Find an equation of the line passing through the point (5,4) and perpendicular to the line 2x+y=3

Step-by-step explanation:

Given equation of line is:

Equating it with we find:

Let the slope of required line be

Required line is parallel to the given line.

Required line passes through (5, 4) and has a slope

Equation of line in slope point form is given as:

Hence, x -2y + 3=0 is the required equation of line.

  x -2y = -3

Step-by-step explanation:

The line perpendicular to

  ax +by = c

through point (h, k) can be written as ...

  b(x -h) -a(y -k) = 0 . . . . . swap x and y coefficients, negate one of them.

Putting this in standard form gives ...

  bx -ay = bh -ak

For a=2, b=1, and (h, k) = (5, 4), this is ...

  x -2y = 5 -2(4) = -3

  x -2y = -3

_____

The graph shows the two lines and the point.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: