If A (1, 2), B(-4, 2). Number of points P in the plane such that APB=90 degrees and Area of APB is 7 sq. units is
Answers
Given : A(1 2) B(-4 2) APB=90° and ∆APB = 7 sq units
To find : no .of points of p in plane
Solution :
A(1 2) B(-4 2)
∠APB = 90°
=> AB would be a diameter of a circle at which P will lie
AB = 5 => Radius = 5/2
Mid point = ( -3/2 , 2)
Equation of circle
(x + 3/2)² + (y - 2)² = (5/2)²
PQ ⊥ AB
Area of ∆APB
(1/2) * 5 * PQ = 7
=> PQ = 2.8
Radius of Circle is 2.5 cm and PQ can not be greater than Radius Hence no possible Solution
Number of point P are Zero .
Assuming Area = 6 sq unit
Area of ∆APB
(1/2) * 5 * PQ = 6
=> PQ = 2.4
AQ * BQ = (PQ)²
AQ = x => BQ = 5 - x
=> x(5-x) = 2.4²
=> x² - 5x + 2.4² = 0
=> x² - 1.8x - 3.2x + 2.4² = 0
=> (x - 1.8)(x - 3.2) = 0
=> x = 1.8 or x = 3.2
Hence
Point Q = ( -2.2 , 2) , ( -0.8 , 2)
PQ = 2.4 & PQ ⊥ AB
so y coordinates 2 ± 2.4 = -0.4 , 4.4
=> P = ( -2.2 , -0.4) , (-2.2 , 4.4) , ( -0.8 , -0.4) , (-0.8 , 4.4)
There are 4 such points if area is 6 sq unit
No Points if Area is 7 sq units
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