How to do q8 part e:

8 (e) Work out the area enclosed by line A, line B and the y-axis. 

Dear Student,



$$\begin{gathered} Given : Two lines (Line A\hspace{0.17em}: y=5x-4 Line B\hspace{0.17em}: 3x+2y=18) and y axis \\ solution : at y-axis x=0 \\ So, put X=0 in the equation of both the lines to obtain points of intersection of these lines with y-axis. \\ Line A\hspace{0.17em}: y=5\times 0-4\Rightarrow y=-4 \\ Line B : 3\times 0+2y=18\Rightarrow y=9 \\ So, point of intersection of line A with y-axis is (0, -4) \\ and point of intersection of line B with y-axis is (0, 9) \\ Now, calculating point of intersection of line A and line B \\ Putting the value of line A (i.e. y=5x-4) into line B \\ \Rightarrow 3x+2(5x-4)=18 \\ \Rightarrow 3x+10x-8=18 \\ \Rightarrow 13x=26 \\ \Rightarrow x=2 \\ putting x=2 in line A\hspace{0.17em}(we can put x=2 in either of the lines) \\ \Rightarrow y=5\times 2-4 \\ \Rightarrow y=10-4 \\ \Rightarrow y=6 \\ So, point of intersection of both the lines is (2, 6) \\ So, we have all three vertices of a Triangle \\ Assuming (0, 9) be A (0, -4) be B and (2, 6) be C. \\ and we can calculate area of △ABC by the formula \\ Area of triangle=\frac{1}{2}\left[\right(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_1-x_1{y}_3\left)\right] \\ Area of triangle=\frac{1}{2}\left[\right(0\times (-4)-0\times \left(9\right))+(0\times 6-2\times (-4))+(2\times 9-0\times 6\left)\right] \\ Area of triangle=\frac{1}{2}\left[\right(0+8+18\left)\right] \\ Area of triangle=\frac{1}{2}\times 26=13 square units \\ or, As you can see in the graph \\ Area can also be calculated as taking AB as base and dropping a perpendicular from C on AB meeting AB at D \\ So, Base AB=9+4=13 \\ and Altitude CD=2 \\ So, Area of △ABC=\frac{1}{2}\times 13\times 2=13 square units \end{gathered}$$


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