Question

What is the area in square units of the right triangle?
R(-8,7)
S(-5,4)
T(-3, 6)

Answer 1

well, first off if we sit the triangle with the right-angle at the bottom, hmmm and say the RS vertical, then we can say that the triangle has a height of RS and a base of ST, hmmm let's check those lengths then

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ R(\stackrel{x_1}{-8}~,~\stackrel{y_1}{7})\qquad S(\stackrel{x_2}{-5}~,~\stackrel{y_2}{4})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ RS=√((~~-5 - (-8)~~)^2 + (~~4 - 7~~)^2) \implies RS=√((-5 +8)^2 + (4 -7)^2) \\\\\\ RS=√(( 3 )^2 + ( -3 )^2) \implies RS=√( 9 + 9 ) \implies \stackrel{height}{RS=√( 18 )} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ S(\stackrel{x_1}{-5}~,~\stackrel{y_1}{4})\qquad T(\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ ST=√((~~-3 - (-5)~~)^2 + (~~6 - 4~~)^2) \implies ST=√((-3 +5)^2 + (6 -4)^2) \\\\\\ ST=√(( 2 )^2 + ( 2 )^2) \implies ST=√( 4 + 4 ) \implies \stackrel{base}{ST=√( 8 )} \\\\[-0.35em] ~\dotfill`

`A=\cfrac{1}{2}(√(18))(√(8))\implies A=\cfrac{√(144)}{2}\implies A=\cfrac{12}{2}\implies {\Large \begin{array}{llll} A=6 \end{array}}`