If the sides of a triangle are given by y = m | vedclass

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(A) The lines are $L_1: y = mx + a$,$L_2: y = nx + b$,and $L_3: x = 0$ (the $y$-axis).The intersection of $L_1$ and $L_3$ is $A = (0, a)$.The intersec

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$\frac{(a - b)^2}{2|m - n|}$

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(A) The lines are $L_1: y = mx + a$,$L_2: y = nx + b$,and $L_3: x = 0$ (the $y$-axis).The intersection of $L_1$ and $L_3$ is $A = (0, a)$.The intersection of $L_2$ and $L_3$ is $B = (0, b)$.The base of the triangle along the $y$-axis is the distance between $A$ and $B$,which is $|a - b|$.To find the third vertex $C$,we solve $mx + a = nx + b$:$(m - n)x = b - a \Rightarrow x = \frac{b - a}{m - n} = \frac{a - b}{n - m}$.The height of the triangle is the absolute value of the $x$-coordinate of $C$,which is $h = |\frac{a - b}{n - m}|$.The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |a - b| 	imes |\frac{a - b}{n - m}| = \frac{(a - b)^2}{2|m - n|}$.

If the sides of a triangle are given by $y = mx + a$,$y = nx + b$,and $x = 0$,then its area is:

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