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.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.)

sufiyantahir96's profile pic

sufiyantahir96 | Student, Grade 9 | eNotes Newbie

Posted on

d^2 = (x2-x1)^2+(y2-y1)^2.
So AB^2 = (3-3)^2+(1-4)^2 = 9
AC^2 = (8-3)^2+(4-4)^2 = =25
BC^2 = (8-3)^3+(4-1)^2 = 34
Therefore AB^2+AC^2 =  9+25 =34 = BC^2. So the triangle ABC is a right angled at A.
Therefore area  of trianle = (1/2) AB*AC = (1/2) (sqrt9)(sqrt25) = 7.5 sq units.
Therefore  Area of triangle ABC = (1/2) BC*AD where AD is the perpendicular from A to BC
Or AD = 2*area of ABC/ BC = 2*7.5/sqrt34 = 15/sqrt34 = 2.5725 units.nealy.


ALL ANSWER IS OF NEELA BUT SHE FORGOT TO WRITE X IN FIRST...d^2 = ({{{{{x}}}}}2-x1)^2+(y2-y1)^2.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

By distance formula the distance d between  points (x1,y1) and (x2,y2)  is  given by:

d^2 = (2-x1)^2+(y2-y1)^2.

So AB^2 = (3-3)^2+(1-4)^2 = 9

AC^2 = (8-3)^2+(4-4)^2 = =25

BC^2 = (8-3)^3+(4-1)^2 = 34

Therefore AB^2+AC^2 =  9+25 =34 = BC^2. So the triangle ABC is a right angled at A.

Therefore area  of trianle = (1/2) AB*AC = (1/2) (sqrt9)(sqrt25) = 7.5 sq units.

Therefore  Area of triangle ABC = (1/2) BC*AD where AD is the perpendicular from A to BC

Or AD = 2*area of ABC/ BC = 2*7.5/sqrt34 = 15/sqrt34 = 2.5725 units.nealy.

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