$Given,thepoint\left(4,3\right)liesonthegraphoftheequation3x-ay=6.Therefore,\phantom{\rule{0ex}{0ex}}thepoint\left(4,3\right)satisfiestheequation.\phantom{\rule{0ex}{0ex}}3\times 4-a\times 3=6\phantom{\rule{0ex}{0ex}}\Rightarrow 12-3a=6\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=6\phantom{\rule{0ex}{0ex}}\Rightarrow a=2\phantom{\rule{0ex}{0ex}}Thus,theequationwillbe3x-2y=6.\phantom{\rule{0ex}{0ex}}Now,putthepoint\left(-2,-6\right)inaboveequation,wehave,\phantom{\rule{0ex}{0ex}}3\times \left(-2\right)-2\times \left(-6\right)=6\phantom{\rule{0ex}{0ex}}\Rightarrow -6+12=6\phantom{\rule{0ex}{0ex}}\Rightarrow 6=6\phantom{\rule{0ex}{0ex}}Thus,thepoint\left(-2,-6\right)liesonthesamegraph.\phantom{\rule{0ex}{0ex}}Whengraphcutsx-axis,puty=0\phantom{\rule{0ex}{0ex}}3x-2\times 0=6\Rightarrow x=2\phantom{\rule{0ex}{0ex}}Thus,thecoordinateis\left(2,0\right).\phantom{\rule{0ex}{0ex}}And,\phantom{\rule{0ex}{0ex}}Whengraphcutsy-axis,putx=0\phantom{\rule{0ex}{0ex}}3\times 0-2\times y=6\Rightarrow y=-3\phantom{\rule{0ex}{0ex}}Thus,thecoordinateis\left(0,-3\right).$

Now, to draw the graph join the point (2,0) and (0,-3), we have,

Now, to draw the graph join the point (2,0) and (0,-3), we have,

Posted by Priyanka Kediaon 24/2/14

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