A triangle has verticies of (1,3) (4,5) and (6,2). Is this a right angle triangle? Solve using algebra.
find the distances between the points (all three, two at a time)
if the distances satisfy .... a^2 + b^2 = c^2
... then it is a right triangle
or:
if right angled, then two lines must be perpendicular
slope between 1st point and 2nd = 2/3
slope between 1st and third = -1/5
slope between 2nd and 3rd = -3/2
notice that 2/3 and -3/2 are negative reciprocals of each other, thus , yes we have a right-angled triangle
To determine if a triangle is a right triangle using algebra, we need to check if the slopes of any two sides are negative reciprocals of each other.
Let's label the three vertices of the triangle as A(1, 3), B(4, 5), and C(6, 2).
Now we can find the slopes of the two sides and see if they are negative reciprocals.
The slope of side AB is:
mAB = (y2 - y1) / (x2 - x1)
= (5 - 3) / (4 - 1)
= 2/3
The slope of side AC is:
mAC = (y3 - y1) / (x3 - x1)
= (2 - 3) / (6 - 1)
= -1/5
The slope of side BC can be found similarly:
mBC = (y3 - y2) / (x3 - x2)
= (2 - 5) / (6 - 4)
= -3/2
Now, let's check the product of the slopes:
mAB * mAC = (2/3) * (-1/5) = -2/15
Since the product of the slopes of AB and AC is not -1, the triangle ABC is not a right triangle.
Therefore, using algebra, we have determined that the triangle with vertices (1, 3), (4, 5), and (6, 2) is not a right angle triangle.