Line L contains points (3, 5) and (7, 9). Points P has coordinates (2, 10). Find the distance from P to L.

Question

This is a question from McGraw-Hill Geometry lesson 3-6 and is also a question I have been struggled about.

Answer 1

Line L:  (3, 5), (7, 9)

Find the equation of the line.

Slope:  m = (change in y)/(change in x) = (9-5)/(7-3) = 4/4 = 1

 

Line L in slope intercept form  y = mx +b is

y = x + b

Using (3,5) for (x,y) find b

5 = 3 + b

b = 2

Line L has equation:  y = x + 2

 

Find the equation of the line perpendicular to y = x+2

going through the point (2, 10)

Perpendicular lines have slopes that are negative reciprocals...

m_⊥ = -1/m ... 

m_⊥ = -1/1 = -1

 

We want the equation of the line with slope m = -1, through (2, 10)

y = -x + b

10 = -2 + b

b = 12

y = -x + 12

 

The original line is:  y = x + 2

Perpendicular line through (2, 10) is:  y = -x + 12

 

Find the crossing point for the two lines:

y = x + 2

y = -x + 12

 

x + 2 = -x + 12

2x = 10

x = 5

y = 7

 

The two lines cross at the point (5, 7)

 

The distance from (2, 10) to (5, 7) will be the distance from (2, 10) to Line L

 

d = √[(x₂-x₁)²+(y₂-y₁)²

d = √[(5-2)²+(7-10)²]

d = √[3² + (-3)²]

d = √18

d = 3√2 ≅ 4.243

 

 

 

 

Answer 2

Find the slope between the two points on line L.

 

m = (9 - 5) / (7 - 3)

m = 4 / 4

m = 1

 

Then the line passing through point P is perpendicular to line L.  Knowing this, we can use the point-slope form of a line.

 

y = -1(x - 2) + 10

y = -x + 12

 

Now we find the equation of line L.

 

L = (x - 3) + 5

L = x + 2

 

Next, we set the lines equal to each other to find the point of intersection.

 

-x + 12 = x + 2

 

Solve for x.

 

-2x = -10

   x = 5

 

Plug in this value of x into any of the equations, we get the point of intersection (5, 7).

 

Finally, find the distance between (2, 10) and (5, 7) using the distance formula.