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Draw the graph of 2x-y-2=0,4x+3y-24=0 and y-4=0 find the area of triangle obtained

Answer» Given equations,\xa0{tex}2 x - y - 2 = 0{/tex}{tex}4x + 3y - 24 = 0{/tex}{tex}y + 4 = 0{/tex}We have, {tex}2x - y - 2 = 0{/tex} or\xa0{tex}x = {{y+2}\\over2} {/tex}When y = 0, we have\xa0{tex}x = {{0+2}\\over2} = 1{/tex}When x = 0, we have y = -2.Thus, we obtain the following table giving coordinates of two point on the line represented by the equation\xa0{tex}2x - y - 2 = 0{/tex}\xa0and its graph is shown below.\tx10y0-2\tNow we have,\xa0{tex}\\begin{array}{l}4x+3y-24=0\\Rightarrow y=\\frac{24-4x}3\\\\\\end{array}{/tex}When y = 0, we have x = 6When x = 0, we have y = 8Thus, we obtain the following table giving coordinates of two points on the line represented by the equation {tex}4x + 3y - 24 = 0{/tex}\xa0and its graph\xa0is shown below.\tx60y08\tAlso we have {tex}y + 4 = 0{/tex}Clearly, y = - 4 for every value of x.So, let E(2, -4) and F(0, -4) be two points on the line represented by y + 4 = 0. Plotting these points on the same graph and drawing a line passing through them, we obtain the graph of the line represented by the equation y + 4 = 0 as shown in Figure.From Fig.\xa0we have\xa0{tex}\\begin{array}{l}\\bigtriangleup PQR\\\\\\end{array}{/tex}\xa0having vertices P(3,4), Q(-1,-4) and R(9, -4).\xa0Also,\xa0PM = 8 and QR = 10.{tex}\\therefore \\quad \\text { Area of } \\triangle P Q R = \\frac { 1 } { 2 } ( \\text { Base } \\times \\text { Height } ){/tex}{tex}\\Rightarrow \\quad \\text { Area of } \\triangle P Q R = \\frac { 1 } { 2 } ( Q R \\times P M ) = \\frac { 1 } { 2 } ( 10 \\times 8 )sq.\\ units{/tex}{tex}\\Rightarrow \\quad \\text { Area of } \\triangle P Q R = 40 \\mathrm { sq } .units.{/tex}


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