Write an equation of a line that is perpendicular to the line below and passes through the given point

Write An Equation Of A Line That Is Perpendicular To The Line Below And Passes Through The Given Point class=

Sagot :

When you see an equation in the format:
y = mx + c.
Know that the slope = m.

y = -1/5 x +  3.

Therefore slope, m = -1/5.  But for the new line to be perpendicular to this, the new line's slope would be the negative reciprocal of -1/5.

That is using the condition for perpendicularity:
m1* m2 = -1.            m1 = -1/5.

Where m2 = new slope of the line.

m2  =  -1 / (-1/5)  = 5.

So the slope of the perpendicular line is 5.

Using the slope and one point format:

y - y1 = m(x - x1)  here  (x1 , y1) is the point given (2, 8),  x1 = 2 ,  y1 = 8.

m = 5, the new gradient here.

y - 8  = 5(x - 2)
y - 8 = 5x - 10
y = 5x - 10 + 8
y = 5x -2.

So that's the equation of the perpendicular line passing through the given point.
a line perpendicular  would have the opposite slope, meaning that if you multilied the line of this slope and the new slope, you would get -1

in y=mx+b   m=slope and b=yintercept
-1/5 times 5/1=-1 so the new slope is 5
so y=-1/5x+3 
the point you want is 2,8 or this equation is true when x=2 and y=8 are true so subsitute and solve for the new 'b'

8=5(2)+b
8=10+b
subtract 10 from both sides
-2=b

the equaqtion is y=5x-2