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প্রদত্ত বিন্দু দুটি হল (1, 2) এবং (3, –2)।

এই বিন্দুগামী সরলরেখার সমীকরণ হবে:

\(\)\frac{x-x_1}{x_1-x_2} = \frac{y-y_1}{y_1-y_2}\(\)

এখানে, \((x_1, y_1) = (1, 2)\) এবং \((x_2, y_2) = (3, -2)\)।

\(\) \frac{x-1}{1-3} = \frac{y-2}{2-(-2)} \(\)

\(\) \frac{x-1}{-2} = \frac{y-2}{4} \(\)

আড়াআড়ি গুণ করে পাই:

\(\) 4(x-1) = -2(y-2) \(\)

\(\) 4x - 4 = -2y + 4 \(\)

পদগুলো একপাশে নিয়ে আসি:

\(\) 4x + 2y = 4 + 4 \(\)

\(\) 4x + 2y = 8 \(\)

সমীকরণটিকে ছেদক আকারে \((\frac{x}{a} + \frac{y}{b} = 1)\) প্রকাশ করার জন্য 8 দ্বারা ভাগ করি:

\(\) \frac{4x}{8} + \frac{2y}{8} = \frac{8}{8} \(\)

\(\) \frac{x}{2} + \frac{y}{4} = 1 \(\)

এখানে, x-অক্ষের ছেদবিন্দু \(a = 2\) এবং y-অক্ষের ছেদবিন্দু \(b = 4\)।

অক্ষ দুটি দ্বারা গঠিত খণ্ডিতাংশের দৈর্ঘ্য হলো \((a, 0)\) এবং \((0, b)\) বিন্দুর মধ্যবর্তী দূরত্ব। দূরত্বের সূত্র অনুযায়ী:

\(\) d = \sqrt{a^2 + b^2} \(\)

\(\) d = \sqrt{2^2 + 4^2} \(\)

\(\) d = \sqrt{4 + 16} \(\)

\(\) d = \sqrt{20} \(\)

\(\) d = \sqrt{4 \times 5} \(\)

\(\) d = 2\sqrt{5} \(\)

সুতর

(1, 2) ও (3, – 2 ) বিন্দুগামী রেখার অক্ষ দুটির মধ্যবর্তী খণ্ডিতাংশের দৈর্ঘ্য কত ?