Pre-Alg Question?

Question

Line 1 in perpendicular to the graph of the equation -3x-5y=2 and has points (2,-6). Find the equation for 1.

Answer 1

slope of the given line is -3/5.... so the slope of the required line will be 5/3 (since product of slopes of prependicular lines is -1)

let the equation of the required line be y = mx + c, where m is slope and c is constant term

we have m = 5/3

so y = 5x/3 + c

we know point (2,-6) is on the line... substituting the values of x and y in the above equation, we get

-6 = 5*2/3 + c

-18 = 10 + c

therefore c = -28

so the required line is given by
y = 5x/3 - 28

3y = 5x - 28

5x - 3y = 28

Answer 2

Rewrite the equation in y=mx+b form. Then m will be the slope of the equation. The slope of the reciprocal will be the negative reciprocal of m, or -1/m.

If you have a slope and a line, plug them into the line equation and solve for b.

Specifically for this problem:

Rewrite the equation as y = (-3/5)x -2/5

Slope of the perpendicular line will be 5/3, so the equation of line 1 will be

y = (5/3)x + b

Substitute

-6 = (5/3)(2) +b

b = -6 - 10/3 = -28/3

So the equation for line 1 is y = (5/3)x - 28/3

Hope that helps.

Answer 3

If you know the formula of a line, you can find that of a perpendicular line by switching the coefficients and taking the negative of one of them.

In your situation, from -3x-5y=2 we make

| 5x - 3y = ???

The only problem is the number ????. However, we know that the formula must be true when x = 2 and y = -6:

| 5*2 - 3*(-6) = 10 - (-18) = 28

Therefore, the equation of line 1 is

| 5x - 3y = 28.

Answer 4

Evidently, I made a grievous error:

If line 1 is perpendicular to the other equation, that means that it's slope is -1/m (the opposite recipricol of the slope) of equation 2. Slope (m) of line 2 is -3/5. The opposite recipricol of this is 5/3(slope of line 1). Since it is at the point (2,-6), plug that into the formula y=mx+b where y= -6, m= 5/3, x= 2, and we are solving for b.

-6 = 5/3(2) + b
-6 = 10/3 + b
-28/3 = b

so the equation of our line 1 is y = 5/3x -28/3

Answer 5

Let's find the slope of the given line first:

-5y=3x+2
y=-3/5 x -2/5

This shows you that the given line has slope -3/5.

Lines are perpendicular when their slopes are negative reciprocals. So, you want an equation of a line with slope 5/3 and passing through the point (2,-6).

Uisng point-slope form, an equation of the desired line is
y-(-6)=5/3 (x-2).

Depending on the form of the answer that you teacher wants, you might need to simply this and rewrite it in slope-intercept form.

Remember, lines are parallel if their slopes are the same; they are perpendicular when their slopes are negative reciprocals of each other.

m1=m2 -> parallel
m1=-1/m2 -> perpendicular

This will help you when you take calculus and differential equations, too.

Keep up the good work and keep working hard on your math problems! :)

Answer 6

Here is the answer:

We can arrange the equation of the line, let's call it L2, which is perpendicular to line 1, L1.

-3x -5y =2, 5y = -3x -2
y = (-3x -2)/5, y = (-3/5)x +(-2/5)

so the gradient of L2 is -3/5.
Gradient of L2 * Gradient of L1 = -1 since those lines are perpendicular.

so the gradient of L1 = -1 *(-5/3) = 5/3

equation of L1: y= mx+c, where m is the gradient and c is the y-intercept. Let's take the point (2, -6) for x and y:

-6 = (5/3)*(2) +c
-6 = (10/3) +c
c= -6 -(10/3) = -9.3333333 = -9 1/3 = -28/3

so the equation of L1:

y = (5/3)x -(28/3) = (5x -28)/3
so : 3y = 5x -28

Answer 7

-3x - 5y = 2
-5y = 3x + 2
y = -(3/5)x - (2/5)
Slope = m = -(3/5)

Slope of perpendicular line is the negative reciprocal of slope of other line:

m1 = - (-5/3) = 5/3

Line 1:

y - (-6) = (5/3)(x - 2)
y + 6 = 5x/3 - 10/3
y = 5x/3 - 10/3 - 6
y = 5x/3 - 10/3 - 18/3
3y = 5x - 10 - 18

3y - 5x = -28

Answer 8

Perpidicular lines have opposite slope.
your equation: y = -3/5x - 2/5

opposite slope = 5/3

perp eq. : y = 5/3x + b

plug your numbers and solve for b.