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Number of possible point C such that area of AABC is 10 square unit where
A(0,6), B(8,0) and angleACB = 90°, is
Answers
Given : area of Δ ABC is 10 square units where A(0,6), B(8,0) and angle ACB = 90°, is
To find : Number of possible point C
Solution:
Let say
C = ( x , y)
A = (0,6) & B = (8,0) , C = ( x , y)
Area of triangle = (1/2) | 0 (0 - y) + 8 (y - 6) + x(6-0)|
= (1/2) | 8y - 48 + 6x |
= | 4y - 24 + 6x |
4y - 24 + 6x = ± 10
=> 4y + 6x = 34 , 14
=> 2y + 3x = 17 , 7
angle ACB = 90°,
AC & CB are perpendicular to each other
=> Slope of AC * Slope of CB = - 1
=> (( y - 6 )/(x - 0) ) * ( ( y - 0)/(x - 8) ) = - 1
=> (y - 6 ) y = - x(x - 8)
=> y² - 6y = -x² + 8x
=> x² + y² = 8x + 6y
Case 1 : 2y + 3x = 7 => y = (7 - 3x)/2
=> x² + ((7-3x)/2)² = 8x + 6 (7 - 3x)/2
=> x² + ( 49 + 9x² - 42x)/4 = 8x + 21 - 9x
=> 4x² + 49 + 9x² - 42x = -4x + 84
=> 13x² - 38x - 35 = 0
Two possible points
Case 2 : 2y + 3x = 17 => y = (17 - 3x)/2
=> x² + ((17-3x)/2)² = 8x + 6 (17 - 3x)/2
=> x² + ( 289 + 9x² - 102x)/4 = 8x + 51 - 9x
=> 4x² + 289 + 9x² - 102x = -4x + 204
=> 13x² - 98x +85 = 0
Two possible points
Hence total 4 possible points C such that area of Δ ABC is 10 square units where A(0,6), B(8,0) and angle ACB = 90°
if we simply understand then AB is diameter of cirlcle
& a circle with Radius 5 & point C lies on a Height of 2 from Diameter will form a right angle triangle with area 10
Hence 4 possible triangles
Learn more:
The vertices of a A ABC are A(-5, -1), B(3.-5), C(5,2).
https://brainly.in/question/8993582
D is mid point of side BC of triangle ABC and E is the mid point of BD ...
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