write an equation that is perpendicular to y=3/5x-1 and passes through the point (9,-14)

Question

HELP THIS IS DUE IN 30 MINS AND IM FAILING THIS CLASS

Answer 1

Hi....

I just saw this...

Remember that PERPENDICULAR lines have the relationship that the product of their slopes will be -1

That is, m₁(m₂) = -1 OR m₁ = 1/m₂ (vice versa)...one slope is the NEGATIVE RECIPROCAL of the other slope)

The slope of the 2nd line will be NEGATIVE RECIPROCAL of the 1st line...m₂ = -⁵/₃

Use the POINT-SLOPE FORMULA to find the equation of the 2nd line knowing a point on that line as

(9, -14); x₁ = 9 and x₂ = -14

y - y₁ = m(x - x₁) ..........POINT-SLOPE FORMULA

y - (-14) = -⁵/₃(x - 9)

y + 14 = -⁵/₃(x) - (-⁵/₃)(9) ...... Keep ALL signs/operations until the end

y + 14 = -⁵/₃(x) - (-15)

y + 14 = -⁵/₃(x) + 15

y = -⁵/₃(x) + 15 - 14 ...... Isolate the y term by subtracting 14 on both sides

y = -⁵/₃(x) + 1 Final Answer

Answer 2

perpendicular equation will have the negative inverse of the x coefficient as its slope = m = -5/3

y= mx +b

y=-5x/3 + b

-14 = -5(9)/3 + b

-14 = -15 + b

1 = b

y= -5x/3 + 1

check the answer, plug in the point (3,-14)

-14 = -5(9)/3 + 1

-14 =-45/3 + 1

-14 =-15 +1

-14 = -14

the perpendicular equation is

y=-5x/3 +1 which can be rewritten to eliminate the fraction to get

3y = -5x +1 or

5x+3y = 1