﻿ A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is : Kaysons Education

# A Point Equidistant From The Lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 And 7x + 24y – 50 = 0 Is

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## Question

### Solution

Correct option is

(0, 0)

42 +32 = 52, 52 + 122 = 132, 72 + 242 = 252

Hence the distance of (0, 0) form the three sides are

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