Question

Write an equation passing through the point (-4,-1) and perpendicular to the line y = 2x - 4

Answer 1

`\displaystyle x + 2y = -6\:or\:y = -(1)/(2)x - 3`

Step-by-step Step-by-step explanation:

Remember, perpendicular equations have OPPOCITE MULTIPLICATIVE INVERCE RATE OF CHANGES, so 2 is altered to −½ as you move forward with plugging the information into the Slope-Intercept Formula:

`\displaystyle -1 = -(1)/(2)[-4] + b \hookrightarrow -1 = 2 + b; -3 = b \\ \\ \boxed{y = -(1)/(2)x - 3}`

Now, suppose you need to write this perpendicular equation in Standard Form. You would follow the procedures below:

y = −½x - 3

+ ½x + ½x

__________

½x + y = −3 [We CANNOT leave the equation this way, so multiply by 2 to eradicate the fraction.]

2[½x + y = −3]

`\displaystyle x + 2y = -6`

With that, you have your equation(s).

I am joyous to assist you at any time.

Answer 2

Hello.

Perpendicular lines have slopes that are opposite reciprocals.

This means we take the slope of the given line, flop it over, and change its sign.

The slope of the given line is 2.

First, we flop it over:

`\mathrm{\displaystyle(1)/(2)}`

Change its sign:

`\mathrm{\displaystyle-(1)/(2) }`

Now, we also have a point that the line passes through, so we can write its equation in Point-Slope Form:

`\mathrm{y-y1=m(x-x1)}`

Plug in the values:

`\mathrm{y-(-1)=-(1)/(2) (x-(-4)}\\\rm{y+1=-(1)/(2) (x+4)`

`\rm{y+1=-(1)/(2) x-2}`

`\rm{y=-(1)/(2) x-3}` (This is our final answer)

I hope it helps.

Have a nice day.

`\boxed{imperturbability}`