Question 626744

Hello, please help me solve:

A first line in the xy-plane is given by y= 2x-1 and a second line is given by 
y= x+1.
a) find the coordinates of the point P of intersection of these two lines.
b) find the distance from the origin (0,0) to the point P.

Thank you so much! 


Since the equations are non-parallel (different slopes) they will intersect each other. We can therefore say that since y = 2x - 1, and y = x + 1, then:


2x - 1 = x + 1
2x - x = 1 + 1
x = 2


y = 2(2) - 1 ----- Substituting 2 for x in equation
y = 4 - 1
y = 3


Since x = 2, and y = 3, then point of intersection of both lines, or P = ({{{highlight_green(2)}}},{{{highlight_green(3)}}})


Drawing a diagram, you will see that the distance from the origin, (0, 0) to P (point of intersection) is represented as the hypotenuse of a right-triangle, with one leg measuring 2 units (on the x-axis), and another leg measuring 3 units (vertical and parallel to the y-axis), on the xy-plane. 


With H being the length of the hypotenuse, we have: {{{H^2 = 2^2 + 3^2}}} ---- {{{H^2 = 4 + 9}}} ----- {{{H^2 = 13}}} ----- {{{H = sqrt(13)}}}


The distance from the origin (0, 0) to P (point of intersection), or coordinate point (2, 3) = {{{highlight_green(sqrt(13))}}}.


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