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

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william1941 | College Teacher | (Level 3) Valedictorian

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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.

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neela | High School Teacher | (Level 3) Valedictorian

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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).

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