Exercise 19 Page 202

Consider the given equation of a line. y= 2/3x To find the slope perpendicular to this line, we need to find the negative reciprocal of 23.

m_1 * m_2 = -1

Substitute

m_1= 2/3

2/3 * m_2 = -1

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Solve for m_2

MoveRightFacToNum

a/c* b = a* b/c

2m_2/3=-1

MultEqn

LHS * 3=RHS* 3

2m_2=-3

DivEqn

.LHS /2.=.RHS /2.

m_2=-3/2

MoveNegNumToFrac

Put minus sign in front of fraction

m_2 = -3/2

We have found that - 32 is the negative reciprocal of 23. Using this information we can write a general equation in point-slope form for the lines perpendicular to the given line. y- y_1 = - 32(x- x_1) Since we know that point P( 6, 6) should be contained in the perpendicular line, we can substitute it for ( x_1, y_1). By doing so, we will be able to write the equation of the line perpendicular to the given one, and containing point P. y- 6 = -3/2(x- 6)