High school geometry
Finding a quadrilateral from its symmetries (example 2)
Two of the points that define a certain quadrilateral are (-4,-2) and (0,5). The quadrilateral has a reflective symmetry over the lines y=x/2 and y=-2x+5. Draw and classify the quadrilateral. Created by Sal Khan.
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- why does the bisectors being perpendicular have anything to do with the side equality of the quadrilateral? please explain this to me(12 votes)
- First off, the property of parallelograms are that opposite sides are parallel and congruent. The property of kites are that diagonals are perpendicular. So the only thing that is both a kite and a parallelogram is a rhombus (since it cannot be a square because the slopes of sides are not perpendicular. We can always check by Pythagorean Theorem:
7^2 + 4^2 = 49+16 = 65 so these sides are √65
8^2 + 1^2 = 64 + 1 = 65 so these sides are also √65
Since all sides are equal, and vertices are not perpendicular, we verify that rhombus is the correct answer.(2 votes)
- You explained it very well its just hard to understand because your not working it out in equations and stuff(9 votes)
- There are little to no equations in finding a quadrilateral from its symmetries.(4 votes)
- This literally doesn't make any sense, any guidance?(9 votes)
- So if the lines of symmetry on any quadrilateral are perpendicular, is the shape always symmetrical?(5 votes)
- If a figure has any line of symmetry, it must be symmetric.(8 votes)
- Why are the diagonals of a kite perpendicular? I thought kites were defined by having 2 pairs of equal adjacent sides. Thanks in advance!(4 votes)
- Draw a kite ABCD. (Actually draw it.) Draw the diagonals, and say they intersect at E.
Because it's a kite, AB=BC. So triangle ABC is isosceles, and angle BAE=angle BCE.
Also, because it's a kite, CD=DA, and BD is equal to itself. So triangles ABD and CBD are congruent by SSS.
So their corresponding angles are congruent. So angle ABE=angle CBE.
So by SAS, triangle AEB is congruent to triangle CEB.
So angle AEB is congruent to angle CEB.
But AEB and CEB are also supplementary. So they must be right angles.
So the diagonals of a kite are perpendicular.(6 votes)
- How did he get the reflection line of y=-2x+5? He got y=0 and x=5(4 votes)
- Ok, I solved this on my own.
Choose x = 0 and solve for y. y = -2x + 5. y = (-2) (0)(multiply -2 by 0 and you get 0) + 5 = 5. Now you are left with y = 5
Which gives us x = 0, and y = 5, which is Sal's answer.
Hope this helps!
- There isn't a comprehensive video on rotational symmetry and no space to ask on the practice tests but the test question reads:
One of the points that defines a certain quadrilateral is (1,1). The quadrilateral has rotational symmetry of 90∘ degrees about the point (−4,−4)
And the first answer/hint reads:
Performing the rotation brings the starting point to (−9,1)
How do people know this? Using what formula did they arrive at this answer? Have I missed a video? Intro to rotational symmetry doesn't answer any of these questions and certainly doesn't give sufficient examples.
Also, on the same practice test, they ask if lines dividing a given geometric figure do so in such a way that creates symmetry and some questions you can eyeball but some are very precise and only off by a slight margin and in the answers it just shows arrows - how do they decide the slope of these arrows to ensure symmetry? How should I do this at home? Should I be calculating the slope of each line segment to ensure all slopes are identical? And if so, was this in a video that I missed somewhere?
P.S. Love Khan Maths program - life changing(5 votes)
- In the Symmetry of 2D shapes exercise you have to calculate the order of rotational symmetry of some regular figures. I was wondering how does one determine the order of rotational symmetry of non-regular figures?(4 votes)
- Hi there, why 2 lines with slopes of m and -1/m would be perpendicular? Thanks.(3 votes)
- Let the lines
𝑦 = 𝑚₁𝑥 + 𝑏₁ and 𝑦 = 𝑚₂𝑥 + 𝑏₂
be perpendicular to each other.
Shifting the lines vertically doesn't change the fact that they are perpendicular,
so it's actually enough to consider the lines
𝑦 = 𝑚₁𝑥 and 𝑦 = 𝑚₂𝑥
– – –
One point on the first line is 𝑃₁ = (1, 𝑚₁)
One point on the second line is 𝑃₂ = (1, 𝑚₂)
Also, the lines intersect at the origin 𝑃₃ = (0, 0)
– – –
Because the two lines are perpendicular, these three points form a right triangle, with the right angle at 𝑃₃.
According to the Pythagorean theorem we then have
|𝑃₂ − 𝑃₁|² = |𝑃₃ − 𝑃₁|² + |𝑃₃ − 𝑃₂|²
– – –
Using the distance formula, we find
|𝑃₂ − 𝑃₁| = √((1 − 1)² + (𝑚₂ − 𝑚₁)²) = √(𝑚₂ − 𝑚₁)²
⇒ |𝑃₂ − 𝑃₁|² = (𝑚₂ − 𝑚₁)²
|𝑃₃ − 𝑃₁| = √((0 − 1)² + (0 − 𝑚₁)²) = √(1 + 𝑚₁²)
⇒ |𝑃₃ − 𝑃₁|² = 1 + 𝑚₁²
|𝑃₃ − 𝑃₂| = √((0 − 1)² + (0 − 𝑚₂)²) = √(1 + 𝑚₂²)
⇒ |𝑃₃ − 𝑃₂|² = 1 + 𝑚₂²
– – –
(𝑚₂ − 𝑚₁)² = 1 + 𝑚₁² + 1 + 𝑚₂²
⇒ 𝑚₂² − 2𝑚₂𝑚₁ + 𝑚₁² = 2 + 𝑚₁² + 𝑚₂²
⇒ −2𝑚₂𝑚₁ = 2
⇒ 𝑚₂ = −1∕𝑚₁(3 votes)
- Sal said that two lines are perpendicular when one of the slopes are the negative reciprocal of the other. How would you tell if they are parallel?(3 votes)
- They have the same slope but different y intercepts. So if you start with y - 2x + 4, parallel lines would look like y = 2x - 3 or y = 2x + 3 and perpendicular lines would look like y = -1/2 x - 3 or y = -1/2 x + 4.(2 votes)
Two of the points that define a certain quadrilateral are negative 4 comma negative 2. So let's plot that. So that's negative 4 comma negative 2. And 0 comma 5. So that's 0 comma 5 right over there. The quadrilateral is left unchanged by a reflection over the line y is equal to x over 2. So what does that line look like? y is equal to x over 2. I'll do that in the blue. y is equal to x over 2. So when x is equal to 0, y is 0. The y-intercept is 0 here. And the slope is 1/2. Every time x increases by 1, y will increase by 1/2. Or when x increases by 2, y will increase by 1. So x increases by 2, y increases by 1. X increases by 2, y increases by 1. Or another way to think about it, y is always 1/2 of x. So when x is 4, y is 2. When x is 6, y is 3. When x is 8, y is 4. So we can connect these. Let me try my best attempt to draw these in a relatively straight line. And then I can keep going. When x is negative 2, y is negative 1. When x is negative 4, y is negative 2. So it actually goes through that point right there. And it just keeps going with a slope of 1/2. So this line, and I can draw it a little bit thicker now, now that I've dotted it out. This is the line y is equal to x over 2. And they also say that the quadrilateral is left unchanged by reflection over the line y is equal to negative 2x plus 5. So the y-intercept here is 5. When x is 0, y is 5. So it actually goes through that point. And the slope is negative 2. Every time we increase by 1-- or every time we increase x by 1, we decrease y by 2. So that would go there. We go there. And we keep going at a slope of negative 2. So it's going to look something like this. It actually goes through that point and just keeps going on and on. So this is my best attempt at drawing that line. So that is y is equal to negative 2x plus 5. Now let's think about it. Let's see if we can draw this quadrilateral. So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over 2. So this is the line y is equal to x over 2. This magenta point, the point negative 4, 2 is already on that line. So it's its own reflection, I guess you could say. It's on the mirror is one way to think about it. But we can easily reflect at this line over here. This line, if we were to drop a perpendicular-- And actually, this line right over here, y is equal to negative 2x plus 5 is perpendicular to y is equal to x over 2. How do we know? Well if one line has a slope of m, then the line that's perpendicular would be the negative reciprocal of this. It would be negative 1 over m. So this first line has a slope of 1/2. Well what's the negative reciprocal of 1/2? Well the reciprocal of 1/2 is 2/1. and you make that negative. So it is equal to negative 2. So this slope is a negative reciprocal of this slope. So these lines are indeed-- I'm trying to erase that-- perpendicular. So we literally could drop a perpendicular, literally go along this line right over here, in our attempt to reflect. And we see that we're going down 2, over 1 twice. So let's go down 2, over 1, down 2, over 1, twice again. Let me do that in that same color. The reflection of this point, across y is equal to x/2 is this point right over there. So now we have three points of our quadrilateral. Let's see if we can get a fourth. So let's go to the magenta point. The magenta point we've already seen. It's sitting on top of y equals x/2, so trying to reflect it doesn't help us much, but we could try to reflect it across y is equal to negative 2x plus 5. So once again these lines are perpendicular to each other. Actually let me mark that off. These lines are perpendicular. So we can drop a perpendicular and try to find its reflection. So we're going to the right 2 and up 1. We're doing that once, twice, three times on the left side. So let's do that once, twice, three times on the right side. So the reflection is right there. We essentially want to go to that line. And however far we were to the left of it, we want to go that same, that bottom left direction, which you want to go in the same direction, to the top right, the same distance to get the reflection. So there you have it. There is our other point. So now we have the four points of this quadrilateral. Four points of this quadrilateral are-- or the four sides, let me actually just draw the quadrilateral. We have our four points. So this is one side right over here. This is one side, right over here. This is another side, right over here. And you can verify that these are parallel. How would you verify that they're parallel? Well they have the same slope. To get from this point to that point, you have to go over, so your run has to be 4, and you have to rise 1, 2, 3, 4, 5, 6, 7. So the slope here is 7/4. Rise over run, or change in y over change in x, is 7/4. And over here, you go 1, 2, 3, 4. So you run 4 and then you rise 1, 2, 3, 4, 5, 6, 7. So the slope here is also 7/4. So these two lines are going to be parallel. And then we could draw these lines over here. So this one at the top. The one at the top right over there. And what's its slope? Well, let's see. We go from x equals 0 to x equals 8. So we go down, our change in y is negative 1 every time we increase x by 8. So this is the slope of-- slope is equal to negative 1/8. And that's the exact same slope that we have right over here. Negative 1/8. So these two lines are parallel as well. This line is parallel to this line as well. So at minimum, we're dealing with a parallelogram. But let's see if we can get even more specific. Because this kind of looks like a rhombus. It looks like a parallelogram, where all four sides have the same length. So there's a couple of ways that you could verify that this parallelogram is a rhombus. One way is you could actually find the distance between the points. We know the coordinates, so you could use the distance formula, which really comes straight out of the Pythagorean theorem. Or even better, you could look at the diagonals of this rhombus-- we could look at the diagonals of this parallelogram. We are trying to figure out if it's a rhombus. And if the diagonals are perpendicular, then you're dealing with a rhombus. And we've already shown that these diagonals, that this diagonal, this diagonal, and this diagonal are perpendicular. They intersect at right angles. And so this must be a rhombus.