How the find the distance between a given point and a line?
up vote
0
down vote
favorite
I would like some help with geometry. I need to find the answer to the following questions
Line L: $y = x + 1$
Point P: = $(-5, -3)$
a: Determine the line that is perpendicular to line L and crosses point P.
b: Calculate the distance between line L and point P. Give your answer with $2$ decimals.
How would one go about calculating this?
geometry
edited Nov 15 at 4:28
PradyumanDixit
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
add a comment |
up vote
0
down vote
favorite
I would like some help with geometry. I need to find the answer to the following questions
Line L: $y = x + 1$
Point P: = $(-5, -3)$
a: Determine the line that is perpendicular to line L and crosses point P.
b: Calculate the distance between line L and point P. Give your answer with $2$ decimals.
How would one go about calculating this?
geometry
edited Nov 15 at 4:28
PradyumanDixit
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like some help with geometry. I need to find the answer to the following questions
Line L: $y = x + 1$
Point P: = $(-5, -3)$
a: Determine the line that is perpendicular to line L and crosses point P.
b: Calculate the distance between line L and point P. Give your answer with $2$ decimals.
How would one go about calculating this?
geometry
edited Nov 15 at 4:28
PradyumanDixit
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
I would like some help with geometry. I need to find the answer to the following questions
Line L: $y = x + 1$
Point P: = $(-5, -3)$
a: Determine the line that is perpendicular to line L and crosses point P.
b: Calculate the distance between line L and point P. Give your answer with $2$ decimals.
How would one go about calculating this?
geometry
geometry
edited Nov 15 at 4:28
PradyumanDixit
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
edited Nov 15 at 4:28
PradyumanDixit
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
edited Nov 15 at 4:28
PradyumanDixit
827214
edited Nov 15 at 4:28
PradyumanDixit
827214
edited Nov 15 at 4:28
PradyumanDixit
827214
827214
asked Nov 15 at 4:05
NovemberDeveloper
31
asked Nov 15 at 4:05
NovemberDeveloper
31
asked Nov 15 at 4:05
NovemberDeveloper
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The line perpendicular to L will have a slope of -1 .Therefore use the general formula of a straight line as $y=mx+c$ .Here $m=1$ calculate c by putting the coordinates of point p .
Next, the distance between a point and a line $(ax+by+c=0)$ is $|(ah+bk+c)|/(√(a^2+b^2)$)
answered Nov 15 at 4:14
Atharva Kathale
809
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The line perpendicular to L will have a slope of -1 .Therefore use the general formula of a straight line as $y=mx+c$ .Here $m=1$ calculate c by putting the coordinates of point p .
Next, the distance between a point and a line $(ax+by+c=0)$ is $|(ah+bk+c)|/(√(a^2+b^2)$)
answered Nov 15 at 4:14
Atharva Kathale
809
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
add a comment |
up vote
0
down vote
accepted
The line perpendicular to L will have a slope of -1 .Therefore use the general formula of a straight line as $y=mx+c$ .Here $m=1$ calculate c by putting the coordinates of point p .
Next, the distance between a point and a line $(ax+by+c=0)$ is $|(ah+bk+c)|/(√(a^2+b^2)$)
answered Nov 15 at 4:14
Atharva Kathale
809
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The line perpendicular to L will have a slope of -1 .Therefore use the general formula of a straight line as $y=mx+c$ .Here $m=1$ calculate c by putting the coordinates of point p .
Next, the distance between a point and a line $(ax+by+c=0)$ is $|(ah+bk+c)|/(√(a^2+b^2)$)
answered Nov 15 at 4:14
Atharva Kathale
809
The line perpendicular to L will have a slope of -1 .Therefore use the general formula of a straight line as $y=mx+c$ .Here $m=1$ calculate c by putting the coordinates of point p .
Next, the distance between a point and a line $(ax+by+c=0)$ is $|(ah+bk+c)|/(√(a^2+b^2)$)
answered Nov 15 at 4:14
Atharva Kathale
809
answered Nov 15 at 4:14
Atharva Kathale
809
answered Nov 15 at 4:14
Atharva Kathale
809
answered Nov 15 at 4:14
Atharva Kathale
809
809
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
add a comment |
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
@NovemberDeveloper the answer should be $y+x+8=0 $ and $1/√2$ ie $0.7071$
– Atharva Kathale
Nov 15 at 4:18
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
Thank you so much!
– NovemberDeveloper
Nov 15 at 4:22
add a comment |
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%2f2999199%2fhow-the-find-the-distance-between-a-given-point-and-a-line%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
