elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41

Hey all, it seems to be not very hard problem,but I have trouble drawing it that's why I couldn't solve it at the first place. Thanks for help.

Correct answer isE

No need to draw in this case. If you know the forumla, you can calculatethe distance between two points

Distance between (x1, y1) and (x2, y2) is SQRT( (y2-y1)^2 + (x2-x1)^2))

Therefore distance between (3,5) and (7,10) is SQRT((5^2) + (4^2) ) = SQRT(41). Choose E.

vk_vinayak wrote:

elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41

Hey all, it seems to be not very hard problem,but I have trouble drawing it that's why I couldn't solve it at the first place. Thanks for help.

Correct answer isE

No need to draw in this case. If you know the forumla, you can calculatethe distance between two points

Distance between (x1, y1) and (x2, y2) is SQRT( (y2-y1)^2 + (x2-x1)^2))

Therefore distance between (3,5) and (7,10) is SQRT((5^2) + (4^2) ) = SQRT(41). Choose E.

Thanks a lot for your reply. I will memorize this formula for my future use.

Graphical Solution :

Let (3,5) be point A and (7,10) be point B. When plotted on a graph, the line AB looks like the one in the attachement below.

From the graph, AC = 4 Units and BC = 5 units. Since triangle ACB is a right angled triangle,
AB^2 = AC^2 + CB^2 (Pythagorean theorem)
AB^2 = 4^2 + 5^2 = 16 + 25 = 41
Distance between the points = AB = √41

√(7-3)^2+(10-5)^2=√16+25=√41

ans (E)

neelgandham wrote:Graphical Solution :

Let (3,5) be point A and (7,10) be point B. When plotted on a graph, the line AB looks like the one in the attachement below.

From the graph, AC = 4 Units and BC = 5 units. Since triangle ACB is a right angled triangle,
AB^2 = AC^2 + CB^2 (Pythagorean theorem)
AB^2 = 4^2 + 5^2 = 16 + 25 = 41
Distance between the points = AB = √41

Thanks for this picture and very easy solution!

elenaelena wrote:In the coordinate plane, what is the distance between points (3,5) and (7,10)?

A) 3
B) 3√3
C) 2√5
D) √37
E) √41

We can use the distance formula:

d = √[(y2 - y1)^2 + (x2 - x1)^2]

d = √[(10 - 5)^2 + (7 - 3)^2]

d = √(5^2 + 4^2)

d = √(25 + 16) = √41

Answer: E