What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?

What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?
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Answer :

If: y = 5x +10 and y = x^2 +4Then: x^2 +4 = 5x +10Transposing terms: x^2 -5x -6 = 0Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x =-1Therefore by substitution endpoints of the line are at: (6, 40)and (-1, 5)Midpoint of line: (2.5, 22.5)Slope of line: 5Perpendicular slope: -1/5Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y= -x+114.75Perpendicular bisector equation in its general form: x+5y-114.75= 0
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Related questions

What are the points of intersection of the line 2x plus 5y equals 4 with the curve y squared equals x plus 4?

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Last Answer : If: 2x +5y = 4 then 25y^2 = 4x^2 -16x +16If: y^2 = x +4 then 25y^2 = 25x +100So: 4x^2 -16x +16 = 25x +100Transposing terms: 4x^2 -41x -84 = 0Factorizing the above: (4x+7)(x-12) = 0 meaning x = -7/4 or x =12By substitution into original equation points of intersection:(-7/4, 3/2) and (12, -4)

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Last Answer : If y = 3x + 1 is a tangent to x² + y² = k (k > 0 since it isa square), then where they meet has a repeated root; they meetat:x² + (3x + 1)² = k→ x² + 9x² + 6x + 1 - k = 0→ 10x² + 6x + (1 - k) = 0This is the ... 3/10→ y = 3 -3/10 + 1 = 9/10 + 1 = 1/10→ point of contact is (-3/10, 1/10) with k = 1/10

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Last Answer : The line meets the circle when:y = x + c→ x² + y² = 4→ x² + (x + c)² - 4 = 0→ x² + x² +2cx + c² - 4 = 0→ 2x² + 2cx + (c² - 4) = 0If the line is a tangent to the circle, this has a repeatedroot. This occurs ... 2 = 0→ (x + √2)² = 0→ x = -√2→ y = x + 2√2= -√2 + 2√2= √2→ point of contact is (-√2, √2)

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Last Answer : NEED ANSWER

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