The line -2x -5y =10 can be written as:
-2x -5y =10
=> -5y = 2x +10
Divide by -5
=> y = (-2/5) x - 2
This is of the form y = mx + c where m is the slope and c is the y-intercept.
So the slope of the given line is -2/5. No the product of the slopes of two perpendicular lines is -1. Therefore the slope of the line perpendicular to -2x -5y =10 is 5/2
So we can write this as y = (5/2) x +c
Now this line passes through (2, -2)
=> -2 = (5/2)*2 +c
=> -2 = 5 +c
=> c = -7
Therefore the equation of the line is y = (5/2) x -7
=> 2y = 5x -14
=> 5x - 2y -14 =0.
The required equation in the standard form is 5x - 2y -14 =0.
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).