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The normal to the curve y = x^3 - 5x2 +3x +1, at the points a(4, - 3) and b(1, 0) meet at point c. a) Find the coordinates of C. b) Find the area of triangle ABC?

Sagot :

sqdancefan

9514 1404 393

Answer:

  a) (-7, -2)

  b) 15 square units

Step-by-step explanation:

(a)

The slope at x is given by the derivative:

  y' = 3x^2 -10x +3

The slope of the normal is the negative reciprocal of this.

at a(4, -3) the slope of the normal is ...

  ma = -1/y' = -1/((3(4) -10)(4) +3) = -1/11

Then the point-slope equation of the line is ...

  y +3 = -1/11(x -4)

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at b(1, 0) the slope of the normal is ...

  mb = -1/y' = -1/((3(1) -10)(1) +3) = -1/-4 = 1/4

Then the point-slope equation of the line is ...

  y = 1/4(x -1)

Solving these two equations will give the coordinates of point C.

  (1/4(x -1) +3) = -1/11(x -4)

  11(x -1) +132 = -4(x -4)

  15x = -105

  x = -7

  y = 1/4(-7 -1) = -2

The coordinates of point C are (-7, -2).

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(b)

There are a few ways to find the area of a triangle that is specified by its vertex coordinates. I like the method that involves finding the determinants of pairs of coordinates around the figure. The area is half the absolute value of their sum. This can be made a little easier by listing the coordinates, repeating the first pair:

  a(4, -3)

  b(1, 0)

  c(-7, -2)

  a(4, -3)

Working down the list, the area will be ...

  A = 1/2|(4(0) -1(-3)) +(1(-2) -(-7)(0)) +((-7)(-3) -4(-2))| = 1/2|(0 +3) +(-2 +0) +(21 +8)|

  A = 1/2|30| = 15 . . . . square units

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The attached graph finds the intersection of the normals quite easily. The attached spreadsheet does the area calculation from the triangle coordinates. These tools are very handy for problems like this.

View image sqdancefan
View image sqdancefan
caylus

Answer:

Step-by-step explanation:

[tex]Curve:\ y=x^3-5x^2+3x+1\\\\y'(x)=3x^2-10x+3\\A=(4,-3)\\B=(1,0)\\y'(4)=3*4^2-10*4+3=11\\slope\ of\ the\ normal\ in \ A: -\frac{1}{11} \\Equation\ normal\ in \ A:\\y+3= -\frac{1}{11}*(x-4)\ or\ -x-11*y=29\ (1)\\\\y'(1)=3*1^2-10*1+3=-4\\slope\ of\ the\ normal\ in \ B: \frac{1}{4} \\Equation\ normal\ in \ B:\\y-0=\frac{1}{4} (x-1)\ or\ -x+4y=-1\ (2)\\\\Coordinates\ of\ C:\\(1)-(2) ==> y=-2\\(2) ==> x=-7\\\\C=(-7,-2)\\[/tex]

Area of the  triangle ABC:

[tex]S=\dfrac{|CB|*|CA|*sin(\widehat{ACB})}{2}\\A=(4,-3),B=(1,0),C=(-7,-2)\\AB^2=(4-1)^2+(-3-0)^2=9+9=18\\AC^2=122\\BC^2=68\\\\Using\ Al'Kashi\ theorem:\\AB^2=CB^2+CA^2-2|CB||CA|*cos(\widehat{ACB})\\\\cos(\widehat{ACB})=\dfrac{68+122-18}{2\sqrt{68*122} } =\dfrac{86}{\sqrt{68*122} } \\\\sin^2(\widehat{ACB})=1-cos^2(\widehat{ACB})=1-\dfrac{86^2}{68*122} \\\\S=\dfrac{\sqrt{68} *\sqrt{122}*\sqrt{\dfrac{900}{68*122} }}{2}=15\\[/tex]

View image caylus
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