what is an equation of the line that passes through the point (-2,5) and is perpinduclar to the line whose equation is y= 1/2x +5

CAN SOMEONE HELP ME SOLVE IT AND TELL ME WHY?

Question

what is an equation of the line that passes through the point (-2,5) and is perpinduclar to the line whose equation is y= 1/2x +5 

CAN SOMEONE HELP ME SOLVE IT AND TELL ME WHY?

calculusfunctions · Answer

If the new line is perpendicular to the given line, then the new line has a slope value that is the negative reciprocal of the slope value of the given line.

anonymous · Answer

The gradient of a line is the value of m when you look at the equation y=mx+c, so for this example it is 1/2. The perpendicular gradient is given as -1/m, so you can work out the perpendicular gradient is -2.

calculusfunctions · Answer

y = mx + b is the equation of any line that has  slope  m  and  y-intercept  b.  Your given line is  y = (1/2)x + 5.  Hence the slope value of the given line is  1/2.  The negative reciprocal of  1/2  is  -2. Thus a line perpendicular to the given line has a slope of  -2.  Hence we have  y = -2x + b.  Now we're told that this line must pass through the point  (-2, 5).  Simply substitute  x = -2  and  y = 5, and solve for b.

anonymous · Answer

(x,y) = (-2,5)
y = -2x +p         =>  5 = -2(-2) +p     => 5 = 4 + p

=>    p = 1      =>

y = -2x +1

anonymous · Answer

Slope of a Perpinduclar Line  = mP
Slope of a Line                     = m

mP = -1/m

your Line :   y = 1/2 x + 5       => slope of line (m=1/2)

Slope of Perpinduclar Line to your Line :     mP = -2

anonymous · Answer

did you get it its >> y = -2x +1 <<