Math, asked by malleshamburra6, 7 months ago

A triangle ABC lying in the first quadrant has two vertices A(1,2)A(1,2) and B(3,1)B(3,1). If ∠BAC=90∘∠BAC=90∘ and area of tringle ABCABC is 55–√55 square units then the abscissa of the vertex CC is​

Answers

Answered by amitnrw
7

Given : A triangle ABC lying in the first quadrant has two  vertices A(1,2) and B(3,1).  ∠BAC = 90° .area of tringle ABC is 55 square units

To Find : abscissa of the vertex C

Solution:

Let say point C  = ( x , y)

A = ( 1, 2)

B = ( 3 , 1)

Slope of BA = ( 1 - 2) /(3 - 1)  = -1/2

=> Slope of AC  = 2

( as slope of BC * slope of AC = - 1)

Slope of AC = ( y - 2)/(x - 1)  = 2

=> y - 2 =  2x    - 2

=> 2x  = y

Hence C = ( x , 2x)

Length of BA = √(2 - 1)² + ( 3 - 1)²  =  √5

(1/2) BA * AC  =  55

=> AC = 110/ √5

=> AC = 22√5

AC = √(x - 1)² + (2x - 2)²      =   (22√5)

=>  |x - 1|√5  =  22√5

=>  x = 23   or x  = - 21

triangle ABC lying in the first quadrant

=> x = 23

Point C  = ( 23  , 46)

abscissa of the vertex C  = 23

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