Find the distance from the point (0,-3) to the line y=x-1

Answer 1

Normal equation of given line (x - y - 1)/√2 = 0; To get distance d substitute coordinates (0, - 3):

d = abs(0 + 3 - 1)/√2 = 2/√2 = √2

Answer 2

y=x-1 has a slope = m = 1, and y-intercept b= -1

A line perpendicular to y=x-1 will have a slope that is the negative inverse of 1 which is -1

that perpendicular line is y=-x + b

b=-3

the perpendicular line is y=-x-3

Find where the perpendicular intersects the original equation, set them equal

y=x-1 = y = -x-3

x-1 = -x-3

2x =-2

x = -1

plug that value into either equation

to determine y

y=-1-1 = -2

that's the point of intersection (-1,-2)

Find the distance from (0-3) to (-1,-2), that's the distance from (0,-3) to the line y=x-1

d^2 = 1 +1 = 2

distance squared = sum of the squares of the differences in x & y coordinates

d^2 = (-1-0)^2 + (-2--3)^2 = 1^2 + 1^2 = 2

d = sqr2 = about 1.414

Or just graph it carefully, and you can see d = the hypotenuse of a right triangle with sides 1 and 1, so d = sqr2. Plot the point and the line, and the perpendicular line from (0,-3)