Linear Equation

[untold ravine]

Need help on how to solve this and how to remember and identify which form I wound need to use. I am barely passing Grade 10 math by 53% and in need of desperation

[plush grove]

You need to post actual questions for us to help with. Linear equations is a very large topic.

[sleek inlet]

@untold ravine

[untold ravine]

Word problem @plush grove

[leaden needle]

The slopes of perpendicular lines are the opposite reciprocals of each other.

The slopes of parallel lines are the exact same.

[tender citrus]

which question?

[untold ravine]

All

[tender citrus]

q 9. Parallel = same gradient

, put x+2y+1 = 0 into y=mx + b

so 2y=-x-1

y= -1/2x -1/2

to find y int, let x = 0

so y = -1/2

point is now (0,-1/2)

y - y1 = m (x - x1)

use that

to find equation

using all the into above

[untold ravine]

@tender citrus wait so if I want it to be X I make what 0

Also it would be better explained if it was written

[velvet breach]

Looks fine to me

what do you mean by “want it to be x”

[untold ravine]

Does he mean to let 0 be the x position for y-y1=m(x-x1)

Or let x be 0 when only finding Y

[velvet breach]

They’re not finding y

That’s not what it means

You can’t find y

You can find the y intercept

Which is the point on the line where x is 0

[untold ravine]

So ur telling me

y=-1/2=m(x-0)

[abstract bolt]

@untold ravine What is your problem with on linear equations?

[rapid bay]

No, the -1/2 is the y intercept, which is where the line intersects the y axis. Point intercept form of a linear equation is (y-y1) = m(x-x1), so if you have the point (0, -1/2) then x1 =0 and y1 = -1/2, and you can just substitute those into the equation.

[tender citrus]

Yes

[tender citrus]

The point (0,-1/2) lays ON the line, to find the eq of a line you need a point which is found here, so 0 is x1 and -1/2 is y1

Using y-y1 = m(x-x1)

You know m = 2/5

Y - (-1/2) = 2/5(x - 0)

Y + 1/2 = 2/5x - 0

Y = 2/5x - 1/2

^^ new eq

[modern harbor]

.close