Equation of straight line..find diagonal plz help me:)!?

Question

Given that quadrilateral ABCD is a rectangle whose vertices are A (-4, 4) B (-4, 2) C (4,4) D(4,2) Find diagonal
I really don't get this I've tried it and got the wrong answer.

Find the eqution of DIAGONAL AD:

What i did was...
m= -2/8 slope = -1/4

then..


y-4=m(x-4)
y-4= -1/4x +1
+4 + 4

y=-1/4x +5

this is wrong and I am unsure how to do it, if you can help it'd be much appreciated!

Thanks but I really don't understand your answers too well..

The answer is supposed to be: x+4y-12=0. How can i get to that point/ thanks

Answer 1

You can't find the diagonal connecting A & D. A & D are connected by a side of the quadrilateral. The two diagonals would be AC and BD. The order ABCD tells you that AB, BC, CD, and DA are the sides of the quad. If a point is next to another point in the name of the shape, then the points are next to each other in the picture of the shape.

Now just find the slope of AC using (y2 - y1)/(x2 - x1) then use the point-slope form y - y1 = m(x - x1) using either point A or C for (x1 , y1) to find the equation of the line in slope-intercept form.

repeat for BD.

Ok here goes, even though AD is not a diagonal, it is a side... I'll find AD for you... its gonna be long..

slope of segment connecting A(-4,4) & D(4,2)

is (2 - 4) / (4 - (-4)) = -2/8 = -1/4

now use point slope equation with either point ill use D cause its easier with no negatives...

y - 2 = -1/4(x - 4) ...... distribute -1/4
y - 2 = (-1/4)x + 1 ...... add 2 to each side
y = (-1/4)x + 3 .... slope-intercept form of line AD

now they must have asked for the line in standard form which is Ax + By + C = 0 so we need to move things around...

y = (-1/4)x + 3
(1/4)x + y - 3 = 0 ..... multiply by 4 to get rid of fraction
x + 4y -12 = 0 ...... standard form of line through AD

** in your work you made a mistake because you plugged in A(4,4) into point slope form instead of A(-4,4)

it should've read y - 4 = (-1/4)(x + 4)
then y - 4 = -1/4x - 1
y = -1/4x + 3

then move to standard the same as above... that's where your mistake was. everything else was right though.

Answer 2

Given that quadrilateral ABCD is a rectangle whose vertices are A(-4, 4); B(-4, 2); C(4,4); and D(4,2). Find the equation of diagonal AD.

First find the slope m.
A(-4, 4) and D(4,2)

m = ∆y/∆x = (2 - 4) / (4 - -4) = -2/8 = -1/4

Now use the point slope form of the line and plug in one of the points. Let's choose D(4, 2).

y - 2 = (-1/4)(x - 4)
y - 2 = (-1/4)x + 1
y = (-1/4)x + 3

If you want to eliminate fractions, multiply thru by 4.

y = (-1/4)x + 3
4y = -x + 12
x + 4y = 12
x + 4y - 12 = 0

Answer 3

There are two diagonals.

We'll find the one through B (-4,2) and C (4,4)

What I do is say that any point on the line through those two points obeys the following slope relation.
I.e, for any point P (x,y) on the line the slope remains constant;
the slope from P (x,y) to B is the same as the slope from B to C:

(y-2)/(x-(-4)) = (4-2)/(4-(-4)) = 2/8

y-2 = (2/8)*(x+4)
y = x/4 +1 + 2
y = x/4 + 3

Now check it:
at x = -4 , y = 2 Ok
at x = +4, y = 4 OK..

Answer 4

You placed up up whilst the pony's outdoors leg is going forward. once you're a sturdy rider, you could experience once you're on the incorrect diagnol. My ideal advice to u is to have a coach watch you experience (in case you do no longer take instructions) and characteristic them permit you already know once you're on the perfect posting diagonal and once you're on the incorrect posting diagonal. once you already know which diagonal is sturdy and it truly is faulty, after some prepare you're able to have the means to experience once you're appropriate or incorrect. then you definitely could have the means to alter it very certainly, so in case you have no longer any partitions and are perplexed, then you definitely will understand appropriate away by using feeling in case you're appropriate or incorrect. i wish i did no longer confuse you! Sorry if I did! wish I helped!!!

Answer 5

well let us expect that the two terminal lines of the rectangle are
rays then we got two rays :[ab]and [bc]. and then we find by the basic rules of adding rays that [ab]+[ac] =[ad]. then we account the ray[ad]from accoutning the points wich means :
ad is a ray . we have the parametrics of two points of it so now we account it :
ad =(xd-xa),(yd-ya)=(4-(-4) ),(2-4)
x=8
y=-2
we exchange it with its values in the equation (by expecting that A is the institute
y=mx
-2=8m => m=-2/8= -1/4
the equation is :
y=-1/4x