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Find the distance between $(x-y, y-x),(x+y,$, $x+y$ )....

Question

Find the distance between $(x-y, y-x),(x+y,$, $x+y$ ).

Find the distance between $(x-y, y-x),(x+y,$, $x+y$ ).



Answers

Fill in the blanks.
The distance between the point $\left(x_{1}, y_{1}\right)$ and the line $A x+B y+C=0$ is given by $d=$ ______.

We have to find the distance between the 0.11 on the given light. They know that the distance between airline on the point is given by a X one plus B y, even plus C. It's marvelous. Divided by wrote off a squared plus B squared when the line occasion off the line is expressed US X Plus B by plus C called a zero, and the point is X one violent. So here we have to express the given line in the standard. For that this X minus y plus one equal to zero. This is the expression off line we have equal to one be equal to minus one and see is one so writing there is sin formula. What? Substituting for distance formula exponents one. So one of the mine, these minus one minus one into one. Let's see, is one its magnitude divided by square root off a squared plus B squared. So that is one plus one. It is equal to one by route to or which is 0.707 This is the distance between the given point and the light

We have our line and the point, and we have to find the distance between the point and the light. Let us consider a line. Eggs, plus B Y plus C is equal to zero on the point X one by one. Then we can write the distance between the line and the point as they magnitude off eggs. One place be very one plus C divided by square it off a square, plus the square. In our question. We had to express the equation off line in the standard form, which is two X minus Y, minus five is equal to zero. Then we can find the distance that is the sequel to magnitude off two in the one plus three minus five, divided by squared off to square plus what which is equal to zero. That means the distance between the point and Linus zero implies the point is playing on like line Yes, corn plane. Okay,

In problems. 74 were given this number line with two points on it. And I have copied down this formula from for the distance between two points from our book. And this formula just says that the distance from two points A and B is equal to the absolute value of B minus A. So our first step in solving this problem is going to be figuring out what are two points are. So our first point, which I'm going to go ahead and call a looks to be right in between Negative too and negative. Three. So that means that it's negative two and 1/2 or negative 2.5. And our second point right here is exactly in between points one and two. So we're going to called B, our second quite equal to 1.5. So now, to find the distance between these two points, we're just going to plug them into our formula right here. So the distance from our point A, which is negative 2.5 to our point B, which is 1.5, is equal to the absolute value of 1.5 minus negative 2.5 and I put these parentheses here so that we make sure that we don't get confused with our our negative signs right there. Because when we solve for what is inside of the absolute value, we're going to get 1.5 plus 2.5 because we're are distributing our negative number here. Negative times negative is a positive. So when we add these two numbers, together we get 1.5 plus 2.5 is equal to four, and then we're going to go ahead and substitute this back into our our absolute value. And the absolute value of four is simply equal to four. So the distance between the two points on the number line is for

So in this question, we need to find a distance between two lines. Let's see if you can use the formula to find the distance between a point and a line to help us up. So first we need to put the equation of this second line or this first line. That's put the question his first line in standard for So when we do that, we'll get why minus two X equals zero. So a equals negative, too. B equals one Now the equation for the formula between a point and a line was absolute value. A x o plus B y o square root of a squared plus B squared. So let's choose the 0.0 comma five. From this equation. That's just the wind. So now we have our information. Let's go ahead and plug it. Solving this is going to get us five over the square root of five just equal to five. Route five over five, which is simply go to school through five. So the distance between the two lines is square roof


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