Distance of x – z + 2 = 0 from origin
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distance is = | 2 | / √[1² + (-1)² ] = 2/√2 = √2
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we know, distance between (r , s, t) and ax + by + cz + d =0 is -------------
distance = |ar + bs + ct + d|/√(a^2 + b^2 + c^2)
now , use this =======
distance = |0 + 0 + 0 + 2|/√(1^2 + (-1)^2) =2/√2 =√2.
distance = |ar + bs + ct + d|/√(a^2 + b^2 + c^2)
now , use this =======
distance = |0 + 0 + 0 + 2|/√(1^2 + (-1)^2) =2/√2 =√2.
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