Write the equation of the line that passes through (1;2) and is perpendicular to y = 6x - 2

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The line passes through the point (1,2)

Then the equation for the line is:

d1:y-y1 = m1 (x-x1)

d1:y- 2 = m1 (x-1)

But we know that the line d1 is perpendicular to the line:

d2: y= 6x - 2

Sinde d1 and d2 are perpendiculr, then the product of...

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The line passes through the point (1,2)

Then the equation for the line is:

d1:y-y1 = m1 (x-x1)

d1:y- 2 = m1 (x-1)

But we know that the line d1 is perpendicular to the line:

d2: y= 6x - 2

Sinde d1 and d2 are perpendiculr, then the product of the slopes equals -1

m1*m2 = -1

m1 * 6 = -1

==> m1 = -1/6

Then the equation for the line d1 is:

d1: y-2 = (-1/6)(x-1)

d1: y= (-1/6)x + 1/6 + 2

d1: y + (1/6)x - 7/6 = 0

Multiply by 6:

d1: 6y + x - 7 = 0

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