Amit solves 12 problems in 2 hours
1 answer:
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Answer:
9.1
Explanation:
Use the distance formula
d=sqrt[(x2-x1)^2 + (y2-y1)^2]
d=sqrt[(3-12)^2 + (4-5)^2]
d=sqrt[(-9)^2 + (-1)^2]
d=sqrt[81 + 1]
d=sqrt[82]
d=9.055≈9.1
Where (x1, y1) is (12,5) and (x2, y2) is (3, 4)
9.1
Explanation:
Use the distance formula
d=sqrt[(x2-x1)^2 + (y2-y1)^2]
d=sqrt[(3-12)^2 + (4-5)^2]
d=sqrt[(-9)^2 + (-1)^2]
d=sqrt[81 + 1]
d=sqrt[82]
d=9.055≈9.1
Where (x1, y1) is (12,5) and (x2, y2) is (3, 4)
Draw an accurate picture of the points in the coordinate axis, as shown in the figure.
D, F have equal x coordinate, so DF is perpendicular to the x-axis.
D, E have equal x coordinate, so DE is perpendicular to the y-axis.
this means DF and DE are perpendicular, so DEF is a right triangle, with m(D)=90°.
Let A be the midpoint of DF. The x coordinate of A is 1 and the y coordinate can be found as follows:
subtract the y coordinate of D from the y coordinate of F, that is 5-1=4.
Divide by 2 and add to the y coordinate of D, that is 4/2+1=2+1=3 is the y coordinate of A.
Let B be the midpoint of DE, follow the previous procedure to find the coordinates of B.
A(1, 3) and B(4, 1).
draw the perpendicular through A and the perpendicular through B, thet meet at point C(4, 3), which is the circumcenter.
Answer: C(4, 3)
D, F have equal x coordinate, so DF is perpendicular to the x-axis.
D, E have equal x coordinate, so DE is perpendicular to the y-axis.
this means DF and DE are perpendicular, so DEF is a right triangle, with m(D)=90°.
Let A be the midpoint of DF. The x coordinate of A is 1 and the y coordinate can be found as follows:
subtract the y coordinate of D from the y coordinate of F, that is 5-1=4.
Divide by 2 and add to the y coordinate of D, that is 4/2+1=2+1=3 is the y coordinate of A.
Let B be the midpoint of DE, follow the previous procedure to find the coordinates of B.
A(1, 3) and B(4, 1).
draw the perpendicular through A and the perpendicular through B, thet meet at point C(4, 3), which is the circumcenter.
Answer: C(4, 3)