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a line passes through a(1,-3) and b(2,7)

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a line passes through a(1,-3) and b(2,7)
write the equation of a line perpendicular to a. how would you go about solving this problem?
asked Mar 14, 2014 in ALGEBRA 1 by johnkelly Apprentice

1 Answer

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The points are a(1, - 3) and b(2, 7).

Line equation is y- y ₁ = m(x - x₁).

Slope  formula : m = (y₂ - y₁)/(x₂ - x₁).

Substitute the values of a(x₁, y₁) = (1, - 3) and b(x₂, y₂) = (2, 7) in Slope  formula.

m = (7 +3)/(2 - 1)

m = 10/1

m = 10.

Substitute the values of a(x₁, y₁) = (1, - 3) and m = 10 in line equation (or) y- y₁ = m(x - x₁).

y + 3 = 10(x - 1)

y = 10(x) - 10 - 3

y = 10(x) - 13.

The line is Perpendicular to y = 10(x) - 13.

So slope of the line = -1/10.

The point is a(1, - 3)

Substitute the values of a(x₁, y₁) = (1, - 3) and m = - 1/10 in line equation (or) y - y ₁ = m(x - x₁).

Perpendicular line equation y + 3 = (- 1/10)(x - 1)

y = (- 1/10)(x - 1) - 3

y  = (- 1/10)x + 1/10 - 3

y  = (- 1/10)x  - (29/10).

So, the perpendicular line is y = (- 1/10)x  - (29/10).

answered Mar 15, 2014 by dozey Mentor

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