Write the equation in slope intercept form of the line passing through?

Question

(-1,-2) that is perpendicular to the line 2x + 5y +8 =0.

Answer 1

The other line has slope:

5y = -2x -8
y = -2/5x - 8/5

m = -2/5

So the new line has slope of 5/2 (negative reciprocal).

Using point-slope equation:

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

Answer 2

A line "x" is perpensicular to line "y" if line x's slope is the negative inverse of line y's slope. If m1 = line x's slope, and m2 = line y's slope then:
m1 = -(1 / m2)

Now find the slope of the equation that we have, we'll put it into slope intercept form (y = mx + b) by solving for y:
2x + 5y + 8 = 0 --- Subtract 8 from both sides...
2x + 5y = -8 --- Subtract 2x from both sides...
5y = (-2x) - 8 --- Divide both sides by 5...
y = -(2/5)x - (8/5)

Now we know that the slope of this equation is -(2/5). We also know that m1 = -(1/m2). Therefore:
m1 = -(1 / (-2/5))
m1 = (5 / 2)

Now that we know the perpendicular line's slope, we can use point slope form: y - y1 = m(x - x1); where (x1, y1) is a known point on the line and m is the slope. We know that one of the points on the line is (-1, -2). We will stick this point, as well as the slope, into this equation:
y - (-2) = (5/2)(x - (-1) --- Simplify...
y + 2 = (5/2)x + (5/2) --- Subtract 2 from both sides...
y = (5/2)x + ½

ANSWER: y = (5/2)x + ½ or y = 4.5x + 0.5

Answer 3

Let 2x + 5y +8 =0 ----- <1>

Finding the slope of <1>:

<1> => 5y = -2x - 8
=> y = (-2/5)x + (-8/5)

slope = -2/5

Then, slope for the desired line = 5/2
(since they are perpendicular)

desired equation:
y = mx + b
-2 = m(-1) + b ==========> b = m - 2

we have the slope = m = 5/2 ======> b = 1/2

so, the line equation: y = (5/2)x + (1/2)


Hope this will be helpful!

Answer 4

the perpendicular line that passes through (-1,-2) is y= (5/2)x+1/2

Answer 5

2x + 5y + 8 = 0
5y = -2x - 8
y = (-2/5)x - 8/5

Slope: -2/5
Perpendicular Slope: 5/2

Eq'n:

(y - y1) = m(x - x1)
y + 2 = (5/2)(x + 1)