What is the equation of a line that passes through.....?

Question

What is the equation of a line that passes through (1, -4) and has a slope of -2?

Answer 1

y = mx + b
m = -2
y = -2x + b
x = 1 and y = -4
-4 = -2(1) + b
- 4 = -2 + b
-2 = b
y = -2x - 2

Answer 2

The basic equation of a straight line is Y = mX. For a line thru a specific point just insert the coordinates of the point and the slope as follows. For the point (1,-4) and slope -2:

(y + 4) = -2(X-1)

At this point you have the right answer on your paper. A little algebraic manipulation will convert it to whatever form the question asks for.

Answer 3

y+4=-2(x-1)
is the equation of line

Answer 4

y = mx +c where m is the slope, and c is the y-intercept
so -4 = -2(1) + c
so c will be -4 +2 = -2

y = -2x -2

Answer 5

A straight away line perpendicular to 3y-2x-27=0 has a equation 2y+3x+ok=0. it is because the gradient of the former line is two/3(m1). So, a line perpendicular to it must have a gradient (-a million)/(2/3) = -3/2. Now, to guage ok, you position the point (-2,-5) in to the equation 2(-5)+3(-2)+ok = 0 ; or, ok = 16; So, the equation is 3x+2y+16 = 0.

Answer 6

Use the point-slope equation:

y - y1 = m(x - x1)

(x1 , y1) is a point on the line, in your case its (1 , -4)
m is the slope, or -2, here

y - (-4) = -2(x - 1)
y + 4 = -2x + 2
y = -2x -2

Answer 7

y = mx + c pt(1,-4) & m = -2
Fill in the information into the equation.
-4 = (-2)(1) + c
-4 = -2 + c
-2 = c

y = mx + c
y = -2x + (-2)
y = -2x -2 (Answer)

Answer 8

y=mx+c; m=-2
y=-2x+c; (1,-4) must satisfy this equation
-4=-2+c; c=-2
y=-2x-2 ;
2x+y+2=0 is the required equation

Answer 9

y = -2x-4

Answer 10

y=-2x-2