Find an equation of a line that is perpendicular to x+2y-12=0 and the same x-intercept of 3x-2y+12=0.

y = 2x + 8

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

x + 2y - 12 = 0 ( subtract x - 12 from both sides )

2y = - x + 12 ( divide terms by 2 )

y = - x + 6 ← in slope- intercept form

with slope m = -

Given a line with slope m then the slope of a line perpendicular to it is

= - = - = 2

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To find the x- intercept let y = 0 in the equation and solve for x

3x - 2(0) + 12 = 0

3x + 12 = 0 ( subtract 12 from both sides )

3x = - 12 ( divide both sides by 3 )

x = - 4

Then x- intercept at (- 4, 0 )

Then

y = 2x + c ← is the partial equation

To find c substitute (- 4, 0) into the partial equation

0 = - 8 + c ⇒ c = 0 + 8 = 8

y = 2x + 8 ← equation of perpendicular line

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47