Show that triangle ABC is a right triangle.
point A: (-7,-2)
point B: (-3,-3)
point C: (-4,-7)
I really don't know how to do this, please help.
you know that the sides of a right triangle obey
c^2 = a^2 + b^2
So, find the lengths of the sides and check them.
AB^2 = (-3-(-7))^2 + (-3-(-2))^2 = 4^2+1^2 = 17
Now find the other two sides (squared) and check that the largest is the sum of the other two.
To show that triangle ABC is a right triangle, we need to check if any of its angles measures 90 degrees. One way to do this is by examining the slopes of the lines connecting the three points.
Step 1: Calculate the slopes of lines AB, BC, and AC.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
slope = (y2 - y1) / (x2 - x1)
For line AB:
point A: (-7, -2)
point B: (-3, -3)
slope(ab) = (-3 - (-2)) / (-3 - (-7))
= (-3 + 2) / (-3 + 7)
= -1 / 4
For line BC:
point B: (-3, -3)
point C: (-4, -7)
slope(bc) = (-7 - (-3)) / (-4 - (-3))
= (-7 + 3) / (-4 + 3)
= -4 / -1
= 4
For line AC:
point A: (-7, -2)
point C: (-4, -7)
slope(ac) = (-7 - (-2)) / (-4 - (-7))
= (-7 + 2) / (-4 + 7)
= -5 / 3
Step 2: Check if any pair of slopes are negative reciprocals of each other.
Two lines are perpendicular (form a right angle) if and only if the product of their slopes is -1 (negative reciprocal).
Let's check for line AB and BC:
slope(ab) * slope(bc) = (-1/4) * (4) = -1
We see that the product of the slopes of AB and BC is -1, which means that these lines are perpendicular, and therefore, angle ABC is a right angle.
Thus, we can conclude that triangle ABC is a right triangle.
To determine if triangle ABC is a right triangle, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
Let's find the lengths of the three sides of triangle ABC:
Side AB:
Distance formula: √((x2 - x1)^2 + (y2 - y1)^2)
AB = √((-3 - (-7))^2 + (-3 - (-2))^2)
= √(4^2 + (-1)^2)
= √(16 + 1)
= √17
Side BC:
BC = √((-4 - (-3))^2 + (-7 - (-3))^2)
= √(1^2 + (-4)^2)
= √(1 + 16)
= √17
Side AC:
AC = √((-4 - (-7))^2 + (-7 - (-2))^2)
= √(3^2 + (-5)^2)
= √(9 + 25)
= √34
Using the Pythagorean theorem, we can check if the triangle is a right triangle:
AC^2 = AB^2 + BC^2
(√34)^2 = (√17)^2 + (√17)^2
34 = 17 + 17
34 = 34
Since AC^2 = AB^2 + BC^2, we can conclude that triangle ABC is a right triangle with vertices at points A, B, and C.