Identify the equation of the line that is perpendicular to =12−7 and runs through point (4,−2). Group of answer choices

12y+x = -20

Step-by-step explanation:

Question restructured

Identify the equation of the line that is perpendicular to y =12x−7 and runs through the point (4,−2).

The equation of a line in point-slope form is expressed as;

y-y0 = m(x-x0)

m is the slope

(x0,y0) is a point on the line

Given the equation y = 12x - 7

Slope = 12

Since the required line is perpendicular to this line, the slope of the required line will be;

m = -1/12

Get the required equation

y-(-2) = -1/12 (x - 4)

y+2= -1/12(x-4)

Cross multiply

12(y+2) = -(x-4)

12y+24 = -x+4

12y + x = 4-24

12y+x = -20

Hence the required equation is 12y+x = -20

NB: The equation of the line used in question was assumed

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.