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Of Chapter 510-18 and the major exercisesAssume that the parallel lines and lar segment MM' Prove have that MM' is the common shortest ' perpendicu: ...

Question

Of Chapter 510-18 and the major exercisesAssume that the parallel lines and lar segment MM' Prove have that MM' is the common shortest ' perpendicu: any point of and any point of /' . (Hint: In segment between first dispose of the case in which showing AA' is MM' perpendicular ' AA' , result about Lambert quadrilaterals to [" by means other case by Exercise 22, Chapter and then take care of the 4.) Again, assume that MM' is the common tween / and

of Chapter 5 10-18 and the major exercises Assume that the parallel lines and lar segment MM' Prove have that MM' is the common shortest ' perpendicu: any point of and any point of /' . (Hint: In segment between first dispose of the case in which showing AA' is MM' perpendicular ' AA' , result about Lambert quadrilaterals to [" by means other case by Exercise 22, Chapter and then take care of the 4.) Again, assume that MM' is the common tween / and ' _ Let A and B be perpendicular segment be- any points of and drop perpendiculars AA such that M + A * B and BB to [' Prove that AA (Hint: Use Proposition 4.13; see Figure 6.16.) < BB' . n1J4 # Figure 6.16 and m: Given points A and B that lie on the Given parallel lines P on L, A and P are on opposite side of m from [; Le, for any point opposite sides f m. Prove opposite sides of m, and B and P are on (This holds in any Hilbert that A and B lie on the same side of L plane_ through P and let X be point limiting parallel ray to Let PY be seem intuitively and (Figure 6.17) . It may ' through X_ but this _ on this ray between P parallel _ ray obvious that XY is & limiting not been justified. have Justify the steps that quires proof:



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Refer to the given figure and complete parts (a)-(h) to prove that if two lines are perpendicular and neither line is parallel to an axis, then the lines have slopes whose product is $-1$ (a) In triangle $O P Q,$ angle $P O Q$ is a right angle if and only if $$[d(O, P)]^{2}+[d(O, Q)]^{2}=[d(P, Q)]^{2}.$$ What theorem from geometry is this? (Check your book to see given figure) (b) Find an expression for the distance $d(O, P)$ (c) Find an expression for the distance $d(O, Q)$ (d) Find an expression for the distance $d(P, Q)$ (e) Use your results from parts (b)-(d) and substitute into the equation in part (a). Simplify to show that the resulting equation is $-2 m_{1} m_{2} x_{1} x_{2}-2 x_{1} x_{2}=0$ (f) Factor $-2 x_{1} x_{2}$ from the final form of the equation in part (e). (g) Use the zero-product property from intermediate algebra to solve the equation in part (f) to show that $m_{1} m_{2}=-1$ (h) State your conclusion on the basis of parts (a)-(g).

Given to match the elements of column 12 elements of column to there can be single or multiple matches. So let's start. So with the first part A Given that P is equal to modelers of five. A whole divided by the square root of X squared plus 25. So that would clearly mean that 25 times a square would be cool to after we square on both the sides, it will be equal to 25. A square would be equal to P square into A Square past 25. So since but a squared plus 25 is equal to the square. Since a square piece square and 25 are in AP. I respect it. But since is square P square quantify you, R N A P A squared plus 25 is equal to two p square. So that wouldn't play To into p. to the power of four is equal to 25 a square Which is equal to 25 you sorry to P sq minus a squared into a square? So that would imply U. to the power of four Plus two. Into p. to the power of four -2 is where be square Plus four equal to four. So this will be and the answer for the first party No. Uh Let us go to the second part B. We have three place Cuba too Equal to two and 2 People pacified by two plus C. No no it would imply two people see two people to see minus Q plus seven equal to zero. So since Y is equal to two X plus C passes through the point become a Q. That were therefore you equal to to people. S C. No it wouldn't play two people is to see minus two P Minus seven equal to zero. So therefore it would mean Sequencing to -7. So this is the answer for a second part. Now let us move to the 3rd part C of the question since the given lines are concurrent so we can take the determinant A B. B. Because self sign alfa minus B saying as far minus course alpha zero would be go to zero because we have been given that they've given lines are concurrent. So that would imply ecowas alpha plus B. Sine alpha Press one equal to zero. Now angle between first two lines is 45 degrees. So using the formula of tan theta. So 10 45 degrees would be equal to -8 by Me plus because Alpha by sine alpha upon one place you might be into council for by signing for. Therefore we would be getting minuses in alpha plus because alpha Equal to plus or -1. No, Let this be a question # two. Uh huh. Uh Let this be equation number one. No what's getting further Proceeding further we will be now squaring and adding the two equations one and 2. So equation one, two square place equation too square. So by doing this we will be getting a squared plus B squared equals two. So this is the answer. Now let us proceed to the 4th party. The X coordinate of the point of intersection of the alliance is given by two X plus mx Plus two equal to seven. So that couldn't play X equal to fire upon and place to. And since X is an integer, that would imply exit. That wouldn't play in place to just a minute and place to would be equal to Plus or -1. Hard Plus or -5. So that would imply M. E. would be equal to -3. Hard minus one or three Or -7. So this is the answer. Thank you friends. I hope you like the video.

So this problem we are trying to prove If there So what I'm going to do here has tried to help you guys gain intuition about this problem before going on to the proofing going throughout the blanks here. So ask for the intuition. Let's say here I haven't X and y axis took some corn a plane. I just draw two lights. I'm one one and line to okay. And in the problem, we're given that their slopes are identical to one another here, So let's call this line one. Like to sew the slope off line one here should equal to slope off line too. Well, with the slopes earned exactly the same here. Then the rise of a run Here I make these two triangles here should be exactly the same. Came here. Right. So these two should be course flying triangles. 7 90 degree angle here. And if they're still sort of state here than the course by angle here between the ex assistant, all by itself should also be the same. So how it relates the angle with line here trying easy example where we have a vertical line. Yeah, this is a 90 degree angle carry. That would be your slope that relate to your slope here whereas and I have another vertical line with 90 degree angle. So these two angles Air correspondent, one other. You want to consider that? But nevertheless, they should have, um, very special have the same angles here. Well, then, these three triangles are therefore similar because by angle angle similarly, here, a piece two angles here should also be corresponding as well. Is it sad? About 180 degrees. Yeah, but they are similar trying between the two. Well, but the slopes of the same here and the angle from the excessive access are the same here. Theun intuitively they should be Carol because of the angles were different. Let's say I have a vertical line and something was so like this right then they should intersect. Or if I have two lines here, right, that's very close to Slope Hope. Harry. So something that looks like this If I extend it long enough here, then they will eventually intersect. The only time when they're not going to intersect is when their perilous one other and that comes to the fact that they have the same angle from the X axis there. That's intuition. Problem. Let's look at the proof. So the proof gives ups the fact that the slopes of the two lines are equivalent to one another and they are non vertical, so we can use some trying to make a try and go. You have taken the point on the line and dropping it down to the X axis to vertical line sense. A little trick here to prove our show I'll go here is trying to prove that the two lines are in fact parallel. So we're given that disclosure same and by the definition of slope, to get those two ratios in which the slope of line L is given as BC over a c and the slope off line and is given as e f o r D f here right by the rise of a run. So this would be here in order to get the next step happy, right? Since we know that the slope circle we can set the proportions equal to one another here. So in order to do that, we need to use substitution, right, because if the slopes are the same and we know the values off the slopes here, Then we essentially substituted for the slips. Now, next up here is to rewrite the proportion and here And if you were to rewrite the proportion and here because rewrite it such that we have PC over e f is equal to a C or D f. And when you sell the portion this way here, well, it shows that we are comparing the corresponding sites here. He knows C, B, C and E f or the vertical lines and A c and D f r the faces over the horizontal lines within the triangles that are formed and trying on ABC and travel D E f. So we're using mathematical manipulation thing in order to get this proportion here. Right, So we can revive the proportion family. Now we have the right angles. Concurrence there, Um, so we need to compare the two right angles here so you can see here that the right angles here our angle on a C B care. That's one bright angle here and since all right angles are congruent to one another here, right? We can say that angle. A CB is concluded to d f. Yeah, Okay, so we have that. And the two triangles are similar. A wise that the case, he say, Well, we have the sites being proportional to one another R B C over e f and a c D e f. A right decides our correspond to widely. And we have an angle in between the two sides here. That's congruent. Bye. So we can use SCS similarity here To show that Triangle B A C A B C is similar to triangle D E f and to similar triangles will have corresponding angles. Angles will be concluded here, right? So angle B A C is groomed to angle e d f well, with their corresponding angles. Here You're right. Everything used to correspond the angles. I'll stay over here, just say that the two lines out and on parallel to one over here. Because if I have transfers here with two parallel lines here, their corresponding angles or equivalent to one another in this picture the transfers is the X axis here that completes

So to be good on this problem, where will first find the coordinates of A and B. So we know that A and B have an x coordinate of one because we're given this 10.10 here to the X coordinate of a is one. Now to find the y coordinate with plug in one for X here. So X equals one. Why equals m one? So one Akama and one is our point for a for being. We also have this x coordinate of war. I'm plugging in one for X here you get why equals two? It's really similar. So our coordinates for B R one comma come to So there we go. This is our part A for this problem. Now we'll do part B. So for part B, we need Teoh. Take a look at what's going on here. We want to find Oh, a squared. So what we noticed here is that we actually have a right triangle, all right, in here. And oh, a is the high pop news. And so we couldn't use the Pythagorean theorem. So we know that this leg here is one unit long so one square and we know that this side here is equivalent to the Y value of a and we know that the value of a is one. So one squared plus and once word well equal are high Potter news. Oh, a squared. So this is our first statement that we just verified now for the second part is going to be really similar because we also have a right triangle here with O. B as the high pot use. So again, we have this leg on. The X axis is equal to one. And then the height of this side or the length of the side will be equivalent to either m two or the absolute value of em too. But either way, when we square m two, it becomes positive. So we had negative. I'm two squared in here. We would still get and two squared a squaring Negative things makes some positive. So either way will have a positive value here. And this is equal to the high poverty new switches O b squared. This is our second statement and then we want to do a B square. So taking a look at this a b so from a to B this length here will be equal to a is why value and one minus bees. My value m two. So a B equals M one minus two and then we want to square that. So if we swear a B were squaring em one by this to so will distribute this out and make a M one squared minus two. I'm want them to plus m to school after combining like terms. So this is our third equation that we have now verified. So next for part C, we're going to be setting Oh, a squared plus B squared equal to 80 square. So O k squared IHS one squared, which is just one plus at one squared O B squared is one squared, which is just one plus m two squared and a B squared is on one squared minus 21 I'm two plus mt expert. So now we're going to simplify this. So let's subtract. I'm once word and I'm two squared from both sides doing that on the left hand side we have to and on the right hand side we have negative too times someone him to So then we'll divide by negative too when we get that M one trumps them. Two is equal to negative, but which is exactly what we wanted to get. So we have gotten our final statement and we are now done.

So to be good on this problem, where will first find the coordinates of A and B. So we know that A and B have an x coordinate of one because we're given this 10.10 here to the X coordinate of a is one. Now to find the y coordinate with plug in one for X here. So X equals one. Why equals m one? So one Akama and one is our point for a for being. We also have this x coordinate of war. I'm plugging in one for X here you get why equals two? It's really similar. So our coordinates for B R one comma come to So there we go. This is our part A for this problem. Now we'll do part B. So for part B, we need Teoh. Take a look at what's going on here. We want to find Oh, a squared. So what we noticed here is that we actually have a right triangle, all right, in here. And oh, a is the high pop news. And so we couldn't use the Pythagorean theorem. So we know that this leg here is one unit long so one square and we know that this side here is equivalent to the Y value of a and we know that the value of a is one. So one squared plus and once word well equal are high Potter news. Oh, a squared. So this is our first statement that we just verified now for the second part is going to be really similar because we also have a right triangle here with O. B as the high pot use. So again, we have this leg on. The X axis is equal to one. And then the height of this side or the length of the side will be equivalent to either m two or the absolute value of em too. But either way, when we square m two, it becomes positive. So we had negative. I'm two squared in here. We would still get and two squared a squaring Negative things makes some positive. So either way will have a positive value here. And this is equal to the high poverty new switches O b squared. This is our second statement and then we want to do a B square. So taking a look at this a b so from a to B this length here will be equal to a is why value and one minus bees. My value m two. So a B equals M one minus two and then we want to square that. So if we swear a B were squaring em one by this to so will distribute this out and make a M one squared minus two. I'm want them to plus m to school after combining like terms. So this is our third equation that we have now verified. So next for part C, we're going to be setting Oh, a squared plus B squared equal to 80 square. So O k squared IHS one squared, which is just one plus at one squared O B squared is one squared, which is just one plus m two squared and a B squared is on one squared minus 21 I'm two plus mt expert. So now we're going to simplify this. So let's subtract. I'm once word and I'm two squared from both sides doing that on the left hand side we have to and on the right hand side we have negative too times someone him to So then we'll divide by negative too when we get that M one trumps them. Two is equal to negative, but which is exactly what we wanted to get. So we have gotten our final statement and we are now done.


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