Lines with slopes -1 and -2 are drawn through the first quadrant point (a,b) forming one triangle with a side on the x-axis and the other with one side on the y-axis. What is the total area of the two shaded triangles? (Write you answers in terms of a and b)

Thanks

SydSu22 May 2, 2019

#1**+3 **

**Lines with slopes -1 and -2 are drawn through the first quadrant point (a,b) forming one triangle **

**with a side on the x-axis and the other with one side on the y-axis.**

**What is the total area of the two shaded triangles? (Write you answers in terms of a and b)**

\(\mathbf{b_1=\ ?}\)

\(\begin{array}{|rcll|} \hline y&=&-x+b_1 \quad | \quad P(a,b) \text{ on line} \\ b&=&-a+b_1 \\ \mathbf{b_1} &=& \mathbf{a+b} \\\\ \mathbf{y} &=& \mathbf{-x+(a+b)} \\ \hline \end{array}\)

\(\mathbf{b_2=\ ?}\)

\(\begin{array}{|rcll|} \hline y&=&-2x+b_2 \quad | \quad P(a,b) \text{ on line} \\ b&=&-2a+b_2 \\ \mathbf{b_2} &=& \mathbf{2a+b} \\\\ \mathbf{y} &=& \mathbf{-2x+(2a+b)} \\ \hline \end{array}\)

\(\mathbf{y=0}\\ \mathbf{c_x=\ ?} \)

\(\begin{array}{|rcll|} \hline 0 &=& -x_1+(a+b) \\ \mathbf{x_1} &=& \mathbf{a+b} \\\\ 0 &=& -2x+(2a+b) \\ 2x &=& 2a+b \\ \mathbf{x_2} &=& \mathbf{a+\dfrac{b}{2}} \\\\ c_x &=& x_1-x_2 \\ &=& a+b - \left(a+\dfrac{b}{2} \right) \\ \mathbf{c_x} &=& \mathbf{\dfrac{b}{2}} \\ \hline \end{array}\)

\(\mathbf{x=0}\\ \mathbf{c_y=\ ?}\)

\(\begin{array}{|rcll|} \hline y_1 &=& -0 +(a+b) \\ \mathbf{y_1} &=& \mathbf{a+b} \\\\ y_2 &=& -2\cdot 0+(2a+b) \\ &=& 2a+b \\ \mathbf{y_2} &=& \mathbf{2a+b} \\\\ c_y &=& y_2-y_1 \\ &=& 2a+b-(a+b) \\ \mathbf{c_y} &=& \mathbf{a} \\ \hline \end{array}\)

\(\mathbf{A_x=\ ?} \)

\(\begin{array}{|rcll|} \hline A_x &=& \dfrac{c_xh_x}{2} \quad | \quad h_x = b,\qquad c_x = \dfrac{b}{2} \\\\ &=& \dfrac{\dfrac{b}{2}\cdot b}{2} \\\\ \mathbf{A_x} &=& \mathbf{\dfrac{b^2}{4}} \\ \hline \end{array}\)

\(\mathbf{A_y=\ ?}\)

\(\begin{array}{|rcll|} \hline A_y &=& \dfrac{c_yh_y}{2} \quad | \quad h_y = a,\qquad c_y = a \\\\ &=& \dfrac{a\cdot a}{2} \\\\ \mathbf{A_y} &=& \mathbf{\dfrac{a^2}{2}} \\ \hline \end{array}\)

\(\mathbf{A=\ ?}\)

\(\begin{array}{|rcll|} \hline A&=& A_x + A_y \\ &=& \dfrac{b^2}{4} + \dfrac{a^2}{2} \\ \mathbf{A} &=& \mathbf{\dfrac{1}{2}\cdot \left( a^2 + \dfrac{b^2}{2} \right) } \\ \hline \end{array} \)

heureka May 2, 2019