A triangle ABC lying in the first quadrant has two
vertices A(1,2) and B(3,1). If ZBAC = 90° and
area of tringle ABC is 55 square units then the
abscissa of the vertex C is
Answers
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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Answer:
This question is from (4th Sep 1st Shift 2020) Jee Main Exam
Step-by-step explanation:
ur answer is there in image
