The vertices of an isosceles triangle are A(3, 6), B(7, 2), and C(4, 3).
What is the equation of the triangle's line of symmetry?
y= x+1
y= x-1
y= x
y = -x+1
y = x - 1
To find the line of symmetry of an isosceles triangle, we can start by finding the midpoints of two sides of the triangle. The line passing through these midpoints will be the line of symmetry.
Let's find the midpoints of sides AB and AC:
Midpoint of AB:
x-coordinate = (x1 + x2) / 2
y-coordinate = (y1 + y2) / 2
Using A(3, 6) and B(7, 2):
Midpoint of AB = ((3 + 7) / 2, (6 + 2) / 2)
= (10 / 2, 8 / 2)
= (5, 4)
Midpoint of AC:
x-coordinate = (x1 + x2) / 2
y-coordinate = (y1 + y2) / 2
Using A(3, 6) and C(4, 3):
Midpoint of AC = ((3 + 4) / 2, (6 + 3) / 2)
= (7 / 2, 9 / 2)
= (3.5, 4.5)
Now, we have the coordinates of the two midpoints: (5, 4) and (3.5, 4.5).
Next, we can find the equation of the line passing through these two midpoints.
First, let's calculate the slope of the line:
slope = (y2 - y1) / (x2 - x1)
= (4.5 - 4) / (3.5 - 5)
= (0.5) / (-1.5)
= -1/3
Now, we have the slope (-1/3) and one point on the line of symmetry (5, 4).
Using the point-slope form of the line: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can substitute the values:
y - 4 = (-1/3)(x - 5)
Simplifying the equation:
y - 4 = (-1/3)x + 5/3
y = (-1/3)x + 5/3 + 4
y = (-1/3)x + 5/3 + 12/3
y = (-1/3)x + 17/3
The equation of the line of symmetry of the isosceles triangle is y = (-1/3)x + 17/3.
So, none of the given options (y = x+1, y = x-1, y = x, y = -x+1) is the equation of the triangle's line of symmetry.