Question

What is an equation of the line that passes through the point (4,-3) and is perpendicular to the line x-2y=2?

Answer 1

Answer: 2x+y+11=0

Explanation:

The equation of given line is x-2y-2=0

Slope of given line is (m1)= - coefficient of x÷coefficient of y

m1= -1÷-2 =1÷2

Let slope of required line be m2.

Since the two lines are perpendicular,

So,

m1×m2= -1

or, 1÷2×m2 = -1

m2 = -2

The required line is passing through (4,-3)

Using two points slope form,

y-y1 = m(x-x1)

or, y-(-3)= -2(x+4)

or, y+3= -2x -8

or, 2x+y+3+8 = 0

2x+y+11 = 0 which is required equation of line.