Answer:
a^2 +b^2=c^2
Given vertices of the triangle are A(4,4),B(3,5) and C(−1,−1)
We know that slope of line passing through the points (x 1,y 1) and (x 2,y 2
) is given by m= x 2−x 1
y
2
−y
1
,x
2
=x
1
Slope of AB i.e.m
1
=
3−4
5−4
=−1
Slope of BC i.e.m
2
=
−1−3
−1−5
=
−4
−6
=
2
3
Slope of CA i.e. m
3
=
4+1
4+1
=
5
5
=1
Clearly, m
1
m
3
=−1
⇒ line segments AB and CA are perpendicular to each other i.e; the given triangle is right angled at A(4,4).
Thus the points (4,4),(3,5) and (1,1) are the vertices of a right angled triangle.
Step-by-step explanation:
It would be d if the markup was that high
Answer:
a) 20 units
b) 2 √10 units
c) 2 √17 units
Step-by-step explanation:
The distance formula is;
D = √(y2-y1)^2 + (x2-x1)^2
a) D = √(-7-9)^2 + (-7-5)^2
D = √256 + 144
D = √400
D = 20
b) D = √(10-8)^2 + (9-2)^2
D = √(2)^2 + 6^2
D = √4 + 36)
D = √40
D = 2 √10 units
c) D = √(1 + 7)^2 + (-8+10)^2
D = √(64 + 4)
D = √68
D = 2 √17 units
1) 0.78
2) 34.28
This is because the number in the 3rd decimal place for both numbers is 5 or above so it is rounded up rather than down.
