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

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

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

Answer 2

$\\ \rm\longmapsto AB=AC$

$\\ \rm\longmapsto \sqrt{(x-5)^2+(1-5)^2}=\sqrt{(x-4)^2+(1+3)^2}$

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

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

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

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

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

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

$\\ \rm\longmapsto x=\dfrac{3}{-2}=-1.5$