How many points of intersection are there between line a and line b if they contain the points listed? line a: (2, 8) and (–2, –4) line b: (4, 10) and (–3, –11)

No intersection

Zero intersections

Step-by-step explanation:

Let's determine the slope first.

If the slopes are different, then there is one solution.

If the slopes are the same, there are 2 possibilities.  The first possibility is that there is no solutions because the lines are parallel.  The second possibility is that there is infinitely many solutions because they are the same line. When I say solution, I'm also referring to intersection.

So I'm going to find the slope by lining up the points and subtracting vertically then putting 2nd difference over 1st difference.

Let's do that for line A:

 (  2  ,   8)

- (  -2 , -4)

  4        12

So the slope is 12/4 or just 3.

Let's do this for line B now:

  (  4  ,  10)

-  ( -3  ,  -11)

    7     21

So the slope is 21/7 or just 3.

So we have more work now. The lines either are the same or parallel.

We are going to use this to determine if they same or parallel, we are going to find the slope-intercept form of the equation for both lines.

That is y=mx+b where m is slope and b is y-intercept.

Let's look at line A:

y=mx+b with m=3 and a point (x,y)=(2,8)

8=3(2)+b

8=6+b

2=b

So the line is y=3x+2

Let's look at line B.

y=mx+b with m=3 and a point (x,y)=(4,10)

10=3(4)+b

10=12+b

-2=b

The equation of this line is y=3x-2

So the lines y=3x+2 and y=3x-2 are not the same, they are parallel which means they intersect zero times.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

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step-by-step explanation: