Draw the graphs of the pair of linear equatio | vedclass

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(D) $1$. For the equation $x + 3y = 6$: If $x = 0$,$3y = 6 \implies y = 2$. If $y = 0$,$x = 6$. Points are $(0, 2)$ and $(6, 0)$.$2$. For the equation

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$(6, 0), (0, -4), (0, 2)$

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(D) $1$. For the equation $x + 3y = 6$: If $x = 0$,$3y = 6 \implies y = 2$. If $y = 0$,$x = 6$. Points are $(0, 2)$ and $(6, 0)$.$2$. For the equation $2x - 3y = 12$: If $x = 0$,$-3y = 12 \implies y = -4$. If $y = 0$,$2x = 12 \implies x = 6$. Points are $(0, -4)$ and $(6, 0)$.$3$. Solving the system: Adding the two equations,$(x + 3y) + (2x - 3y) = 6 + 12 \implies 3x = 18 \implies x = 6$. Substituting $x = 6$ into $x + 3y = 6$,we get $6 + 3y = 6 \implies y = 0$. The intersection point is $(6, 0)$.$4$. The triangle is formed by the lines and the $Y$-axis. The vertices are the intersection of the two lines $(6, 0)$,the $Y$-intercept of the first line $(0, 2)$,and the $Y$-intercept of the second line $(0, -4)$.$5$. Thus,the vertices are $(6, 0), (0, 2), (0, -4)$.

Draw the graphs of the pair of linear equations $x+3y=6$ and $2x-3y=12$. Determine the coordinates of the vertices of the triangle formed by these linear equations and the $Y$-axis.

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