Question

three points have coordinates A(2, 6), B(8, 9) and C (6, 0). the perpendicular bisector of AB meets the line BC at D. Find: coordinates of D

Answer 1

@nnpatricia do you know vector geometry?

Answer 2

Wether or not, I recommend you to do the following, first draw the picture - it's not very hard because it's two dimensional, so all your points will be in the ordinary xy-plane.

Then define algebraically, or vector-geometrically the Line AB and BC, you then know it's slope and therefore you know the slope of the perpendicular. You can figure out the bisection value by taking the average of the endpoints A and B.

Then you only need to define the line that goes through said midpoint, lets call it X for the purpose, by using the slope normal to AB, let this new line g cross the line BC and you will get the point D.

Answer 3

wait i guess i do. yeah, but this one involves gradient and equation of the line. is the answer (91/31, 125/13)? just want to make sure i did it right ._.

Answer 4

oops, I haven't done it myself yet, well the equation of a line is easy, gradient you know how to find too, let me give it a try.

Answer 5

thankss :D

Answer 6

The equation of the line AB is

\[ g: y= \frac{1}{2}x + 5\] so the normal slope to this line -2, follow so far?

Answer 7

the Midpoints of AB are (10, 7.5)

Answer 8

so we have this equation we want to cross with BC

\[h : y= -2x+27.6\]

Answer 9

wait the mid point of AB isnt (5, 8)? where did you get (10, 7.5)?

Answer 10

27.5 of course, not .6

you want to cross equation with the line BC

\[ \large L_{BC}: y= \frac{9}{2}x-27\]

Answer 11

wait wait oh damn. i'm giving the wrong one B supposes to be B (8, 10) I'm sorry for being so sloppy

Answer 12

ohh, well this changes a lot then haha (-:

Answer 13

so the midpoint indeed is 5/8

Answer 14

So this gives the new equation

\[ y=-2x+18\] then, for the one perpendicular to the bisector of AB, right?

Answer 15

and for the line BC we get new

\[ y=5x-30 \]

Answer 16

In fact, I will have to start from the beginning, sorry, but I didn't change the point B for the initial equation, I just noticed, it won't change the general procedure of this problem, but it will indeed change the solution.

Answer 17

yes! lol it isnt 3y+2y=31? the equation of perpedicular of AB?

Answer 18

ah right. take ur time :)

Answer 19

The slope of AB is 2/3, so the normal - 3/2

Answer 20

So the line perpendicular to AB is

\[ y= -\frac{3}{2}x+15.5\]

Answer 21

when you multiply that by two you get 2y+3x=31

Answer 22

so then we do it with elimination since we got 2 equations there, yes?

Answer 23

just set them equal to each other, when you have them in slope intersect form.

\[- \frac{3}{2} x +15.5 = 5x-30 \]

Answer 24

x=7

Answer 25

therefore y=5

Answer 26

it's proven that my answer is wrong lol! thanks :)

Answer 27

you're welcome (-: