Question

find the equation of the line passing through the point intersection d lines x+3y+2=0 and x-2y-4= 0 and perpendicular the line 2y+5x-9=0​

Answer 1

Answer:

The line perpendicular to rx+sy+t=0 through a given point (x₀,y₀) is sx-ry+ry₀-sx₀=0.

The intersection of ax+by+c=0 and dx+ey+f=0 is ((b×f-c×e)/(a×e-b×d), (c×d-a×f)/(a×e-b×d)) where the lines are not parallel.

The equation of line passing through the point of intersection of the lines ax+by+c=0 and dx+ey+f=0, and perpendicular to the line rx+sy+t=0 is:

sx - ry + (r(c×d-a×f) - s(b×f-c×e))/(a×e-b×d) = 0

The equation of line passing through the point of intersection of the lines x+3y-1=0 and x-2y+4=0, and perpendicular to the line 3x+2y=0 is 2x - 3y + (3(-1×1 - 1×4) - 2(3×4 - -1×-2))/(1×-2 - 3×1) = 0 which simplifies to 2x - 3y + 7 = 0.

Answer 2

Answer:

Let the equation of line which is perpendicular to

5x−2y=7 is 2x+5y=λ ...... (i)

The given lines are 2x+3y=1 ...... (ii) and 3x+4y=6 ..... (iii)

Solving (ii) and (iii), we get

x=14,y=−9

∴ The point of intersection of given lines is (14,−9)

Since, the Eq. (i) is passing through the point (14,−9)

∴2(14)+5(−9)=λ⇒λ=−17

∴ Eq. (i) becomes

2x+5y+17=0