[Guest]

A point will be chosen at random from triangle ABC where the vertices are A = (0,0), B = (0,2), C = (0,10). So this is a right triangle with sides of length 2 and 10. What is the probability that the randomly chosen point will be closer to the origin (0,0) than it will to the point (5,-1)?

[Guest]

100% / 0%

I found it by finding the midpoint of 0,0 to 5,-1 and finding the perpendicular line which represents all the points where the distance is equal. the triangle lies entirely in the 0,0 side.

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[Guest]

dont u mean 0,0 10,0 and 0,2? or am i missing something

[Glycereine]

dont u mean 0,0 10,0 and 0,2? or am i missing something

Unless final is correct, the 2nd post is right, which doesn't make a whole lot of sense to me.

[Guest]

assuming im right about the correction

i dont remember geomety much but im pretty sure i did this right i got two similar triangles the second having the right angle at vertice 1.4,3 and each side of about 71% the length of the biggen. anyway i got an answer of 51% but assuming i rounded or did something wrong the answer is prolly 50%

[Guest]

Goodness. The third vertex is indeed supposed to be (10,0). Sorry for the confusion guys.

[Guest]

Taking ratios of areas where a point chosen will be closer to the origin, the prob of a random chosen point being closer to the origin than to point (5,-1) is 47.55%

[Guest]

i think i did it backwards so

tid bit under 49 not quite jumpin on deegee's train tho

[Guest]

47.35% chance that the point is closer the (0,0) than (5,-1).

[Guest]

yah u guys are right i un-rounded the math and got a 72.56297% difference in size which squared gives you 47.34615385% difference in area

didnt think i rounded that much but i guess .01 difference squared is a bit too much

[Guest]

didnt think i rounded that much but i guess .01 difference squared is a bit too much

yah u guys are right i un-rounded the math and got a 72.56297% difference in size which squared gives you 47.34615385% difference in area

I get exactly 1231/26 %

[Guest]

Good job guys!

The equation of the perpendicular bisector between the two points (0,0) and (5,-1) is the line y = 5x - 13. It is important to realize that each point on this line is equidistant from our two points (0,0) and (5,-1). All points above this line are closer to (0,0), and all points below this line are closer to (5,-1).

Thus, we need to find the part of our triangle which is higher than the line y = 5x - 13. The ratio of the part higher than the line to the total area of the traingle (which is 10) is the final answer.

The line intersects the triangle at the points (2.6,0) and (75/26,37/26). The lower part is a triangle with length 10-2.6 = 7.4 and height 37/26. Thus, the area of this lower triangle is 0.5(7.4)(37/26) = 1369/260 and the area of the upper part is 10 - 1369/260 which equals 1231/260.

The area of the upper part to the whole triangle is (1231/260)/10 = 1231/2600 or about 47.35%.

[Guest]

Good job guys!

The equation of the perpendicular bisector between the two points (0,0) and (5,-1) is the line y = 5x - 13. It is important to realize that each point on this line is equidistant from our two points (0,0) and (5,-1). All points above this line are closer to (0,0), and all points below this line are closer to (5,-1).

Thus, we need to find the part of our triangle which is higher than the line y = 5x - 13. The ratio of the part higher than the line to the total area of the traingle (which is 10) is the final answer.

The line intersects the triangle at the points (2.6,0) and (75/26,37/26). The lower part is a triangle with length 10-2.6 = 7.4 and height 37/26. Thus, the area of this lower triangle is 0.5(7.4)(37/26) = 1369/260 and the area of the upper part is 10 - 1369/260 which equals 1231/260.

The area of the upper part to the whole triangle is (1231/260)/10 = 1231/2600 or about 47.35%.

Thats the exact same method I used .