The area of the triangle abc in which bd is altitude and having coordinates a(-6,0),b(0,0),c(0,6),d(3,-3) is 18 square units.
Given the coordinates of abcd are a (-6,0) b (0,0) c (o,6) and d (3,-3).
We are required to find the area of the triangle when bd is the altitude of the triangle.
When we draw the triangle we will find that that the base is ac and altitude is given bd.
ac=[tex]\sqrt{(6-0)^{2} +(0+6)^{2} }[/tex]
=[tex]\sqrt{36+36}[/tex]
=[tex]\sqrt{72}[/tex]
=6[tex]\sqrt{2}[/tex]
bd=[tex]\sqrt{(-3-0)^{2} +(3-0)^{2} }[/tex]
=[tex]\sqrt{9+9}[/tex]
=[tex]\sqrt{18}[/tex]
=3[tex]\sqrt{2}[/tex]
Area=(Base*Height)/2
=(6[tex]\sqrt{2}[/tex]*3[tex]\sqrt{2}[/tex])/2
=(18*2)/2
=18 square units.
Hence the area of the triangle abc in which bd is altitude and having coordinates a(-6,0),b(0,0),c(0,6),d(3,-3) is 18 square units.
Learn more about area at https://brainly.com/question/25965491
#SPJ1
