What is the equation of the line that is perpendicular to the given line and passes through the point (2, 6)?

x = 2

x = 6

y = 2

y = 6

The equation of the line that is perpendicular to the given line and passes through the point (2, 6) is y = 6: Option D is correct.

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

y - y0 = m(x-x0)

m is the slope of the line

(x0, y0) is any point on the line

Get the required equation:

Recall that y - y0 = m(x-x0)

y - 6 = 0(x-2)

y - 6 = 0

y = 6

Hence the equation of the line that is perpendicular to the given line and passes through the point (2, 6) is y = 6

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  x = 2

Step-by-step explanation:

The given line is a horizontal line, so the perpendicular line will be something of the form x=constant, a vertical line.

In order for that line to go through a point with x-coordinate 2, the value of the constant must be 2:

  x = 2

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m