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Question

Reduce the following equations to the normal form and find p and α in each case:
(i) x+3y-4=0
(ii) x+y+2=0
(iii) x-y+22=0
(iv) x − 3 = 0
(v) y − 2 = 0.

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Solution

(i) x+3y-4=0

x+3y=4x12+32+3y12+32=412+32 Dividing both sides by coefficient of x2+coefficient of y2x2+3y2=2

This is the normal form of the given line, where p = 2, cosα=12 and sinα=32α=π3.

(ii) x+y+2=0

-x-y=2-x-12+-12-y-12+-12=2-12+-12 Dividing both sides by coefficient of x2+coefficient of y2-x2-y2=1

This is the normal form of the given line, where p = 1, cosα=-12 and
sinα=-12α=225 The coefficent of x and y are negative.So, α lies in third quadrant

(iii) x-y+22=0

-x+y=22-x-12+12+y-12+12=22-12+12 Dividing both sides by coefficient of x2+coefficient of y2-x2+y2=2

This is the normal form of the given line, where p = 2, cosα=-12 and
sinα=12α=135 The coefficent of x and y are negative and positive respectively.So, α lies in second quadrant.

(iv) x − 3 = 0

x=3x+0×y=3x12+02+0×y12+02=312+02 Dividing both sides by coefficient of x2+coefficient of y2x+0×y=3

This is the normal form of the given line, where p = 3, cosα=1 and sinα=0α=0.

(v) y − 2 = 0

y=20×x+y=20×x02+12+y02+12=202+12 Dividing both sides by coefficient of x2+coefficient of y20×x+y=2

This is the normal form of the given line, where p = 2, cosα=0 and sinα=1α=90.

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