what- Isosceles triangle SIX has congruent sides SI and IX and vertices S at (x, 5), I at(-2, 2), and X at (4, -1), where x > 0.What is the value of x?

what- Isosceles triangle SIX has congruent sides SI and IX and vertices S at (x, 5), I at(-2, 2), and X at (4, -1), where x > 0.What is the value of x?
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4
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A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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Last Answer : Let A ≡ (2, - 2), B ≡ (-2, 1), C ≡ (5, 2 ). Then,AB = \(\sqrt{(-2-2)^2+(1+2)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5BC = \(\sqrt{(5+2)^2+(2-1)^2}\) = \(\sqrt{49+1}\) = \(\sqrt{50}\) = \( ... of ΔABC = \(rac{1}{2}\) x base x height = \(rac{1}{2}\) x AB x AC = \(rac{1}{2}\)x 5 x 5 = 12.5 sq. units.

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Last Answer : The two given lines are ax + by + c1 = 0 and ax + by + c2 = 0. Any line parallel to these two lines and midway between them is ax + by + c = 0 ...(i) Putting x = 0, y = \(-rac{c}{b}\) is ... c1 = c - c2 ⇒ c = \(rac{c_1+c_2}{2}\)∴ Required equation is ax + by + \(rac{c_1+c_2}{2}\) = 0.

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Last Answer : (c) \(rac{b+k}{f+h}\)Let the slope of the lin passing through the points (-k, h) and (b, - f) be m1. Then m1 = \(rac{-f-h}{b+k}\) = \(-\bigg(rac{f+h}{b+k}\bigg)\)\(\bigg[Slope = rac{y_2-y_1}{x_2-x_1}\bigg]\) ... \(-rac{1}{m_1}\)= \(rac{-1}{-\big(rac{f+h}{b+k}\big)}\) = \(\bigg(rac{b+k}{f+h}\bigg)\)

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Last Answer : (c) 60ºSince p is the length of perpendicular from origin on the straight line ax + by - p = 0.p = \(rac{|a.0+b.0-p|}{\sqrt{a^2+b^2}}\)⇒ 1 = \(\sqrt{a^2+b^2}\) ⇒ 1 ... 60° + y sin 60° = p Hence required angle is 60°, which is the angle between the perpendicular and the positive direction of x-axis.