Write the standard form of the equation of the line passing through the point (2,-2) and perpendicular to the line -2x-5y=10

The standard form of and equation of a line is of the form
ax+by+c = 0.

An equation of  a line perpependicular to
ax+by +c = 0 is got by interchanging the coefficient of x and y with a minus sign to one
of the coefficients.Thus ax-by + k = 0 is a perpendicular line to ax+by+c = 0, where k
is aconstant to be determined by some other  given
condition.

Therefore -2x-5y = 10 has the equation
perpependicular line like -(-2x)-5y + k = 0. Or 2x-5y -k =
0.

Now this line passes through (2 , -2). So it should
satisfy 2x-2y-k = 0. So 2(2)-5(-2) -k = 0.

4+10-k = 0 .
Therefore k = 14. Substitute k=14 in 2x-5y-k = 0. So we get: 2x-5y -14 =
0 which is the standard form of the equation of the perpendicular line 
to -2x-2y = -10 and passing through (2 ,-2).