Andrew M. answered • 07/25/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
We need to find the 3 points of the triangle:
1) The lines AB, y = 2x and BC, y = 3x cross at the point (0,0) so the origin is one point of the triangle
and will be point B since B is the point common to both lines. B = (0,0)
2) Let's find the other two points where AB meets CA to find common point A
AB: y = 2x
CA: x+y=8
substitute 2x for y into CA ... x + 2x = 8
3x = 8
x = 8/3
Since y = 2x we have A = (8/3, 16/3) which is the common point for AB and CA
3) Let's find the last point, intersection of BC and CA to find point C
BC: y = 3x
CA: x+y=8
substitute 3x in place of y into CA
x + 3x = 8
4x = 8
x = 2
y = 3x = 6
point C = (2,6)
We need the point through C perpendicular to AB
point thru (2,6) perpendicular to y = 2x
the slope of the perpendicular line will have m = -1/2 since perpendicular slopes are
negative reciprocals of each other.
Thus, we are looking for the line through (2,6) with slope m = -1/2
Use the point slope formula y - y1=m(x-x1)
y-6 = (-1/2)(x-2)
Multiply both sides by 2
2y-12 = -(x-2)
2y-12 = -x + 2
x + 2y = 14
The perpendicular line through C is x + 2y = 14
We want to find the x intercept where y=0 and the y intercept where x=0
x intercept: (x,0) x+2(0) = 14 y-intercept: (0,y) 0+2y=14
x = 14 2y=14
P = (14,0) y=7
Q = (0,7)
To find the ratio of PC to CQ we need the length of the line segments
Distance formula D = √[(x2-x1)2 + (y2-y1)2]
PC = distance from (14,0) to (2,6)
PC = √[(14-2)2+(0-6)2 = √180 = 6√5
CQ = distance from (2,6) to (0,7)
= √[(2-0)2+(6-7)2] = √5
PC/CQ = 6√5/√5
= 6/1
The ratio of PC/CQ = 6:1
