Question 388495
1. Kite ABCD has 2 of its vertices at A(-3,8) and B(7,14). 
1(a). Given that the equation of the longer diagonal BD is y=4x-14, find the equation of the short diagonal AC expressing your answer in the form ax+by+c=0. 
1(b). Find the co-ordinates of E, which is the point of intersection. 
1(c). Hence establich the co-ordinates of C. 
Can you please provide solutions with full answers please. 
Many Thanks,
Andy.

<pre>
We just help you with the first problem. Post again for the others.

{{{drawing(400,400,-5,18,-8,15, circle(1.75,-7,.2),locate(1.75,-7,"D(?,?)"),
graph(400,400,-5,18,-8,15,4x-14), circle(5,6,.2), locate(5.5,6.7,"E(?,?)"),
circle(-3,8,.2), circle(7,14,.2),
green(line(-3,8,13,4)),  locate(13,4,"C(?,?)"),
locate(-4,9,"A(-3,8)"), locate(7,14,"B(7,14)")   )}}} 

1(a). Given that the equation of the longer diagonal BD is y=4x-14, find the
equation of the short diagonal AC expressing your answer in the form ax+by+c=0.
 
To write the equation of AC we must have its slope and a point it goes through.
We already have a point it goes through, namely A(-3,8). Now all we need is it
slope.  We know that AC is perpendicular to BD, and we know that to find the
slope of a second line perpendicular to a first line, we invert the slope of
the first line and change its sign.  

The long diagonal BD has equation y = 4x-14.  We compare that to y=mx+b
and see that BD has slope m=4.  Since the short diagonal AC is perpendicular
to the long diagonal BD, then AC has slope {{{-1/4}}}.  Next, to find its
equation we use the point-slope form: 

    y - y<sub>1</sub> = m(x - x<sub>1</sub>)

      y - 8 = {{{expr(-1/4)}}}[x - (-3)]

      y - 8 = {{{expr(-1/4)}}}(x +3)

Multiply through by 4:

    4y - 32 = -1(x + 3)

    4y - 32 = -x - 3

x + 4y - 29 = 0

That's the equation of AC in the form ax+by+c=0

1(b). Find the co-ordinates of E, which is the point of intersection.

To do that we solve the system of equations:

          y = 4x - 14
x + 4y - 29 = 0

by substitution and get (x,y) = (5,6), so we can label point E as E(5,6)

{{{drawing(400,400,-5,18,-8,15, circle(1.75,-7,.2),locate(1.75,-7,"D(?,?)"),
graph(400,400,-5,18,-8,15), circle(5,6,.2), locate(5.5,6.7,"E(5,6)"),
circle(-3,8,.2), circle(7,14,.2), line(-12,-62,15,46),
green(line(-3,8,13,4)),  locate(13,4,"C(?,?)"),
locate(-4,9,"A(-3,8)"), locate(7,14,"B(7,14)")   )}}} 





 
1(c). Hence establich the co-ordinates of C. 

To establish the coordinates of point C we realize that E(5,6) is

8 units to the right of point A(-3,8), since 5-(-3) = +8, and also
that E is 2 units below point A((-3,8) since 6-9 = -2.

So therefore point C must be 8 units right and 2 units below point
E(5,6).  Since 5+8=13, and 6-2 = 4, then point C must be (13,4), so
we can fill in the coordinates of point C:

{{{drawing(400,400,-5,18,-8,15, circle(1.75,-7,.2),locate(1.75,-7,"D(?,?)"),
graph(400,400,-5,18,-8,15), circle(5,6,.2), locate(5.5,6.7,"E(5,6)"),
circle(-3,8,.2), circle(7,14,.2), line(-12,-62,15,46),

green(line(-3,8,13,4)),  locate(13,4,"C(13,4)"),
locate(-4,9,"A(-3,8)"), locate(7,14,"B(7,14)")   )}}} 

The kite is the red figure below, although we are not given enough information
to find the coordinates of D, so we can only leave that point as just D(?,?). 

{{{drawing(400,400,-5,18,-8,15, circle(1.75,-7,.2),locate(1.75,-7,"D(?,?)"),
graph(400,400,-5,18,-8,15), circle(5,6,.2), locate(5.5,6.7,"E(5,6)"),
circle(-3,8,.2), circle(7,14,.2), line(-12,-62,15,46),
red(line(-3,8,7,14),line(7,14,13,4),line(13,4,1.75,-7),line(1.75,-7,-3,8)),

green(line(-3,8,13,4)),  locate(13,4,"C(13,4)"),
locate(-4,9,"A(-3,8)"), locate(7,14,"B(7,14)")   )}}} 

Edwin</pre>