Question

Two vertices of a right triangle exist at (3,2) and (6,7). Find the x-coordinate of the third vertex given that the y-coordinate of the third vertex is -8.

I got that the x-coordinate is 59/3 is that right?

Answer 1

@phi help!

Answer 2

you can check by finding the slopes of the two legs, they should be negative reciprocals

or find the lengths of all 3 sides and see if pythagoras theorem works.

Answer 3

@phi but it says there are two solutions and i can only get x=31

Answer 4

i did the negative reciprocal thing and it only gives x=31 as a solution

Answer 5

i found the slope to be 5/3

Answer 6

the slope of the line through the points
(59/3, -8) and (3,2) is -3/5

the slope of the line through the points
(3,2) and (6,7) is 5/3

they are negative reciprocals, so those two lines intersecting at (3,2) for a right angle

Answer 7

then i did:\[-\frac{ 3 }{ 5 }=\frac{ -8-7 }{ x-6 }\] and solved for x

Answer 8

but the problem says 2 solutions?

Answer 9

we could have the right angle at the other point (6,7)

Answer 10

wait but when i solve this equation i get 31 as a solution

Answer 11

for the xcoordinate

Answer 12

yes, you put the right angle at point 6,7

Answer 13

How do you do it if the right angle is at (6,7)?

Answer 14

oh so the two solutions are: 31 and 59/3?

Answer 15

if you do the same thing but at (3,2) you will get the other solution

Answer 16

so i did it all myself without knowing it

Answer 17

so i am right?

Answer 18

yes, (31,-8) is one of the solutions

Answer 19

thanks for the help

Answer 20

I get confused sometimes

Answer 21

here is a graph of the problem

Answer 22

the slope between (3,2) and (6,7) is 5/3

you want the slope from (3,2) to (x,-8) to be -3/5
or
\[ \frac{-8 - 2}{x-3} = \frac{-3}{5} \\ \frac{-10}{x-3}= \frac{-3}{5} \\
\frac{10}{x-3}= \frac{3}{5}\]
flip both fractions and then multiply by 10
\[\frac{x-3}{10}= \frac{5}{3} \\
x-3 = \frac{50}{3}\\
x= 3+\frac{50}{3}= \frac{59}{3}
\]
is the other solution, (which you already found)