+0  
 
0
107
2
avatar

A line with slope equal to $1$ and a line with slope equal to $2$ intersect at the point $P(1,6),$ What is the area of $\triangle PQR?$

 Jun 30, 2020
edited by Guest  Jun 30, 2020
 #1
avatar+26811 
+1

You need to tell us where  Q   and  R are located.

 Jun 30, 2020
 #2
avatar+21953 
+1

The line RP is steeper than the line QP, so the slope of RP must be 2.

Using the point-slope form for a line, the equation of RP  =  y - 6  =  2(x - 1)   --->   y - 6  =  2x - 2

     --->     y  =  2x + 4

     --->   when y = 0:  0  =  2x + 4   --->   -4  =  2x   --->   x  =  -2

 

The slope of PR is 1   --->   the equation of PR  =  y - 6  =  1(x - 1)   --->   y - 6  =  x - 1

     --->     y  =  x + 5

     --->     when y = 0:  0  =  x + 5   --->   x  =  -5

 

The distance from Q to R  is  3.

 

The base of triangle(PQR) = 3; its height = 6     --->     area  =  ½·3·6  =  9

 Jun 30, 2020

14 Online Users