Respuesta :

mhanifa

Answer:

  • x = - 1.5

Step-by-step explanation:

Given vertices:

  • A(x, 1), B(5,3) and C(4,-3)

Use distance formula to find the side lengths:

  • AB = [tex]\sqrt{(5 - x)^2+(3-1)^2} = \sqrt{x^2-10x + 25 + 4} = \sqrt{x^2-10x + 29}[/tex]
  • AC = [tex]\sqrt{(4-x)^2+(-3-1)^2} = \sqrt{x^2-8x+16+16} =\sqrt{x^2-8x + 32}[/tex]

Since AB = AC, compare the squares and solve for x:

  • x² - 10x + 29 = x² - 8x + 32
  • 10x - 8x = 29 - 32
  • 2x = - 3
  • x = - 1.5

[tex]\\ \rm\longmapsto AB=AC[/tex]

[tex]\\ \rm\longmapsto \sqrt{(x-5)^2+(1-5)^2}=\sqrt{(x-4)^2+(1+3)^2}[/tex]

[tex]\\ \rm\longmapsto (x-5)^2+(-2)^2=(x-4)^2+4^2[/tex]

[tex]\\ \rm\longmapsto (x-5)^2+4=(x-4)^2+16[/tex]

[tex]\\ \rm\longmapsto (x-5)^2-(x-4)^2=16-4=12[/tex]

[tex]\\ \rm\longmapsto x^2-10x+25-x^2+8x-16=12[/tex]

[tex]\\ \rm\longmapsto -2x+9=12[/tex]

[tex]\\ \rm\longmapsto -2x=12-9=3[/tex]

[tex]\\ \rm\longmapsto x=\dfrac{3}{-2}=-1.5[/tex]

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