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The points of intersection of three lines $2\text{X} + 3\text{Y} – 5 = 0, 5\text{X} – 7\text{Y} + 2 = 0$ and $9\text{X} – 5\text{Y} – 4= 0$

  1. form a triangle
  2. are on lines perpendicular to each other
  3. are on lines parallel to each other
  4. are coincident
in Quantitative Aptitude 13.3k points 306 2250 2467
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1 Answer

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Given that,

  • $2x+3y-5=0 \longrightarrow(1)$
  • $5x-7y+2=0 \longrightarrow(2)$
  • $9x-5y-4=0 \longrightarrow(3)$

Take equation $(1)$ and $(2).$

            $2x+3y-5=0 \longrightarrow(1)\times7$

            $5x-7y+2=0 \longrightarrow(2)\times3$

            $14x+21y-35=0$

            $15x-21y+6=0$

           ----------------------------------

            $29x-29=0$

            $\boxed{x=1}$

Put the value of $’x’$ in equation $(1)$,we get.

$2(1)+3y-5=0$

$\Rightarrow 3y=3$ 

$\Rightarrow \boxed{y=1}$ 

Take equation $(1)$ and $(3).$

         $2x+3y-5=0 \longrightarrow(1)\times5$

         $9x-5y-4=0 \longrightarrow(3)\times3$

         $10x+15y-25=0$

         $27x-15y-12=0$

         ----------------------------------

            $37x-37=0$

            $\Rightarrow\boxed{x=1}$

Put the value of $’x’$ in equation $(3)$,we get.

$9(1)-5y-4=0$

$\Rightarrow -5y+5=0$ 

$\Rightarrow \boxed{y=1}$ 

Take equation $(2)$ and $(3).$

         $5x-7y+2=0 \longrightarrow(2)\times5$

         $9x-5y-4=0 \longrightarrow(3)\times7$

         $25x-35y+10=0$

         $63x-35y-28=0$

         ----------------------------------

            $-38x+38=0$

            $\Rightarrow\boxed{x=1}$

Put the value of $’x’$ in equation $(2)$,we get.

$5(1)-7y+2=0$

$\Rightarrow -7y+7=0$ 

$\Rightarrow \boxed{y=1}$ 

We can say that, these three lines meet at point $M(x,y)=M(1,1).$

$\therefore$ These three lines are coincident.

Correct Answer $;\text{D}$

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