Question

A(x, 1), B(5,3) and C(4,-3) are the vertices of triangle ABC. If AB=AC, find the value of x​

Answer 1

`\\ \rm\longmapsto AB=AC`

`\\ \rm\longmapsto √((x-5)^2+(1-5)^2)=√((x-4)^2+(1+3)^2)`

`\\ \rm\longmapsto (x-5)^2+(-2)^2=(x-4)^2+4^2`

`\\ \rm\longmapsto (x-5)^2+4=(x-4)^2+16`

`\\ \rm\longmapsto (x-5)^2-(x-4)^2=16-4=12`

`\\ \rm\longmapsto x^2-10x+25-x^2+8x-16=12`

`\\ \rm\longmapsto -2x+9=12`

`\\ \rm\longmapsto -2x=12-9=3`

`\\ \rm\longmapsto x=(3)/(-2)=-1.5`

Answer 2

• x = - 1.5

Explanation:

Given vertices:

• A(x, 1), B(5,3) and C(4,-3)

Use distance formula to find the side lengths:

• AB =
  `√((5 - x)^2+(3-1)^2) = √(x^2-10x + 25 + 4) = √(x^2-10x + 29)`
• AC =
  `√((4-x)^2+(-3-1)^2) = √(x^2-8x+16+16) =√(x^2-8x + 32)`

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