Geometry help

As Subhotosh originally said, you should find the length between the two given points first. Please do that by using the distance formula.
 
There is an easier way without using the slope or the distance or the equation of the line.
Draw a diagram! Use graph paper so you can see more clearly what is going on.
Join A and B.
Form a right angled triangle with AB the hypotenuse. Call the point where the right angle is C.
Now find the point which is 4/5 of the way along AC and the point which is 4/5 of the way along CB.
Can you see how that will help you find the point you want?
 
rachelmaddie said:
View attachment 15431

I need help understanding this question.
Click to expand...
Lets look at the vector from \(\displaystyle A\to B\) is \(\displaystyle <13-3,-15-(-5)>=<10,-10>\)
The line segment \(\displaystyle \overline{AB}=<3,-5>+t<10,-10>,~0\le t\le 1\) note for \(\displaystyle t=0\) we get \(\displaystyle A\) for \(\displaystyle t=1\) we get \(\displaystyle B\).
To get the point you want we simply let \(\displaystyle t=\frac{4}{5}\). What do we get?
 
pka said:
Lets look at the vector from \(\displaystyle A\to B\) is \(\displaystyle <13-3,-15-(-5)>=<10,-10>\)
The line segment \(\displaystyle \overline{AB}=<3,-5>+t<10,-10>,~0\le t\le 1\) note for \(\displaystyle t=0\) we get \(\displaystyle A\) for \(\displaystyle t=1\) we get \(\displaystyle B\).
To get the point you want we simply let \(\displaystyle t=\frac{4}{5}\). What do we get?
Click to expand...
Not sure if Rachelmaddie has studied vectors yet.
 
Harry_the_cat said:
Not sure if Rachelmaddie has studied vectors yet.
Click to expand...
Oh I quite agree with you. But That is no reason not to show the ease of using them.
I first taught high school geometry in 1964. Damn that was awful. I tried to introduce paragraph proofs only to get into trouble with the Head Teacher who had a MED in Mathed. Needless to say I fled to a PhD program as soon as that year was done. After many years of teaching a undergrad/grad course in axiomatic course I reviewed Mel Hausner's textbook A VECTOR SPACE APPROACH TO GEOMETERY and never looked back. I am absolutely convinced that teaching geometry any other way is a total waste of effort.
 
Jomo said:
As Subhotosh originally said, you should find the length between the two given points first. Please do that by using the distance formula.
Click to expand...
Find the length between the two points first using the distance formula.
A(3, -5)
B(13, -15)
d^2 = (x2 - x1)^2 + (y2 - y1)^2
d^2 = (13 - 3)^2 + ((-15) - (-5))^2
d^2 = (10)^2 + (-10)^2
d^2 = 100 + 100
d^2 = 14.14

Is this right?
 
rachelmaddie said:
Find the length between the two points first using the distance formula.
A(3, -5)
B(13, -15)
d^2 = (x2 - x1)^2 + (y2 - y1)^2
d^2 = (13 - 3)^2 + ((-15) - (-5))^2
d^2 = (10)^2 + (-10)^2
d^2 = 100 + 100
d^2 = 14.14
Is this right?
Click to expand...
The last line should be \(\displaystyle d=\sqrt{200}\simeq~ 14.1421\)\[ \]
 
Rachel,

Did you "study" response #10 & 11?

It is already worked out for you!
 
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