Factor
(x-8)(x+3)
Check:
-8+3=-5
-8*3=-24
Use ZPP
x-8=0
x=8
x+3=0
x=-3
Final answer: x=-3, x=8
We will have to use the distance formula in order to determine the lengths of each side of the triangle.
Distance formula:
![\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28x_%7B2%7D%20-%20x_%7B1%7D%29%5E%7B2%7D%20%2B%20%28y_%7B2%7D%20-%20y_%7B1%7D%29%5E%7B2%7D%20%7D%20)
Let's calculate AB first:
A (1, 7) and B (-2, 2)
A: x1 = 1 and y1 = 7
B: x2 = -2 and y2 = 2
so
![\sqrt{(-2 - 1)^{2} + (2 - 7)^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-2%20-%201%29%5E%7B2%7D%20%2B%20%282%20-%207%29%5E%7B2%7D%20%7D%20)
![\sqrt{(-3)^{2} + (-5)^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-3%29%5E%7B2%7D%20%2B%20%28-5%29%5E%7B2%7D%20%7D%20)
![\sqrt{9 + 25 }](https://tex.z-dn.net/?f=%20%5Csqrt%7B9%20%2B%2025%20%7D%20)
AB =
![\sqrt{34}](https://tex.z-dn.net/?f=%20%5Csqrt%7B34%7D%20)
or (rounded to the nearest tenth) ≈ 5.8
Now let's do BC:
B: x1 = -2 and y1 = 2
C: x2 = 4 and y2 = 2
So
![\sqrt{(4 - -2)^{2} + (2 - 2)^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%284%20-%20-2%29%5E%7B2%7D%20%2B%20%282%20-%202%29%5E%7B2%7D%20%7D)
![\sqrt{(6)^{2} + (0)^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%286%29%5E%7B2%7D%20%2B%20%280%29%5E%7B2%7D%20%7D)
BC =
![\sqrt{36 }](https://tex.z-dn.net/?f=%5Csqrt%7B36%20%7D)
or 6
Now let's do CA
C: x1 = 4 and y1 = 2
A: x2 = 1 and y2 = 7
So
![\sqrt{(1 - 4)^{2} + (7 - 2)^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%281%20-%204%29%5E%7B2%7D%20%2B%20%287%20-%202%29%5E%7B2%7D%20%7D%20)
![\sqrt{(-3)^{2} + (5)^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-3%29%5E%7B2%7D%20%2B%20%285%29%5E%7B2%7D%20%7D%20)
![\sqrt{9 + 25}](https://tex.z-dn.net/?f=%20%5Csqrt%7B9%20%2B%2025%7D)
CA =
![\sqrt{34}](https://tex.z-dn.net/?f=%5Csqrt%7B34%7D)
or (rounded to the nearest tenth) ≈ 5.8
So let's recap:
AB ≈ 5.8
BC = 6
CA ≈ 5.8
So AB and AC are the same length while BC is .2 units longer which means this is an
isosceles triangle.
Answer:
5.099
Step by step explanations:
Input Data :
Point 1(xA,yA)(xA,yA) = (4, 3)
Point 2(xB,yB)(xB,yB) = (3, -2)
Objective :
Find the distance between two given points on a line?
Formula :
Distance between two points = √(xB−xA)2+(yB−yA)2(xB-xA)2+(yB-yA)2
Solution :
Distance between two points = √(3−4)2+(−2−3)2(3-4)2+(-2-3)2
= √(−1)2+(−5)2(-1)2+(-5)2
= √1+251+25
= √2626 = 5.099
Distance between points (4, 3) and (3, -2) is 5.099