IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

vyppi

Find the equation of the line passing through the point (2,3) & making intercepts of length 2 units between the lines y + 2x = 3

& y + 2 x = 5

isomniac

assume a general line passing thru 2,3 with a unknown slope m

y-3=m(x-2)

now intersect it with 2 known lines get pts of intersection

now the distance bw these pts will be 2

get the pts and ull hav the line

try it if u dont get it ill solve it

vyppi

buddy its not so easy...u wont get the 2 pts....its an IIT[1991] ques.....try harder buddy.....by the way gud try...

vyppi

cmon guys yaar ye mera doubt hai...plz clear it..

hary

i think this is da sol but im not sure

see da given lines are parallel right and distance bet them is 2

n da line also makes itercept of 2 so da req line must be perpendicular to both lines so slope of req line is 1/2

passes through (2,3)

so da eq is x-2y+4=0

is it correct??

hary

is it correct???

vyppi

nahi yaar hary....who said that the distance b\w them is 2.....i mean there is a line(it may or may not be perpendicular)...(in this case it is not)..(it is slanting) and it is covering a distance of 2units b\w them...this is the interpretation of the ques...chalo ny ways atleast u tried..ill rate u

allamraju

Let k be the angle made by the line with x-axis.Draw a rough figure and observe

Tan^-1(1/2) is the angle made by the perpendicular between the lines[since it's slope is 1/2],as the line makes an intercept of 2 units,perpendicular distance=2/rt5,hence the angle between perpendicular and given line is Tan^-1(2).Hence,k=Tan^-11/2+Tan^-12= $\Pi$ /2.So,it is a vertical line and it's eqn is x=2

hary

@allamraju i didnt get ur sol can u b more clear

sriram.a

I got x=2 or y=3

sriram.a

plot the graph of y+2x=3&y+2x=5

From the equation its clear that distance between their X-coordinates&Y-cordinates is 2

the given distance is 2,so the line should be parallel to X-axis or Y-axis.

So,x-2=0 or y-3=0

Hope you got it

Greatdreams

1 method of solving these type of problems :

Given straight lines : y + 2x = 3 and y + 2x - 5 = 0

They pass through the point , (2,3)

r = - [ ax1 + by1 + c / (acos $\theta$ + b sin $\theta$ ]

where (x1 , y1) = 2,3

Hence r = - [ 2(2) + 1(3) - 3 / 2 cos $\theta$ + sin $\theta$ ] = - ( 4 / 2 cos $\theta$ + sin $\theta$ )

Now its given that it makes an intercept of length 2 units , hence

r+2 = - [ 2(2) + 1(3) - 5 / 2 cos $\theta$ + sin $\theta$ ] = - ( 2 / 2 cos $\theta$ + sin $\theta$ )

Hence subtracting the two we get ,

r+2 - r = 2 = 2 / (2cos $\theta$ + sin $\theta$ )

Hence 2 cos $\theta$ + sin $\theta$ = 1

or 4 cos² $\theta$ = ( 1 - sin $\theta$ ) ²

So 4(1 - sin $\theta$ ) (1 + sin $\theta$ ) - (1 - sin $\theta$ ) ² = 0

or (1 - sin $\theta$ ) [ 4 (1 + sin $\theta$ ) - [ 1 - sin $\theta$ ] = 0

Hence sin $\theta$ = - 3/5

Thus cos $\theta$ = 4/5

or sin $\theta$ = 1 and cos $\theta$ = 0

So two cases are ,

1.) x-2 / 4/5 = y-3 / (-3/5)

Hence , -3(x-2) = 4(y - 3)

or 4y + 3x = 18

2.) x-2/ 0 = y-3/1

or x-2/y-3 = 0

Hence x-2 = 0

Thus the 2 equations are ,

x = 2 and 4y + 3x = 18 ................ans

allamraju

The ans given by sir is perfect.I forgot to find the other line which lies below whose angle with the X-axis is given by Tan^-11/2-Tan^-12=Tan^-1(-3/4) and so,it's eqn. is 3x+4y=18.

studyid

Aliter :

As in the figure , Let AC (the intercept) = 2 , and AB = the perpendicular distance = $\frac{2}{\sqrt{5}}$

Hence by pythagoras theorem , we get BC = $\frac{4}{\sqrt{5}}$

Also Let <ACB = $\theta$

Hence $tan (\theta) = \frac{1}{2}$

Now the slope of the line y + 2x = 3 is -2.

hence by the formula

$tan (\theta) = |\frac{{m}_{1}-{m_{2}}}{1-{m_{1}}{m}_{2}}|$

Here ${m}_{1} = -2 \;and \;the\;slpoe\;of\;the\;req\;line\;is \;{m}_{2}=m$

Hence $\frac{1}{2} = |\frac{2-m}{1+2m}|$

Solving we get $m = \frac{-3}{4} and m = 0$

now the line passes through the point (2,3)

Hence we get the eqn of line as

$y-3 = \frac{-3}{4}(x-2)$

So the line is 3x + 4y=18 and the other line has no slope as it is vertical and it is x = 2. ............ (Ans.)

Ask & Discuss Questions with Community & Experts

Let k be the angle made by the line with x-axis.Draw a rough figure and observe

The ans given by sir is perfect.I forgot to find the other line which lies below whose angle with the X-axis is given by Tan-11/2-Tan-12=Tan-1(-3/4) and so,it's eqn. is 3x+4y=18.

The ans given by sir is perfect.I forgot to find the other line which lies below whose angle with the X-axis is given by Tan^-11/2-Tan^-12=Tan^-1(-3/4) and so,it's eqn. is 3x+4y=18.