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Straight Lines and Pair of Straight Lines question

2024 · 29 Jan · Shift 2 · Q32
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  5. /2024 · 29 Jan · Shift 2 · Q32

Straight Lines and Pair of Straight Lines question

2024 · 29 Jan · Shift 2 · Q32

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let A\mathrm{A}A be the point of intersection of the lines 3x+2y=14,5x−y=63 x+2 y=14,5 x-y=63x+2y=14,5x−y=6 and B\mathrm{B}B be the point of intersection of the lines 4x+3y=8,6x+y=54 x+3 y=8,6 x+y=54x+3y=8,6x+y=5. The distance of the point P(5,−2)P(5,-2)P(5,−2) from the line AB\mathrm{AB}AB is
  1. A
    132\frac{13}{2}213​
  2. B
    8
  3. C
    52\frac{5}{2}25​
  4. D
    6
View written solutionFree

Correct answer: D

  1. Find point AAA as the intersection of

    \qquad 5x-y=6.$$ From $5x-y=6$, we get $$y=5x-6.$$ Substitute into the first equation: $$3x+2(5x-6)=14$$ $$3x+10x-12=14$$ $$13x=26$$ $$x=2.$$ Then $$y=5(2)-6=4.$$ So, $$A=(2,4).$$

  2. Find point BBB as the intersection of

    \qquad 6x+y=5.$$ From $6x+y=5$, we get $$y=5-6x.$$ Substitute into the first equation: $$4x+3(5-6x)=8$$ $$4x+15-18x=8$$ $$-14x=-7$$ $$x=\frac{1}{2}.$$ Then $$y=5-6\left(\frac12\right)=5-3=2.$$ So, $$B=\left(\frac12,2\right).$$

  3. Equation of line ABABAB

    Slope of ABABAB is m=2−412−2=−2−32=43.m=\frac{2-4}{\frac12-2}=\frac{-2}{-\frac32}=\frac{4}{3}.m=21​−22−4​=−23​−2​=34​.

    Using point-slope form through A(2,4)A(2,4)A(2,4): y−4=43(x−2).y-4=\frac43(x-2).y−4=34​(x−2).

    Multiply by 333: 3y−12=4x−83y-12=4x-83y−12=4x−8 4x−3y+4=0.4x-3y+4=0.4x−3y+4=0.

    Hence the line ABABAB is 4x−3y+4=0.4x-3y+4=0.4x−3y+4=0.

  4. Distance of P(5,−2)P(5,-2)P(5,−2) from line ABABAB

    Distance from point (x1,y1)(x_1,y_1)(x1​,y1​) to line ax+by+c=0ax+by+c=0ax+by+c=0 is d=∣ax1+by1+c∣a2+b2.d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.d=a2+b2​∣ax1​+by1​+c∣​.

    Here, line is 4x−3y+4=0,4x-3y+4=0,4x−3y+4=0, so a=4a=4a=4, b=−3b=-3b=−3, c=4c=4c=4.

    For P(5,−2)P(5,-2)P(5,−2): d=∣4(5)−3(−2)+4∣42+(−3)2d=\frac{|4(5)-3(-2)+4|}{\sqrt{4^2+(-3)^2}}d=42+(−3)2​∣4(5)−3(−2)+4∣​ =∣20+6+4∣16+9=\frac{|20+6+4|}{\sqrt{16+9}}=16+9​∣20+6+4∣​ =305=6.=\frac{30}{5}=6.=530​=6.

  5. Check options

    The distance is 6.6.6. So the correct option is D.

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