Using trigonometric formulas to prove that $m_1m_2$ = -1 for perpendicular lines? Announcing...
Generate an RGB colour grid
Most bit efficient text communication method?
How to tell that you are a giant?
How to write this math term? with cases it isn't working
What is a fractional matching?
How often does castling occur in grandmaster games?
Amount of permutations on an NxNxN Rubik's Cube
Using audio cues to encourage good posture
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
How to react to hostile behavior from a senior developer?
What was the first language to use conditional keywords?
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
How does Python know the values already stored in its memory?
Can a new player join a group only when a new campaign starts?
Drawing without replacement: why is the order of draw irrelevant?
How do living politicians protect their readily obtainable signatures from misuse?
Why is it faster to reheat something than it is to cook it?
A term for a woman complaining about things/begging in a cute/childish way
Taylor expansion of ln(1-x)
Selecting user stories during sprint planning
What is the topology associated with the algebras for the ultrafilter monad?
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
How do I find out the mythology and history of my Fortress?
SF book about people trapped in a series of worlds they imagine
Using trigonometric formulas to prove that $m_1m_2$ = -1 for perpendicular lines?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a way to solve for $x$ in $cos^{-1}(ax) / cos^{-1}(bx) = c$?Angular distribution of lines passing through two squares.Radial velocityIs it true that: $frac {1 } {10 }(sin(y_1+y_2)-sin(x_1+x_2)+y_2-x_2)^2+(cos(x_1+x_2)-cos(y_1+y_2)+x_1-y_1)^2) le (y_1-x_1)^2+(y_2-x_2)^2$?Help with simplifying a trig equationFinding $2$ points from angle.Find the control point of quadratic Bezier curve having only the end-pointsGeometric proof of dot product and projectionSimplifying a parametric equation for the formula of a circle given to point on opposite sidesProving rules for real numbers hold for complex numbers
$begingroup$
How do you use trigonometric formulas (or identities) to prove that the product of the gradients of two perpendicular lines is -1?
If
$y = m_1x + c_1 text{ and } y = m_2x + c_2,$
I thought finding an angle would help to incorporate one of the identities, and hence get somewhere. But how do I find an angle? Constructing two vectors in terms of the y-intercepts and x-intercepts of the two equations like below?
($y_1$ is the y-intercept of the first equation, $x_1$ is the x-intercept of the first equation, and so on.)
$
rightarrow left(begin{array}{c}x_1\ y_1end{array}right) text{•}
left (begin{array}{c}x_2\ y_2end{array}right) \~\
rightarrow left(sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}hspace{0.08cm} right) cos theta = x_1x_2 + y_1y_2 \
theta = cos^{{-}1} left[frac{x_1x_2 + y_1y_2}{sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}} right]
$
But this is taking me nowhere!
Thanks in advance.
trigonometry alternative-proof
asked Mar 25 at 9:38
RamanaRamana
17210
$endgroup$
add a comment |
$begingroup$
How do you use trigonometric formulas (or identities) to prove that the product of the gradients of two perpendicular lines is -1?
If
$y = m_1x + c_1 text{ and } y = m_2x + c_2,$
I thought finding an angle would help to incorporate one of the identities, and hence get somewhere. But how do I find an angle? Constructing two vectors in terms of the y-intercepts and x-intercepts of the two equations like below?
($y_1$ is the y-intercept of the first equation, $x_1$ is the x-intercept of the first equation, and so on.)
$
rightarrow left(begin{array}{c}x_1\ y_1end{array}right) text{•}
left (begin{array}{c}x_2\ y_2end{array}right) \~\
rightarrow left(sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}hspace{0.08cm} right) cos theta = x_1x_2 + y_1y_2 \
theta = cos^{{-}1} left[frac{x_1x_2 + y_1y_2}{sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}} right]
$
But this is taking me nowhere!
Thanks in advance.
trigonometry alternative-proof
asked Mar 25 at 9:38
RamanaRamana
17210
$endgroup$
3
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04
add a comment |
$begingroup$
How do you use trigonometric formulas (or identities) to prove that the product of the gradients of two perpendicular lines is -1?
If
$y = m_1x + c_1 text{ and } y = m_2x + c_2,$
I thought finding an angle would help to incorporate one of the identities, and hence get somewhere. But how do I find an angle? Constructing two vectors in terms of the y-intercepts and x-intercepts of the two equations like below?
($y_1$ is the y-intercept of the first equation, $x_1$ is the x-intercept of the first equation, and so on.)
$
rightarrow left(begin{array}{c}x_1\ y_1end{array}right) text{•}
left (begin{array}{c}x_2\ y_2end{array}right) \~\
rightarrow left(sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}hspace{0.08cm} right) cos theta = x_1x_2 + y_1y_2 \
theta = cos^{{-}1} left[frac{x_1x_2 + y_1y_2}{sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}} right]
$
But this is taking me nowhere!
Thanks in advance.
trigonometry alternative-proof
asked Mar 25 at 9:38
RamanaRamana
17210
$endgroup$
How do you use trigonometric formulas (or identities) to prove that the product of the gradients of two perpendicular lines is -1?
If
$y = m_1x + c_1 text{ and } y = m_2x + c_2,$
I thought finding an angle would help to incorporate one of the identities, and hence get somewhere. But how do I find an angle? Constructing two vectors in terms of the y-intercepts and x-intercepts of the two equations like below?
($y_1$ is the y-intercept of the first equation, $x_1$ is the x-intercept of the first equation, and so on.)
$
rightarrow left(begin{array}{c}x_1\ y_1end{array}right) text{•}
left (begin{array}{c}x_2\ y_2end{array}right) \~\
rightarrow left(sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}hspace{0.08cm} right) cos theta = x_1x_2 + y_1y_2 \
theta = cos^{{-}1} left[frac{x_1x_2 + y_1y_2}{sqrt{(x_1^2 + y_1^2)(x_2^2 + y_2^2)}} right]
$
But this is taking me nowhere!
Thanks in advance.
trigonometry alternative-proof
trigonometry alternative-proof
asked Mar 25 at 9:38
RamanaRamana
17210
asked Mar 25 at 9:38
RamanaRamana
17210
asked Mar 25 at 9:38
RamanaRamana
17210
asked Mar 25 at 9:38
RamanaRamana
17210
asked Mar 25 at 9:38
RamanaRamana
17210
17210
3
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04
add a comment |
3
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04
3
3
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $theta$ is the angle made by the first line with $x-$ axis then the slope of this line is $tan (theta)$. The slope of the second line is $tan (pi /2+theta)=-cot(theta)$. Since $tan (theta)$ $(-cot (theta))=-1$ we are done.
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
add a comment |
Your Answer
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%2f3161568%2fusing-trigonometric-formulas-to-prove-that-m-1m-2-1-for-perpendicular-lines%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
$begingroup$
If $theta$ is the angle made by the first line with $x-$ axis then the slope of this line is $tan (theta)$. The slope of the second line is $tan (pi /2+theta)=-cot(theta)$. Since $tan (theta)$ $(-cot (theta))=-1$ we are done.
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
add a comment |
$begingroup$
If $theta$ is the angle made by the first line with $x-$ axis then the slope of this line is $tan (theta)$. The slope of the second line is $tan (pi /2+theta)=-cot(theta)$. Since $tan (theta)$ $(-cot (theta))=-1$ we are done.
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
add a comment |
$begingroup$
If $theta$ is the angle made by the first line with $x-$ axis then the slope of this line is $tan (theta)$. The slope of the second line is $tan (pi /2+theta)=-cot(theta)$. Since $tan (theta)$ $(-cot (theta))=-1$ we are done.
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
If $theta$ is the angle made by the first line with $x-$ axis then the slope of this line is $tan (theta)$. The slope of the second line is $tan (pi /2+theta)=-cot(theta)$. Since $tan (theta)$ $(-cot (theta))=-1$ we are done.
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
answered Mar 25 at 9:41
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
75.9k53270
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
add a comment |
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
$begingroup$
Thanks a lot, sir.
$endgroup$
– Ramana
Mar 25 at 9:43
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%2f3161568%2fusing-trigonometric-formulas-to-prove-that-m-1m-2-1-for-perpendicular-lines%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
3
$begingroup$
If you know vectors, you can recall that the lines have direction vectors $begin{pmatrix}1\ m_1end{pmatrix}$ and $begin{pmatrix}1\ m_2end{pmatrix}$ respectively. Then note that the lines are perpendicular if and only if their directions vectors are perpendicular (i.e. their dot product is $0$). This will get you the result.
$endgroup$
– Minus One-Twelfth
Mar 25 at 9:43
$begingroup$
Minus One-Twelfth.Right your comment as an snsert , I delete mine.OK?
$endgroup$
– Peter Szilas
Mar 25 at 10:04