Show that the points A(3,4), B(3,1) & C(8,4) are the vertices of a right-angled triangle. Find the length of the perpendicular from A to BC.

We have A(3,4) , B(3,1), and
C(8,4)

are vertices of a
triangle.

Let us calculate the length of each
side:

lABl = sqrt(3-3)^2 + (1-4)^2]= sqrt9 =
3

lACl = sqrt(8-3)^2 + (4-4)^2]= sqrt25=
5

lBCl= sqrt[(8-3)^2 + (4-1)^2]= sqrt(25+9)=
sqrt(34)

Then BC is the longest
side.

if ABC is a right angle, then
:

BC^2 = AC^2 + AB^2

34= 5^2 +
3^2

34 = 25+9

34=
34

Then ABC is a right angle triangle where BC is the
hypotenuse.

Now to measure the line from A to
BC

Let D be a point on BC such that AD is perpendicular to
BC

==> let AD = y

let  
 BD = x

==> CD = sqrt34-
x

AB^2 = BD^2 + AD^2

9 = x^2 +
y^2........(1)

AC^2 = CD^2 +
AD^2

25 = (sqrt34-x)^2  +
y^2.........(2)

Let us subtract (1) from
(2):

==> 16 = (sqrt34-x)^2 -
x^2

==> 16 = 34 -(2sqrt34)x + x^2
-x^2

==> (2sqrt34)x =
18

==> x= 18/2sqrt34= 9/sqrt34= 1.54
(approx.)