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ordinate
▪ I. ordinate, a. and n. (ˈɔːdɪnət) Also 4–7 -at. [ad. L. ordināt-us, pa. pple. of ordināre to ordain.] A. ppl. a. and adj. Now Obs. or rare. I. † 1. a. Construed as pa. pple. Ordered, arranged, disposed; ordained, destined, appointed. Obs.1398 Trevisa Barth. De P.R. v. v. (1495) 108 The curtelles o... Oxford English Dictionary
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Abscissa and ordinate
The distance of a point from the x-axis scaled with the y-axis is called the ordinate or y coordinate of the point. The use of the word “ordinate” is related to the Latin phrase “linea ordinata appliicata”, or “line applied parallel”. wikipedia.org
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Sketch of the ordinate set of $f$ Let $f$ be defined on $[0,1] \times [0,1]$ as follows: $f(x,y)= \begin{cases} x+y \mbox{ if } x^2 \leq y \leq 2x^2 \\\ 0 \mbox{ otherwise} \end{cases}$ I want to make a sketch of th...
Comment by OP: > If $f$ is nonnegative, the set $S$ of points $(x,y,z)$ in 3-space with $(x,y)$ in $[0,1]^2$ and $0\le z\le f(x,y)$ is called the ordinate
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Find the least positive value of ordinate of $C$ Let $A(0,2),B$ and $C$ are points on curve $y^2=x+4,$ and such that $\angle CBA=\frac{\pi}{2}.$Then find the least positive value of ordinate of $C$. * * * $y^2...
Let $B(b^2-4,b),C(c^2-4,c)$ where $b\not=\pm 2$. Then, the line (call it $L$) that passes through $B$ and is perpendicular to the line $AB$ is given by $$L : y-b=-\frac{b^2-4}{b-2}(x-(b^2-4))\iff y=-(b+2)x+K$$ where $K=-(b+2)(-b^2+4)+b$. Here, note that the intersection point of $L$ with $y^2=x+4$ o...
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Abscissa, Ordinate and ?? for z-axis? Like x-axis is abscissa, y-axis is ordinate what is z-axis called? It is one of basic doubts from my childhood.
> For 3D diagrams, the names "abscissa" and "ordinate" are rarely used for x and y, respectively. The words abscissa, ordinate and applicate are sometimes used to refer to coordinate axes rather than the coordinate values.
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Co-ordinate of extremities of major axis Ellipse has a focus (3,4), a directrix x+y−1=0 and an eccentricity of 1/2. Using this information I find the equation of ellipse, but I can't find the co-ordinate of the extre...
The line perpendicular to the directrix and passing through the focus, intersects the ellipse at the points you are looking for.
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Co-ordinate Geometry(Incenter) Is there any way of finding the incenter of a triangle, when the equation of all three lines of the triangle are given, without finding out the vertices?
Solving $$ \frac{a_{1} X+b_{1}Y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} =\pm \frac{a_{2} X+b_{2}Y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}} =\pm \frac{a_{3} X+b_{3}Y+c_{3}}{\sqrt{a_{3}^{2}+b_{3}^{2}}} $$ gives one in-centre and three ex-centres.
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Application of derivative - to find interval in which abscissa changes faster than then ordinate Problem : On the curve $x^3=12y$ find the interval of values of x for which the abscissa changes at a faster rate than t...
Note: I am simply following the work the OP has already done if you look at $$ \frac{\mathrm{d}x}{\mathrm{d}y}=\frac{12}{3x^2} $$ then you'll be able to see that $$ \left\lvert \frac{\mathrm{d}x}{\mathrm{d}y} \right\rvert = \frac{12}{3x^2} = 1 \text{ when } x = \pm 2 $$ Looking outside the interval ...
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Area of triangle(Co-ordinate Geometry) Here's the question: A straight line passing through P(3,1) meet the co-ordinate axes at 'A' & 'B'. It is given that distance of this straight line is maximum from origin. Area o...
This line intersects the co-ordinate axes at $(\frac{10}{3},0)$ and $(0,10)$, which are the values of $A$ and $B$ (in no particular order).
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rotating 90 degrees around a circle on a co-ordinate plane !enter image description here I thought the answer would be square root of 3. It would seem that the x co-ordinate of Q would just be the opposite of the x c...
A simple answer to your question is that when you rotate by $90$ degrees (as indicated by the right angle symbol), you swap the $x$ and $y$ coordinates and then negate one or the other, depending on which direction you rotated it. In your case, you had $(-\sqrt{3}, 1)$, which became $(1, -\sqrt{3})$...
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Co-ordinate Geometry (Triangles)(Orthocenter) Is there any way to find out the equation of orthocenter of a triangle, when equation of the three sides of a triangle are given, without finding out the vertices of the t...
If the sides are $L_1 = 0, L_2 = 0, L_3 = 0$, we can find the altitude through the vertices as follows: Find $k$ such that $L_1 + kL_2 = 0$ is perpendicular to $L_3 = 0$. This will give the altitude through the vertex that is the intersection of $L_1 = 0$ and $L_2 = 0$. Similarly find one more altit...
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Finding circumcenter In a triangle A(1,2) B(2,3) C(3,1) and $\angle A = cos^{-1}(4/5)$, $\angle B = \angle C = cos^{-1}(\frac{1}{\sqrt{10} }) $ Ordinate of circumcentre of the $\triangle$ is ? I have tried s...
The angles were probably given because the circumcentre's Y co-ordinate can also be expressed as: $$Y_c=\frac{y_1sin2A+y_2sin2B+y_3sin2C}{sin2A+sin2B+
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Finding Volume of Cube from Co-Ordinate Points If I am given two 3-D points of a cube,how do I find the volume of that Cube?where $(x_1, y_1, z_1)$ is the co-ordinate of one corner and $(x_2, y_2, z_2)$ is the co-ordi...
Compute first the diagonal length $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. An edge of the cube has length $s=d/\sqrt3$ and $V=s^3=d^3/(3\sqrt3)$.
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Why are there only two answers for this co-ordinate geometry question? A co-ordinate geometry question reads "Find value of $p$ such that area of triangle $A (p,2p), B(-1,6), C (3,1)$ is $10$ sq. units. Only 2 values ...
Let $h$ be the altitude of the triangle to the base $\overline{BC}$. Since $BC=\sqrt{41}$, then we must have $h = \dfrac{20}{\sqrt{41}}$. The point $A = (p,2p)$ lies on the line $y=2x$, and there can only be two points on that line that are a distance of $h$ from the line $\overleftrightarrow{BC}$. ...
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Finding arc length of a circle from co-ordinate points I am given three co-ordinate points of a circle O(Ox,Oy) as a center. Then two Points other points as A(Ax,Ay) & B(Bx,By). Now I have to find the arc length of th...
We know the radius of the circle is $OA = \sqrt{(A_x-O_x)^2 + (A_y-O_y)^2}$. We can also find $AB = \sqrt{(A_x-B_x)^2 + (A_y-B_y)^2}$. Since we know $OA$, $OB$ (which is simply $OA$) and $AB$, we can find $\angle AOB$ using the Cosine Rule. Once we have $\angle OAB$, then the arc length $ACB$ is equ...
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