Question

Which equation describes a line that passes through (-6,8) and is perpendicular to the line described by y=2x-4

Answer 1

The equation of the line that passes through (-6, 8) and is perpendicular to y = 2x - 4 is y = -1/2x + 5.

Step-by-step explanation:

To find the equation of a line that is perpendicular to a given line, we need to determine the negative reciprocal of the slope of the given line. The slope of the given line y = 2x - 4 is 2. Therefore, the slope of the perpendicular line would be -1/2.

Using the point-slope formula, y - y1 = m(x - x1), where (x1, y1) is the given point (-6, 8) and m is the slope (-1/2), we can substitute the values and simplify the equation to find the equation of the perpendicular line.

Thus, the equation of the line that passes through (-6, 8) and is perpendicular to y = 2x - 4 is y = -1/2x + 5.

Answer 2

Step-by-step explanation:

Hey, there!!

The given is a point (-6,8) through which a line passes. And is perpendicular to the line y = 2x-4

The equation for point (-6,8) is,

(y-8)= m1(x+6)...........(i)

and given equation is y = 2x-4............(ii)

Now, from equation (ii).

slope (m2)= 2 { as equation (ii) is in the form of y= mx+c where m is a slope}.

Now, For perpendicular,

m1×m2= -1

m1×2= -1

`m1 = ( - 1)/(2)`

Therefore, m1 = -1/2.

Putting, the value of m1 in equation (i).

(y-8) = -1/2×(x+6)

2(y-8)= -1(x+6)

2y - 16 = -x -6

x+2y-10 = 0......... is the required equation.

Hope it helps...