let E be the point where this perpendicular line through c intersects AB. Calculate the cordinates of E .( C =(4,-2) , A=(3,6) B=(-6,0).
Also which is shorter distance. [CD] or [CE]? Find this distance.
I have no clue what to do .
Find slope & eqt of AB.
find eqt of perpendicular line through c. Use perpendicular slope slope to AB
E = pt of intersection of the two lines.
No pt D ???
You found D in an earlier part of the question - I think it was the midpoint part of the question.
Hi Vicky if you post the exact question, even a picture of it i can help you.
you use fomula?
this is in the exam papers yes what year and question
This is what you do:
Find the slope of AB
The slope of AB is:
y2 - y1
x2 - x1
(0) - (6)
(-6) - 3
That means the perpendicular slope is -3/2
Now you have to find the equation of the line with point C (4, -2) and a slope of -3/2
This is how:
y - y1 = m(x - x1)
y - (-2) = -3/2(x - (4))
y + 2 = -3/2x + 6
y = -3/2x + 4
That is the equation of the line with point C and a slope of -3/2
Now you have to find the intersection point of the lines [AB] and C to locate E.
First, we need to find the equation of the line [AB]
y - y1 = m(x - x1)
y - (6) = 2/3(x - (3))
y - 6 = 2/3x - 2
y = 2/3x + 4
Then use your two equations to solve simultaneous equations to find the point E
Rearrange the two equations so that the variables (numbers with letters) are on one side, and a number on the other side
3/2x + y = 4
-2/3x + y = 4
You can now solve these simultaneous equations. Just subtract the equations.
You end up getting:
13/6x = 0
x = 0
Substitute it back in to one of the equations:
3/2(0) + y = 4
y = 4
The point E is located at (0,4)
There you go!
By the way, you never gave information about point D. But anyway, here's the distance of [CE]
[CE] = square root((x2 - x1)^2 + (y2 - y1)^2)
= square root((4 - 0)^2 + (-2 - 4)^2)
= square root(4^2 + (-6)^2)
= square root(16 + 36)
= square root(52)
= 2(square root(13))