i need help with this problem.

Question

Find the distance between a line and a point not on the line. Line p contains points at (4,3) and (-2,0). Find the distance between line p and the point (3,10).

Answer 1

Use the slope formula to find the slope of the line . Let (x₁,y₁) = (4, 3) and (x₂,y₂) = (–2, 0). 

m = (y₂-y₁)/(x₂-x₁) = (0-3)/(-2-4) = -3/-6 = 1/2

Use the slope and any one of the points to write the equation of the line. 

y-y₁ = m(x-x₁) = y-0 = 1/2(x-(-2))

y= 1/2(x)+1

The slope of an equation perpendicular to will be – 2. So, write the equation of a line perpendicular to and that passes through (3, 10). 

y-y₁ = m(x-x₁) = y-10= -2 (x-3) = y=-2x+16

Solve the system of equations to determine the point of intersection.

The left sides of the equations are the same. So, equate the right sides and solve for x. 

1/2 (x) +1= -2x+16

1/2 (x) + 2x = 16-1

5/2(x) = 15

x = 6

Use the value of x to find the value of y. 

y= -2x+16

y=-2(6) +16

y=4

So, the point of intersection is (6, 4). 

Use the Distance Formula to find the distance between the points (3, 10) and (6, 4). 

d=√(x₂-x₁)²+ (y₂-y₁)²

d=√(6-3)²+ (4-10)²

d=√9+36

d=√45

d=3√5

The distance between the line and the point is 3√5