If a(1 2) b(-4 2) the no .of points of p in plane such that APB=90° and ∆apb=6squnits is
Answers
Given : a(1 2) b(-4 2) APB=90° and ∆apb=6squnits is
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 = 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
=> p = ( -2.2 , -0.4) , (-2.2 , 4.4) , ( -0.8 , -0.4) , (-0.8 , 4.4)
There are 4 such points
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