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Consider the equation $y-y_{1}=mleft(x-x_{1}ight)$. In this equation, if $m$ and $x_{1}$ are fixed and different lines are drawn for different values of $y^{1}$, th...

Question

Consider the equation $y-y_{1}=mleft(x-x_{1}ight)$. In this equation, if $m$ and $x_{1}$ are fixed and different lines are drawn for different values of $y^{1}$, then,(A) the lines will pass through a single point(B) there will be one possible line only(C) there will be a set of parallel lines(D) none of these

Consider the equation $y-y_{1}=mleft(x-x_{1} ight)$. In this equation, if $m$ and $x_{1}$ are fixed and different lines are drawn for different values of $y^{1}$, then, (A) the lines will pass through a single point (B) there will be one possible line only (C) there will be a set of parallel lines (D) none of these



Answers

Which of the following is an equation of line $\ell$ in
the $x y$ -plane above?
\begin{equation}
\begin{array}{ll}{\text { A) }} & {x=1} \\ {\text { B) }} & {y=1} \\ {\text { C) }} & {y=x} \\ {\text { D) }} & {y=x+1}\end{array}
\end{equation}

Okay, so this one asks. Which of the following statements is true concerning the lines whose equations are two X plus three wise equal to four. I'm gonna go on and write that and then four x for six y is equal to eight. So the So let's go and go through the answer traces really quit. So first of all, if they're the same line, that basically means that we could just take a constant thatjust a and multiply that by the first equation. And that should give us the second equation. Let's go in and try that. So we know going from two to four or three to six or afford ate. It looks like we're multiplying. But that by two, if we did that to the entire equation two times two X is for X two times three wise six y two times four is eight. So it seems like this is true. That seems like a is the answer. The lines are the same line. Well, let's just go through the other ants traces, to be sure, if they are distinct parallel lines, that means they have the same slope saying Slope a different intercepts Well, we know that's not true, because this one, they hunt same slope, but also the same intercept. So that doesn't work. If they're perpendicular, that means a slope of a, um of a So then from be and should be negative one over m because we have to take the opposite reciprocal. And we know that the sport, because the slopes are actually the same. If the lines intersect but are not perpendicular, that just means that they have different slopes and different intercepts, which we know is not true because they have the same of both.

Hi there. In this problem, we have a plane, which is plain X, so try to stretch it out a little bit This way. So this is plain X. That, of course, is going off continuing on in these direct in all directions here. But this is plain X and we have a line em that is six inches away from X. So this is lying m and the distance between this line and the plane. He's six inches. Yeah, And what we want to know he is if we have a set of points that are six inches away from line em, but one inch away from X. So let's think about that if we think flying them. And we're trying to do a set of points that air six inches away from it, that is going to give us a cylinder. An infinite cylinder that stretches on Okay, Yeah. Or if we were to look at this from an end view. Let's look at this from an end view. This is our plain X This dot represents our line em. It is going straight into and out of the paper. So if I draw my cylinder such that the radius is six inches. This is what my cylinder would look like of all of those points at her six inches from line em. Remember Line, um is going straight in and out of the paper. So it's six inches from our plane. So if we think about this, there are going to be two points on this cylinder where these points actually form lines that are going into the paper themselves. But they're going to be two points on this cylinder here and here that are six inches from M, and yet they are one inch from X. And again, since these are not isolated points thes are also going into and out of the paper That is going to give us two lines that are six inches from M and one inch from X. And these two lines are going to be parallel to him. So that would be if we look at our answer choices here. Yeah, that would be answered. Choice be all right. So we're gonna have to lines parallel to em at our six inches from m yet one inch from B. Yeah, that would be your answer. Thank you so much for watching

Hello, everyone. I hope all is well. Today. I will be helping you with the 16th problem of the chapter on not chapter of the model for test. So this question is asking which of the following equations has graphs or have graphs consisting of two perpendicular lines. So you have the 1st 1 which is X Y equals zero. Then you have the 2nd 1 which is the, um, absolute value of why eagles absolute value of X and then you have the 3rd 1 which is the absolute value of X Y equals one. So if you graft one, um, Graf won consist of the Line X equals zero necks and y equals zero, which are coordinate axes and therefore perpendicular. So we know that one is and then we know graft to consist of the y equals, um why equals the absolute value of X and since we're trying to evaluate it, we also know that there's another graph out. Why equals the negative absolute value of X, which are, um, plus or minus, like 45 degree angles to one another and are therefore perpendicular. So we know one and two are correct, and then we or go along with what the question is asking. And then we have the third graf, which consist of hyper bolas. And so it cannot be that that's why it is D and only one in two are perpendicular. So I hope you found the shovel. And I hope you have a great day. Thank you.

Hello, everyone. I hope all is well today. I'll be helping you with the 17th problem of the model for tests. And this question is asking, Ah, a line em is paralleled to a plain X. So if we had some sort of plain so say this was a plane in some sort of access, there would be line em, which would be like this and then, um and it's six inches from X, right? So there's, ah, six inch difference and the set of points. Thou are six inches from M and one inch ah, from ah, from ex form. What? Right. So a 0.6 inches from M form a cylinder with the M axis, which is tangent to the plain X right, because it's a parallel points one inch from x R. Two planes parallel to ex above, um, above and one below X uh, the cylinder intersex on Lee, one of the planes in to ah, two lines parallel toe em. Right. So that is why it would be beat because it intersect at two parallel lines. Tow them. So I hope you found the shuffle. And I hope you have a great day. Thank you.


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