The perimeter of the triangle is the sum of its side lengths
The possible x-coordinates of point L are -7 and 1
The vertices of the triangle are given as:
[tex]\mathbf{J = (-3,-9)}[/tex]
[tex]\mathbf{K = (-3,6)}[/tex]
[tex]\mathbf{L = (x,-1.5)}[/tex]
Start by calculating distances JK, JL and KL using the following distance formula
[tex]\mathbf{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}[/tex]
So, we have:
[tex]\mathbf{JK = \sqrt{(-3 - -3)^2 + (-9 - 6)^2}}[/tex]
[tex]\mathbf{JK = \sqrt{0 + 225}}[/tex]
[tex]\mathbf{JK = \sqrt{225}}[/tex]
[tex]\mathbf{JK = 15}[/tex]
[tex]\mathbf{JL = \sqrt{(-3 - x)^2 + (-9 - -1.5)^2}}[/tex]
[tex]\mathbf{JL = \sqrt{(-3 - x)^2 + 56.25}}[/tex]
[tex]\mathbf{KL = \sqrt{(-3 - x)^2 + (6 -- 1.5)^2}}[/tex]
[tex]\mathbf{KL = \sqrt{(-3 - x)^2 + 56.25}}[/tex]
The triangle is said to be an isosceles triangle.
By comparison:
[tex]\mathbf{JL = KL}[/tex]
The perimeter is represented as:
[tex]\mathbf{P = JK + JL + KL}[/tex]
Substitute 32 for P
[tex]\mathbf{JK + JL + KL = 32}[/tex]
Substitute 15 for JK
[tex]\mathbf{15 + JL + KL = 32}[/tex]
Subtract 15 from both sides
[tex]\mathbf{JL + KL = 17}[/tex]
Substitute JL for KL
[tex]\mathbf{JL + JL = 17}[/tex]
[tex]\mathbf{2JL = 17}[/tex]
Divide both sides by 2
[tex]\mathbf{JL = 8.5}[/tex]
Recall that:
[tex]\mathbf{JL = \sqrt{(-3 - x)^2 + 56.25}}[/tex]
Substitute 8.5 for JL
[tex]\mathbf{8.5 = \sqrt{(-3 - x)^2 + 56.25}}[/tex]
Square both sides
[tex]\mathbf{72.25 = (-3 - x)^2 + 56.25}[/tex]
Subtract 56.25 from both sides
[tex]\mathbf{16 = (-3 - x)^2 }[/tex]
Take square roots of both sides
[tex]\mathbf{\pm 4 = -3 - x}[/tex]
Add 3 to both sides
[tex]\mathbf{ 3\pm 4 = - x}[/tex]
Rewrite as:
[tex]\mathbf{ -x = 3\pm 4 }[/tex]
Multiply both sides by -1
[tex]\mathbf{ x = -3\pm 4 }[/tex]
Split
[tex]\mathbf{ x = (-3- 4),(-3+4) }[/tex]
Solve
[tex]\mathbf{ x = (-7),(1) }[/tex]
Hence, the possible x-coordinates of point L are -7 and 1
Read more about coordinates at:
https://brainly.com/question/10364988
