,

P=(−15,25).P=\left(-\frac{1}{5},\frac{2}{5}\right).P=(−51​,52​).
  1. Equation of the line through C(2,2)C(2,2)C(2,2) and P(−15,25)P\left(-\frac15,\frac25\right)P(−51​,52​)

Slope:

m=2−252−(−15)=85115=811.m=\frac{2-\frac{2}{5}}{2-\left(-\frac{1}{5}\right)} =\frac{\frac{8}{5}}{\frac{11}{5}}=\frac{8}{11}.m=2−(−51​)2−52​​=511​58​​=118​.

So the line through (2,2)(2,2)(2,2) is

y−2=811(x−2).y-2=\frac{8}{11}(x-2).y−2=118​(x−2).

Multiply by 111111:

11y−22=8x−1611y-22=8x-1611y−22=8x−16 8x−11y+6=0.8x-11y+6=0.8x−11y+6=0.

Thus the required line is

8x−11y+6=0.8x-11y+6=0.8x−11y+6=0.
  1. Check the options

We test each point in 8x−11y+6=08x-11y+6=08x−11y+6=0.

  • A (−9,−7)(-9,-7)(−9,−7):
8(−9)−11(−7)+6=−72+77+6=11≠08(-9)-11(-7)+6=-72+77+6=11\neq 08(−9)−11(−7)+6=−72+77+6=11=0

Not on the line.

  • B (9,7)(9,7)(9,7):
8(9)−11(7)+6=72−77+6=1≠08(9)-11(7)+6=72-77+6=1\neq 08(9)−11(7)+6=72−77+6=1=0

Not on the line.

  • C (7,6)(7,6)(7,6):
8(7)−11(6)+6=56−66+6=−4≠08(7)-11(6)+6=56-66+6=-4\neq 08(7)−11(6)+6=56−66+6=−4=0

Not on the line.

  • D (−9,−6)(-9,-6)(−9,−6):
8(−9)−11(−6)+6=−72+66+6=08(-9)-11(-6)+6=-72+66+6=08(−9)−11(−6)+6=−72+66+6=0

This lies on the line.

  1. Conclusion

The line passing through CCC and PPP also passes through

(−9,−6).\boxed{(-9,-6)}.(−9,−6)​.
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