DanielGrayson
DanielGrayson DanielGrayson
  • 23-05-2021
  • Mathematics
contestada

The vertices of a triangle are A(2, a), B(-3, 1) and C(-8, -2) right-angled at A. Find the possible values of a. ​

Respuesta :

sreedevi102
sreedevi102 sreedevi102
  • 23-05-2021

Answer:

[tex]BC = \sqrt{(-3+8)^2 +(1+2)^2} = \sqrt{34}\\\\AB =\sqrt{(-3-2)^2+(1-a)^2} = \sqrt{25+(1-a)^2} \\\\AC = \sqrt{(2+8)^2+(a+2)^2} = \sqrt{100+(a+2)^2} \\\\AB^2 + AC^2 = BC^2\\\\ (\sqrt{25+(1-a)^2})^2+ (\sqrt{100+(a+2)^2})^2= (\sqrt{34})^2 \\\\25+(1-a)^2+100+(a+2)^2=34\\\\125+1+a^2-2a+a^2+4+2a=34\\\\130+2a^2 =34\\\\2a^2+96=0\\\\a^2+48=0\\\\a = \sqrt{-48}[/tex]

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