Line ℓ has equation y=x. Find the distance between ℓ and the point R( - 6,2)

Question

Can someone help me,please and thank you, on this I haven't been able to solve it for the longest time.

Answer 1

 

The vertical distance between R(-6,2) and the

line at point (-6,-6) is 8

Answer 2

L is line y=x which has slope m=1

 

Point (-6,2)

 

We want the perpendicular distance from (-6,2) to y=x.

Find the equation of the line through (-6, 2) perpendicular to y = x.

The perpendicular line will have slope m = -1/1 = -1

 

Equation of line through (-6, 2) with slope m = -1:

y = -x + b

Plugging in (-6, 2) as (x, y):

2 = -(-6) + b

2 = 6 + b

b = -4

The line is: y = -x - 4

 

Find the crossing point of the two lines:

y = x

y = -x - 4

 

x = -x - 4

2x = -4

x = -2

y = -2

 

The lines cross at (-2, -2)

 

The distance you are looking for is the distance from (-6, 2) to (-2, -2)

 

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

 

d = √[(-2-(-6))²+(-2-2)²] = √[4² + (-4)²] = √32 = 4√2