Find the perpendicular distance between the lines 9x + 40y – 20 = 0 and 9x + 40y + 21 = 0. -Maths 9th

Find the perpendicular distance between the lines 9x + 40y – 20 = 0 and 9x + 40y + 21 = 0. -Maths 9th
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Given line: x – 2y = 3 ⇒ y = \(rac{x}{2}\) - \(rac{3}{2}\)            ....(i)∴ Its slope = m1 = \(rac{1}{2}\)Let m2 be the slope of line through (3, 2). Since this line is inclined at 45º to line (i),tan 45º = \(\bigg|rac{m_1-m_2}{1+m_1m_2}\bigg|\) = \(\bigg|rac{rac{1}{2}-m_2}{1+rac{1}{2}m_2}\bigg|\)⇒\(rac{rac{1}{2}-m_2}{1+rac{1}{2}m_2}\) = ± 1 ⇒ \(rac{1-2m_2}{2+m_2}=±\,1\) ⇒ 1 – 2m2 = 2 + m2 or 1 – 2m2 = –2 – m2⇒ 3m2 = –1 or –m2 = –3 ⇒ m2 = \(-rac{1}{3}\) or m2 = 3Since the line passes through (3, 2), the equation of the required line is(y – 2) = \(-rac{1}{3}\)(x – 3) or (y – 2) = 3 (x –3)⇒ 3y – 6 = – x + 3 or y – 2 = 3x – 9 ⇒ 3y + x – 9 = 0 or y – 3x + 7 = 0.
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