Question

the equation of the line passing through the point of intersection of 2x+3y-5=0 and 7x-5y-2=0 and parallel to the lines 2x-3y+14=0 is​

Answer 1

Solution:-Since the line passing throw the intersection is parallel to 2x-3y+14,this shows that slope of line of this eqaution will be same of the slope of line of intersection

slope of line 2x-3y+14= -2/-3=2/3

slope of the required line=2/3

let the eqaution of required line =2x-3y+c=0

there will be one point (x,y) which will satisfy both the equation 2x+3y-5=0 and 7x-5y-2=0

2x+3y-5=0.....i)

7x-5y+3=0.....ii)

multiplying the eqaution i) by 7 and other by 2 we get

14x+21y-35=0.....iii)

14x-10y-6=0........iv)

subtract iv) from iii)

31y-29=0

y=29/31

putting this value in iv)

14x-290/31-6=0

14x=(290+186)/31

x=(476)/434=238/217

putting this value in equation of required line

2×238/217 -87/31+c=0

c=-476/217 +87/31

c=(-476+609)/217

c=133/217

so putting the value of c we get

equation of required line=2x-3y+133/217=0

{hope it helps you }