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In the first quadrant, what is the \(y\)-intercept of the triangle formed by \(x+3y\le 12\)?प्रथम चतुर्थांश में \(x+3y\le 12\) से बने त्रिभुज का \(y\)-अवरोध कौन-सा है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \((0,4)\)\((0,4)\)
To find the \(y\)-intercept, put \(x=0\). Then \(3y=12\), so \(y=4\). Therefore, the line meets the \(y\)-axis at \((0,4)\). The point \((4,0)\) lies on the \(x\)-axis, so it is not the \(y\)-intercept. Exam tip: always set \(x=0\) to find a \(y\)-intercept.\(y\)-अवरोध ज्ञात करने के लिए \(x=0\) रखते हैं। तब \(3y=12\), अतः \(y=4\)। इसलिए रेखा \(y\)-अक्ष को \((0,4)\) पर काटती है। \((4,0)\) \(x\)-अक्ष पर स्थित बिंदु है, इसलिए वह \(y\)-अवरोध नहीं है। परीक्षा टिप: \(y\)-अवरोध के लिए हमेशा \(x=0\) रखें।
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