Question

show that the points A (3,4),B (3,1) and C (8,4) are the vertices of right angle triangle.Find the length of perpendicular from A to BC.

Answer 1

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.)