Question

The distance of the line 2x-3y=4 from the point(1,1) in the direction of the line x+y=1 is

Answer 1

solve the two equations and get their intersection point..... use distance formula to find the req distance

Answer 2

they are not intersecting

Answer 3

Solve 2x-3y=4 and x+y=1 Find x and y ........ After Use Distance Formula

Answer 4

as wat kunal said

Answer 5

no their intersection point is 7/5,-2/5

Answer 6

(-2,2) is not the intersection point

Answer 7

now im geting
\[\LARGE \sqrt{\frac{53}{25}}\]

Answer 8

\[\LARGE \sqrt{ ( \frac{-2}{5}-1 )^2+(\frac{7}{5}-1)^2}\]

Answer 9

yes it is....

Answer 10

Its wrong sorry

Answer 11

then whats the answer??

Answer 12

\[\LARGE \sqrt{2}\]

Answer 13

they dont even intersect lol what are u guys saying..we cant solve them together

Answer 14

they both are parallel lines..and parallel lines NEVER intersect

Answer 15

is this from analytic geometry?

Answer 16

co-ordinate

Answer 17

do you know the formula for the distance of a point from a line?

Answer 18

see... the equations are 2x-3y=4 and x+y=1 their slope are not the same therefore they got to intersect

Answer 19

i know the one for perpendicular one..

Answer 20

okay. we'll use that.

first, what is the slope of the given line \(2x-3y=4\)?

Answer 21

2/3

Answer 22

right. so the line \(\bot\) has slope __ ?

Answer 23

-3/2

Answer 24

good.
now you have the slope of the \(\bot\) line and a point on that line, \((1,1)\). Can you find the equation of that line?

Answer 25

\[\large (y-1)=\frac{-2}{3}(x-1)\]

Answer 26

3y+2x-5=0

Answer 27

close. it should be \[(y-1)=\frac{-3}{2}(x-1)\]or\[3x+2y-5=0\]

Answer 28

ah sorry.!

Answer 29

now we have two intersecting lines, \(2x-3y=4\) and \(3x+2y-5=0\). can you find the point of intersection of the two lines?

Answer 30

y=-22/14
or
\[\LARGE y= \frac{-\pi}{2}\] :P

Answer 31

where pi=22/7

Answer 32

@DLS
see first of all find the eq of line parallel to x+y=1 and passing through (1,1)

the line passing through 1,1 and parallel to x+y=1
is x+y=2
now we have to find the point of intersection of 2x-3y=4 and x+y=2
which is (2,0)
now we need to find the distance between (1,1) and (2,0) using distance formula
thus reqd distance =sqt(2)
ans is sqrt(2) units ...

Answer 33

@DLs
first check is the ans sqrt(2)

Answer 34

yes

Answer 35

I had a hint on this question..which can be possibly shorter..

Answer 36

ooh, i didn't get that part, "in the direction of the line".

Answer 37

Any point on the line through(1,1) and parallel to x+y=1 is
\[p(1+ rcos \theta,1+r \sin \theta) where \tan \theta=-1.\]
Use the fact that P is on the line
\[2x-3y=4\]
and find |r|.

Answer 38

what does it mean :/

Answer 39

@DLS
then follow the method as i have done ...
i am certain abt it

Answer 40

the question is that there is a line parallel to x+y=1, passing through (1,1) , and the line intersects 2x-3y =4 somewhere .
we need to find the distance between (1,1) and that common co-ordinate..

Answer 41

sorry i didnt notice @matricked has already explained that part..

Answer 42

why did we find the point of intersection? how do we know it will intersect

Answer 43

since the lines are not parallel, ofcorse they'll intersect.. !

Answer 44

i see..thanks! everyone