P and Q are two point on the line x-y+1=0and are at a distance of units from the origin. Find the area of triangle POQ.
Answers
Given : P and Q are two point on the line x-y+1=0and are at a distance of units from the origin.
To find : the area of triangle POQ.
Solution:
P and Q are two point on the line x-y+1=0and are at a distance of units from the origin
=> y = x + 1
P = ( a , a + 1) Q = (b , b + 1)
OP = OQ = 1
=> √a² + (a + 1)² = √b² + (b + 1)²
=> a² + a² + 2a + 1 = b² + b² + 2b + 1
=> 2a² + 2a = 2b² + 2b
=> a² + a = b² + b
=> (a² - b²) + (a - b) = 0
=> (a + b)(a - b) + (a - b) = 0
=> (a - b)(a + b + 1) = 0
=> a = b or a + b = - 1
if a = b then P & Q will be same point
=> a + b = - 1
taking unit distance
√a² + (a + 1)² = 1
=> 2a² + 2a + 1 = 1
=> a(a + 1) = 0
=> a = 0 a = - 1
a = 0 & a + b = - 1
=> a = 0 , b = - 1
a = -1 => b = 0
Hence P & Q
(0 , 1) & (-1,0)
=> P & Q are perpendicular to each other
Area of triangle = (1/2) (1)(1) = 1/2 sq unit
This is different from Question :
Considering a difference case if P & Q are at unit distnce on given line
Lets Draw OM⊥ PQ
Area of Δ POQ = (1/2) PQ * OM
PQ = 1 unit need to find OM
OM ⊥ PQ
PQ is line x - y + 1 = 0
=> y = x + 1
=> Slope = 1
Slope of OM = -1 ( as OM ⊥ PQ)
OM passes through (0,0) origin
=> Equation of line OM
y - 0 = - 1 (x - 0)
=> y = - x
intersection point of OM & PQ
x + 1 = - x
=> 2x = -1
=> x = -1/2
y = 1/2
M = (-1/2 , 1/2)
OM = √(-1/2 - 0)² +(1/2 - 0)² = 1/√2
Area of Δ POQ = (1/2) 1 * 1/√2 = 1/2√2 sq unit
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