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उस सरल रेखा का समीकरण ज्ञात कीजिए जो मूलबिंदु से होकर जाती है तथा जो रेखा ` 3x+ 4y =24` के अक्षों से कटे खंड को समद्विभाजित करती है |

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Correct Answer - `3x-4y =0`
`3x+4y =24` का अन्तः खंड रूप है| `(x)/(8) +(y)/(6) =1`
यदि समीकरण (i ),अक्षों से अन्तः खंड को द्विभाजित करती है|
तब ` " "` द्विभाजन बिंदु `=( (8+0)/(2), ( 0+6)/(2)) =(4,3)`
`therefore ` बिंदु (0,0) व ( 4 ,3 ) से जाने वाली रेखा का समीकरण `y-0=(3-0)/(4-0) (x-0)`
`" "rArr " "3x- 4y =0 `

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