OABC is a square,O is the origin and the points A and B
are(3,0) and (p,q).If OABC lies in the first quadrant,find the values of p and q.Also write down the equations of AB and BC.
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Step-by-step explanation:
OA
O - 0 , 0
A - 3 , 0
Slope = (0-0)/(3 - 0) = 0
=> Line is Parallel to x axis
and as it passes through origin so
OA lies on x axis
OA = √(3-0)² +(0-0)² = 3
OABC is a square
=> AB ║ y axis
& AB = 3
AB = √(P - 3)² + (Q - 0)²
=> √(P - 3)² + (Q - 0)² = 3
=> (P - 3)² + Q² = 9
line is parallel to y axis => P = 3
=> 0² + Q² = 9
=> Q = ±3
but it lies in 1st Quadrant so
Q = 3
P = 3 , Q = 3
AB is x = 3
BC || OA
=> BC is at distnace of 3
=> BC is y = 3
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