the vertices of triangle ABC are A(1,7)B(9,3)C(3,1)
a.prove that the triangle is a right triangle
b.which angle is the right angle?
c.which side is the hypotnuse?
d.what are the coordinatesof the midpoint of the hypotnuse?
e.what is the equation of the median from the vertex of the right angle to the hypotnuse?
f.what is the equation of the altitude from the vertex of the right angle to the hypotnuse?
g.Is the triangle an isoceles right triangle?justify your answer using parts e and f
a. Compute the distances AB, AC and BC and show that the square of the largest number equals the squares of the two others. Most of the other questions will be obvious after you have done that. The hypotenuse will be the longest side.
The length of AB is sqrt[(9-1)^2 + (3-7)^2] = sqrt(64 + 16) = sqrt 80
The length of BC is sqrt[6^2 + 2^2] = sqrt 40.
The length of AC is sqrt[2^2 + 6^2] = sqrt 40
The triangle is a right triangle and the longest side is AB. The right angle will be opposite that side, at C.
Obviously the two short legs are equal in length. That tells you the answer to g.
a. To prove that triangle ABC is a right triangle, we can use the Pythagorean Theorem. According to the Pythagorean Theorem, if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Let's calculate the lengths of the three sides using the distance formula:
- Side AB:
AB = sqrt((9-1)^2 + (3-7)^2) = sqrt(64 + 16) = sqrt(80)
- Side BC:
BC = sqrt((3-9)^2 + (1-3)^2) = sqrt(36 + 4) = sqrt(40) = 2√10
- Side AC:
AC = sqrt((3-1)^2 + (1-7)^2) = sqrt(4 + 36) = sqrt(40) = 2√10
Now, let's compare the squares of the sides:
(AC)^2 = (2√10)^2 = 4(10) = 40
(AB)^2 + (BC)^2 = (sqrt(80))^2 + (2√10)^2 = 80 + 40 = 120
Since (AC)^2 = (AB)^2 + (BC)^2, the Pythagorean Theorem holds, and therefore triangle ABC is a right triangle.
b. To determine which angle is the right angle, we can examine the slopes of the sides.
The slope of side AB (m_AB) = (3 - 7)/(9 - 1) = -4/8 = -1/2
The slope of side BC (m_BC) = (1 - 3)/(3 - 9) = 2/-6 = -1/3
The slope of side AC (m_AC) = (1 - 7)/(3 - 1) = -6/2 = -3
Since the product of the slopes of sides AB and BC is -1 [(m_AB) * (m_BC) = (-1/2) * (-1/3) = 1/6], the two sides are perpendicular to each other and the right angle is at vertex B.
c. The hypotenuse is the longest side of a right triangle and is opposite the right angle. In this case, the hypotenuse is side AC.
d. To find the coordinates of the midpoint of the hypotenuse, we can calculate the average of the x-coordinates and the average of the y-coordinates of points A and C.
Midpoint_x = (1 + 3)/2 = 2
Midpoint_y = (7 + 1)/2 = 4
Therefore, the coordinates of the midpoint of the hypotenuse are (2, 4).
e. The equation of the median from the vertex of the right angle to the hypotenuse can be found by finding the midpoint of side AC and the slope of side BC.
Using the midpoint formula:
Midpoint_x = (1 + 3)/2 = 2
Midpoint_y = (7 + 1)/2 = 4
The midpoint coordinates of AC are (2, 4).
The slope of BC (m_BC) = -1/3.
Using the point-slope form of linear equation, we have:
y - 4 = (-1/3) * (x - 2)
Simplifying the equation, we get:
3y - 12 = -x + 2
Rearranging the equation, we get:
x + 3y = 14
Therefore, the equation of the median from the vertex of the right angle to the hypotenuse is x + 3y = 14.
f. The equation of the altitude from the vertex of the right angle to the hypotenuse can be found by considering the perpendicular slope to side BC.
The slope of BC (m_BC) = -1/3.
The perpendicular slope will have the negative reciprocal of -1/3, which is 3.
Using the point-slope form of linear equation, we have:
y - 1 = 3(x - 3)
Simplifying the equation, we get:
y - 1 = 3x - 9
Rearranging the equation, we get:
3x - y = 8
Therefore, the equation of the altitude from the vertex of the right angle to the hypotenuse is 3x - y = 8.
g. To determine if the triangle is an isosceles right triangle, we need to examine the lengths of the sides.
From part a, we found that:
AB = sqrt(80)
BC = 2√10
AC = 2√10
Since AB ≠ BC and AB = AC, the triangle is not an isosceles right triangle.
Hope this helps!
To prove that the triangle is a right triangle, we need to show that the measures of the three angles satisfy the Pythagorean theorem. Here are the steps to find the answers to the given questions:
a. To prove that the triangle is a right triangle, we will check if any of the sides satisfy the Pythagorean theorem. The Pythagorean theorem states that for any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Using the given coordinates, we can find the lengths of the sides using the distance formula:
- Length of side AB:
AB = √[(x2 - x1)^2 + (y2 - y1)^2]
AB = √[(9 - 1)^2 + (3 - 7)^2]
AB = √[64 + 16]
AB = √80
AB = 4√5
- Length of side AC:
AC = √[(x2 - x1)^2 + (y2 - y1)^2]
AC = √[(3 - 1)^2 + (1 - 7)^2]
AC = √[4 + 36]
AC = √40
AC = 2√10
- Length of side BC:
BC = √[(x2 - x1)^2 + (y2 - y1)^2]
BC = √[(3 - 9)^2 + (1 - 3)^2]
BC = √[36 + 4]
BC = √40
BC = 2√10
Now, we check if the Pythagorean theorem is satisfied:
AB^2 + AC^2 ≟ BC^2
(4√5)^2 + (2√10)^2 ≟ (2√10)^2
80 + 40 ≟ 40
120 ≟ 40
Since 120 is not equal to 40, the triangle ABC is not a right triangle.
b. Since the triangle is not a right triangle, it does not have a right angle.
c. Since the triangle is not a right triangle, it does not have a hypotenuse.
d. Since there is no hypotenuse, there is no midpoint of the hypotenuse to find.
e. Since there is no right angle, there is no median from the vertex of the right angle to the hypotenuse.
f. Since there is no right angle, there is no altitude from the vertex of the right angle to the hypotenuse.
g. The triangle ABC is not an isosceles right triangle because it neither has a right angle nor equal side lengths. Parts e and f are not applicable in this case.
In conclusion, the triangle ABC is not a right triangle, does not have a right angle, hypotenuse, midpoint of hypotenuse, median from the vertex of the right angle to the hypotenuse, or altitude from the vertex of the right angle to the hypotenuse. It is also not an isosceles right triangle.