Answer:
D) y = x + 3
Step-by-step explanation:
The line of symmetry of an isosceles triangle is a line that divides the triangle into two congruent halves by passing through the midpoint of the base and being perpendicular to it.
To find the line of symmetry of an isosceles triangle, first identity which side is its base (the side that is not the same length as the other two sides).
Calculate the lengths of the three sides AB, BC and AC using the distance formula.
Distance AB:
[tex]AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}\\\\AB = \sqrt{(-4 - (-8))^2 + (-5 - (-1))^2}\\\\AB = \sqrt{(-4 + 8)^2 + (-5 + 1)^2}\\\\AB = \sqrt{4^2 + (-4)^2} \\\\AB = \sqrt{16 + 16}\\\\AB = \sqrt{32}[/tex]
Distance BC:
[tex]BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2}\\\\BC = \sqrt{(-5 - (-4))^2 + (-2 - (-5))^2} \\\\BC = \sqrt{(-5 + 4)^2 + (-2 + 5)^2}\\\\BC = \sqrt{(-1)^2 + 3^2}\\\\BC = \sqrt{1 + 9}\\\\BC = \sqrt{10}[/tex]
Distance AC:
[tex]AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \\\\AC = \sqrt{(-5 - (-8))^2 + (-2 - (-1))^2} \\\\AC = \sqrt{(-5 + 8)^2 + (-2 + 1)^2} \\\\AC = \sqrt{3^2 + (-1)^2} \\\\AC = \sqrt{9 + 1}\\\\ AC = \sqrt{10[/tex]
As sides BC and AC have the same length, this means that AB is the base of the triangle.
Now, find the midpoint of base AB:
[tex]\textsf{Midpoint $AB$} =\left(\dfrac{x_B+x_A}{2},\dfrac{y_B+y_A}{2}\right)\\\\\\\textsf{Midpoint $AB$} =\left(\dfrac{-4+(-8)}{2},\dfrac{-5+(-1)}{2}\right)\\\\\\\textsf{Midpoint $AB$} =\left(\dfrac{-12}{2},\dfrac{-6}{2}\right)\\\\\\\textsf{Midpoint $AB$} =\left(-6,-3\right)[/tex]
Since the line of symmetry is perpendicular to base AB, its slope will be the negative reciprocal of the slope of AB.
Find the slope of base AB:
[tex]\textsf{Slope of $AB$}=\dfrac{y_B-y_A}{x_B-x_A}\\\\\\\textsf{Slope of $AB$}=\dfrac{-5-(-1)}{-4-(-8)}\\\\\\\textsf{Slope of $AB$}=\dfrac{-5+1}{-4+8}\\\\\\\textsf{Slope of $AB$}=\dfrac{-4}{4}\\\\\\\textsf{Slope of $AB$}=-1[/tex]
The slope of the triangle's line of symmetry is the negative reciprocal of -1, which is m = 1.
Now, we have the slope of the line of symmetry and a point on it (the midpoint of AB), we can find the equation of the line using the point-slope formula:
[tex]y-y_1=m(x-x_1)\\\\y-(-3)=1(x-(-6))\\\\y+3=x+6\\\\y=x+3[/tex]
Therefore the equation of the triangle's line of symmetry is:
[tex]\LARGE\boxed{\boxed{y=x+3}}[/tex]
