Question

Given triangle GHI with G(4,-3),H(-4,2), and I(2,4), find the perpendicular bisector of line HI in standard form

Answer 1

we have to write the equation of the line HI, first

Answer 2

ok i dont even know how to so that

Answer 3

hint:
the slope of the stright line HI, is:
\[\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 2}}{{2 - \left( { - 4} \right)}} = ...?\]

Answer 4

please write the result

Answer 5

i dont understand to even do next

Answer 6

what am i even doing
do i simplify the top and bottom and then divide

Answer 7

I suppose that:
\[\left( {{x_1},{y_1}} \right) = \left( { - 4,2} \right)\quad \left( {{x_2},{y_2}} \right) = \left( {2,4} \right)\]
now, next step is:
\[m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 2}}{{2 - \left( { - 4} \right)}} = \frac{2}{6} = ...?\]
please simplify

Answer 8

wouldnt 2-(-4) be 8

Answer 9

we have: \(2-(-4)=6\) so the slope \(m\) is:
\[\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 2}}{{2 - \left( { - 4} \right)}} = \frac{2}{6} = \frac{1}{3}\]

Answer 10

i dont understand i am getting 8 for the bottom not 6

Answer 11

it is an algebraic sum, not a multiplication

Answer 12

ok

Answer 13

so the slope is 1/3

Answer 14

where do i go from there

Answer 15

from the official theory, we know that the slope \(m'\) of the line perpendicular to the line HI, is:
\[\Large m' = - \frac{1}{m} = - \frac{1}{{\left( {\frac{1}{3}} \right)}} = ...?\]
please continue

Answer 16

i really dont understand if you can explain how i do it then i can try

Answer 17

it is simple, you have to multiply the first fraction for the inverse of the second fraction:
\[\Large m' = - \frac{1}{m} = - \frac{1}{{\left( {\frac{1}{3}} \right)}} = - \frac{1}{1} \times \frac{3}{1} = ...?\]

Answer 18

\[\frac{ -3 }{ -1 }\]

Answer 19

we have only \(one\) minus sign

Answer 20

so?

Answer 21

please retry

Answer 22

hint:
\[\Large m' = - \frac{1}{m} = - \frac{1}{{\left( {\frac{1}{3}} \right)}} = - \frac{1}{1} \times \frac{3}{1} = - 3\]
am I right?

Answer 23

\[\frac{ -3 }{ 1 }\]

Answer 24

yes you are right i just wrote mine in fraction

Answer 25

yes! bthat's right! Please think about the point \(G=(4,-3)\)
the perpendicular line with respect to the line \(HI\), which passes at point \(G\), is given by the subsequent equation:
\[\Large y - \left( { - 3} \right) = - 3\left( {x - 4} \right)\]

Answer 26

oops.. that's*

Answer 27

sorry internet is slow

Answer 28

ok! no problem. We have to simplify such equation.
Here is the next step:

\[\Large y + 3 = - 3x + 12\]

Answer 29

ok so \[y+3=-3x+12\] to\[y=-3x+15\] ?

Answer 30

is that right

Answer 31

please, we have to subtract\(3\) to both sides, so we get:
\[\Large y + 3 - 3 = - 3x + 12 - 3\]

Answer 32

oh yeah i dont know what i was thinking \[y=-3x+9\]

Answer 33

that's right!

Answer 34

ok is there anything else

Answer 35

please note that, such line is perpendicular to the line \(HI\), nevertheless it doesn't pass at mid point of the segment \(HI\)

Answer 36

ok thank you so much

Answer 37

furthermore, if we add \(3x-9\) to both sides, we can write:
\[\Large 3x + y - 9 = 0\]
which is another way to write such perpendicular line