A triangle has vertices at A(3,4),B(-3,2),C(-1,-4). Is the triangle a right triangle? Explain.
the slopes of the line segments are:
AB: 1/3
AC: 2
BC: -3
So, AB and BC are perpendicular
You could also calculate the lengths of the sides, and see whether
a^2+b^2 = c^2
for one of the pairs of sides.
Too much work
To determine whether a triangle is a right triangle, we need to check if any of the three angles of the triangle is a right angle (an angle of 90 degrees).
To do this, we can use the slope formula to find the slopes of the three sides of the triangle, and then use the perpendicular property of slopes to check if any two sides are perpendicular (have slopes that are negative reciprocals of each other). If we find a pair of perpendicular sides, then the triangle is a right triangle.
Let's calculate the slopes of the sides AB, BC, and CA using the following formula:
Slope = (y2 - y1) / (x2 - x1)
For side AB:
x1 = 3, y1 = 4
x2 = -3, y2 = 2
Slope_AB = (2 - 4) / (-3 - 3) = -2 / -6 = 1/3
For side BC:
x1 = -3, y1 = 2
x2 = -1, y2 = -4
Slope_BC = (-4 - 2) / (-1 - (-3)) = -6 / 2 = -3
For side CA:
x1 = -1, y1 = -4
x2 = 3, y2 = 4
Slope_CA = (4 - (-4)) / (3 - (-1)) = 8 / 4 = 2
Now, let's check if any two slopes are negative reciprocals of each other.
Slope_AB and Slope_BC:
The slopes 1/3 and -3 are not negative reciprocals of each other, so side AB and side BC are not perpendicular.
Slope_BC and Slope_CA:
The slopes -3 and 2 are not negative reciprocals of each other, so side BC and side CA are not perpendicular.
Slope_CA and Slope_AB:
The slopes 2 and 1/3 are not negative reciprocals of each other, so side CA and side AB are not perpendicular.
Since none of the sides are perpendicular, the triangle formed by the points A(3,4), B(-3,2), and C(-1,-4) is not a right triangle.