Question
Download Solution PDFनिम्न समीकरणों से घिरे हुए क्षेत्र का क्षेत्रफल (वर्ग इकाई में) क्या है?
2x – 3y + 6 = 0, 4x + y = 16 और y = 0?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
2x – 3y + 6 = 0
4x + y = 16
y = 0
प्रयुक्त सूत्र:
त्रिभुज का क्षेत्रफल = \(\frac{1}{{2}} \) ∣x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)∣
गणना:
2x – 3y + 6 = 0
⇒ 2x – 3y = -6 ....(1)
4x + y = 16 ....(2)
समीकरण (2) में 3 से गुणा करने पर,
⇒ 12x + 3y = 48 ...(3)
अब,
(1) और (3) को जोड़ने पर हमें प्राप्त होता है,
⇒ (2x + 12x) = (48 – 6)
⇒ 14x = 42
⇒ x = 3
अब, x का मान समीकरण (1) में रखते हैं,
⇒ 2 × 3 – 3y = -6
⇒ 6 – 3y = -6
⇒ -3y = -1
⇒ y = (-1 – 3) = 4
इसलिए, (x1, y1) = (3, 4)
अब,
y = 0 का मान समीकरण (2) में रखने पर,
⇒ 4x + 0 = 16
⇒ 4x = 16
⇒ x = 4
इसलिए, (x2, y2) = (4, 0)
फिर से,
हम y = 0 मान को समीकरण (1) में रखते हैं,
⇒ 2x – 0 = -6
⇒ 2x = -6
⇒ x = -3
इसलिए, (x3, y3) = (-3, 0)
अब,
त्रिभुज का क्षेत्रफल = \(\frac{1}{{2}} \) ∣x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)∣
⇒ \(\frac{1}{{2}} \)∣3(0 – 0) + 4(0 – 4) + (-3)(4 – 0)∣
⇒ \(\frac{1}{{2}} \)∣0 + (-16) + (-12)∣
⇒ \(\frac{1}{{2}} \)∣-16 – 12∣
⇒ \(\frac{1}{{2}} \)∣-28∣
⇒ 14 [मापांक में ऋणात्मक चिह्न हमेशा धनात्मक लिया जाता है]
∴ अभीष्ट मान 14 है।
Last updated on Jun 25, 2025
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