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| 1. |
3x+y=11 and x-y=1 show it graphically and find the area of triangle formed |
| Answer» On a graph paper, draw a horizontal line XOX\' and a vertical line YOY\' as the x-axis and they-axis respectively.{tex}3x + y - 11\xa0= 0{/tex}{tex}\\Rightarrow{/tex}{tex}y = (11 -3x){/tex}. ...(i)Putting {tex}x = 2{/tex} in (i), we get {tex}y = 5{/tex}.Putting {tex}x = 3{/tex} in (i), we get {tex}y = 2{/tex}.Putting {tex}x = 5{/tex} in (i), we get {tex}y = -4{/tex}.\tx235y52-4\tOn the graph paper, plot the points {tex}A (2, 5), B(3, 2)\\ and\\ C(5, -4).{/tex}Join AB and BC to get the graph line ABC.Thus, the line ABC is the graph of the equation {tex}3x + y - 11\xa0= 0{/tex}.{tex}x - y - 1\xa0= 0{/tex}{tex}\\Rightarrow{/tex}{tex}y = (x - 1){/tex} . ...(ii)Putting {tex}x = -3{/tex} in (ii), we get {tex}y = -4{/tex}.Putting {tex}x = 0{/tex} in (ii), we get {tex}y = -1{/tex}.Putting {tex}x = 3{/tex} in (ii), we get {tex}y = 2{/tex}.\tx-303y-4-12\tOn the same graph paper as above, plot the points {tex}P(-3, -4)\\ and\\ Q(0, -1){/tex}. The third point {tex}B(3,2){/tex} is already plotted.Join PQ and QB to get the graph line PQB.Thus, line PQB is the graph of the equation {tex}x - y - 1 = 0{/tex}.The two graph lines intersect at the point {tex}B(3,2){/tex}. {tex}x = 3, y = 2{/tex} is the solution of the given system of equations.The region bounded by these lines and the y-axis has been shaded.On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the points Q(0, -1) and R(0,11).Area of triangle =\xa0{tex}\\frac 12 \\times 12 \\times 3 = 18{/tex}\xa0sq units. | |