Mixing
Equation of pair of reflected straight lines given the equation of pair of incident straight lines
$begingroup$
If $ax^2 + 2bxy + by^2 = 0$ represents a pair of lines, then find the combined equation of lines that can be obtained by reflecting these lines about the x-axis.
I know that this can be done by separating the lines from the given equation and then finding the reflected rays and then combining the reflected rays again, but I was wondering if there is a short way of solving this problem?
analytic-geometry coordinate-systems
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
$endgroup$
add a comment |
$begingroup$
If $ax^2 + 2bxy + by^2 = 0$ represents a pair of lines, then find the combined equation of lines that can be obtained by reflecting these lines about the x-axis.
I know that this can be done by separating the lines from the given equation and then finding the reflected rays and then combining the reflected rays again, but I was wondering if there is a short way of solving this problem?
analytic-geometry coordinate-systems
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
$endgroup$
add a comment |
$begingroup$
If $ax^2 + 2bxy + by^2 = 0$ represents a pair of lines, then find the combined equation of lines that can be obtained by reflecting these lines about the x-axis.
I know that this can be done by separating the lines from the given equation and then finding the reflected rays and then combining the reflected rays again, but I was wondering if there is a short way of solving this problem?
analytic-geometry coordinate-systems
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
$endgroup$
If $ax^2 + 2bxy + by^2 = 0$ represents a pair of lines, then find the combined equation of lines that can be obtained by reflecting these lines about the x-axis.
I know that this can be done by separating the lines from the given equation and then finding the reflected rays and then combining the reflected rays again, but I was wondering if there is a short way of solving this problem?
analytic-geometry coordinate-systems
analytic-geometry coordinate-systems
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
asked Jun 24 '14 at 15:12
kartikeykant18kartikeykant18
2511315
2511315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reflection about the x-axis can be rapidly obtained substituting $y$ with $-y$. So you obtain $ax^2-2bxy+by^2$. Also note that, in this case, the new lines are symmetrical to the initial Iines with respect to the y-axis as well.
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
$endgroup$
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f845966%2fequation-of-pair-of-reflected-straight-lines-given-the-equation-of-pair-of-incid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The reflection about the x-axis can be rapidly obtained substituting $y$ with $-y$. So you obtain $ax^2-2bxy+by^2$. Also note that, in this case, the new lines are symmetrical to the initial Iines with respect to the y-axis as well.
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
$endgroup$
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
add a comment |
$begingroup$
The reflection about the x-axis can be rapidly obtained substituting $y$ with $-y$. So you obtain $ax^2-2bxy+by^2$. Also note that, in this case, the new lines are symmetrical to the initial Iines with respect to the y-axis as well.
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
$endgroup$
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
add a comment |
$begingroup$
The reflection about the x-axis can be rapidly obtained substituting $y$ with $-y$. So you obtain $ax^2-2bxy+by^2$. Also note that, in this case, the new lines are symmetrical to the initial Iines with respect to the y-axis as well.
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
$endgroup$
The reflection about the x-axis can be rapidly obtained substituting $y$ with $-y$. So you obtain $ax^2-2bxy+by^2$. Also note that, in this case, the new lines are symmetrical to the initial Iines with respect to the y-axis as well.
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
answered Jun 24 '14 at 18:08
AnatolyAnatoly
11.9k21538
11.9k21538
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
add a comment |
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
why not substituting $x$ with $-x$
$endgroup$
– kartikeykant18
Jun 27 '14 at 7:37
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
$begingroup$
In general, to reflect about the x-axis, you have to substitute $y$ with $-y$, not $x$ with $-x$. Your function is an exception because is symmetrical, so that changing the sign to either $x$ or $y$ gives the same result.
$endgroup$
– Anatoly
Jun 27 '14 at 22:32
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f845966%2fequation-of-pair-of-reflected-straight-lines-given-the-equation-of-pair-of-incid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
