graphing???

Question

whats the equation of the line perpendicular to the line y = 1/2(x) - 3 and passes through the point (1, 5)?

Answer 1

its perpendicular to 1/2(x) - 3

so its gradient must be -2 (negative reciprocal)

using the y = mx + b general form of an equation

we get

5 = -2*1 + b

b = 7

so the equation if y = -2x + 7

Answer 2

y = 1/2(x)

m = 1/2
+m = -2

Y-Y{sub1} = m (X-X{sub1})
Y-5 = -2 (X-1)
Y-5 = -2X + 2
+5 + 5
Y = -2X + 7

Answer 3

y = 1/2(x) - 3
the slope of this line is 1/2

product of the slope of the given line & perpendicular line should be -1

1/2*m=-1
m=-1*2
m=-2

so slope of the perpendicular line is -2
so we have the slope and point. using poin slope formula we can find the equation of the line

slope =-2 point (1, 5)
the point slope formula is
(y-y1)=m(x-x1)
(y-5)=-2(x-1)
y-5=-2x+2
y=-2x+2+5
y=-2x+7

Answer 4

The slope of that line would be the negative reciprocal of the line y=1/2(x). So the slope for the perpendicular line is -2

You plug in the given point in the point-slope equation y-y1 = m(x-x1):
y - 5 = -2 (x - 1)

Working everything out:

y - 5 = -2x + 2
y = -2x + 7 (Answer).

Answer 5

m1 = 1/2
m2 = - 2

y - 5 = (- 2) (x - 1)
y = (- 2) x + 2 + 5
y = (- 2) x + 7