(i) Find the direction in which a straight line must be drawn torough the points (1,2) so that its point of intersection with the line `x+y=4` may be the a distance `sqrt((2)/(3))` from this point.
(ii) A straight line through `P(-15,-10)` meets the straight lies `x-y-1=0,x+2y=5and x+3y=7` respectively at A,B and C, if `(12)/(PA)+(40)/(PB)=(52)/(PC),` then prove that the straight line passes through origin.

Explanation

(i) Let the straight line through `P(1,2)` cut the given straight line `x+y=4` at Q and the straight line is inclined at an angle `theta` with positive direction of x-axis. Then its equation is `(x-1)/(costheta)=(y-2)/(sintheta)=r,PQ=r.`
The coordinates of the point Q are `(1+r cos theta, 2+r sin theta)`
Which lies on `x+y=4`
`impliessqrt((2)/(3))(cos theta+sin theta)=1,r=sqrt((2)/(3)),` given
`impliescos theta+sin theta=sqrt((3)/(2))`
`implies(1)/(sqrt(2))costheta+(1)/(sqrt2)sin theta=(1)/(sqrt2)sqrt((3)/(2))=(sqrt3)/(2)`
`impliescos(theta-(pi)/(4))=cos""(pi)/(6)`
`impliestheta-(pi)/(4)=(pi)/(6)or -(pi)/(6)`
`impliestheta=(pi)/(4)+(pi)/(6)or (pi)/(4)-(pi)/(6)`
`impliestheta=(5pi)/(12)or =(pi)/(12)`

(ii) Let the equation of straight line passing throug `P(-15,-10)` be `(x+15)/(costheta)=(y+10)/(sin theta)`
which cuts the given straight lines
`x-y-1=0,x+2y=5andx+3y=7` at A, B

and C respectively.
Any point on this line may be takes as
`(-15+rcostheta,-10+rsintheta)`
For the coordinates of the point `A,(-15rcostheta,-10+r sin theta)` lies on `x-y-1=0`
`implies-15+r cos theta+10-r sin theta-1=0`
`impliesr (cos theta-sin theta)=6`
`implies(6)/(PA)= cos theta-sin theta`
Also for `B,l15+r cos theta-20+2rsin theta=5`
`impliesr(cos theta+2sin theta)=40`
`implies(40)/(PB=cos theta2 sin theta`
and for the point `C,15+r cos theta-30+3rsin theta=7`
`impliesr(cos theta+2sin theta)=52`
`impliescos theta+3sin theta=(52)/(PC)`
Now from the given condition,
`(12)/(PA)+(40)/(PB)=(52)/(PC)`
`implies2(cos theta-sin theta)+costheta+2sintheta=costheta+3sintheta`
`implies2cos theta=3sinthetaimplies(cos theta)/(3)=(sin theta)/(2)`
Hence the required equation of the straight line is
`(x+15)/(3)=(y+10)/(2)`
`implies2x+30=3y+30`
`implies2x-3y=0`
which passes through origin.