Line l contains (-2,0) and (4,8 ). Point P has coordinates (5,1) What is the distance from P to L? Guest 0 mins ago

Question

Please help me slove this step by step

Answer 1

The distance from a point not on a line to the the line is the perpendicular distance from the point to the line.  

 

The slope of the line is (8-0)/(4-(-2)) = 4/3

 

An equation of the line is y-0 = 4/3(x-(-2)) 

 

                                       y = (4/3)x + 8/3

 

We need to find a point, Q = (x,y), on the line so that PQ is perpendicular to the line.  

 

So slope of PQ = -3/4

 

   We get (y-1)/(x-5) = -3/4

 

                      y-1 = (-3/4)(x-5)

 

                        y = (-3/4)x + 19/4

 

  But, since (x,y) is on the line, y = (4/3)x + 8/3

 

  Therefore, (-3/4)x + 19/4 = (4/3)x + 8/3

 

   Multiply by 12 to obtain -9x + 57 = 16x + 32

 

                                             25 = 25x

 

                                               x = 1   y = 4

 

So, Q = (1,4)

 

The distance from P to the line is the distance, D, from P to Q.

 

   D = √[(4-1)² + (1-5)²] = 5