find the point where a line with slope 1 and y-intercept (0, 4) intercepts a line through P(4, 8) and Q(-2, -1). What is the point of intersection

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find the point where a line with slope 1 and y-intercept (0, 4) intercepts a line through P(4, 8) and Q(-2, -1). What is the point of intersection

Mathematics
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`the point where a line with slope 1 and y-intercept (0, 4)` y=mx+b slope intercept form where m is slope and b is y-intercept now plugin m and b value
I get that, but why do I need to know points P and Q. Why are they even in the problem.
you have to find 2nd equation through those (p and q) points

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y = mx + b y = 1x + 4
then we can graph both equations to find out the intersection point |dw:1443817214823:dw|
right can you write the equation that passes through p and q points ??
P(4, 8) and Q(-2, -1) y = 1x + 4 8 = 4 + 4 = 8 -1 = 1(-2) + 4 -1 = 2? I don't know
hmm nope alright let's write the equation first we need to find the slope \[\huge\rm m=\frac{ y_2 -y_1 }{ x_2 -x_1 }\]now substitute x's and y's values from this order pair (4,8)(-2,-1)
\[\rm (\color{ReD}{4},\color{blue}{8})(\color{orange}{-2},\color{green}{-1})\] \[\rm (\color{ReD}{x_1},\color{blue}{y_1})(\color{orange}{x_2},\color{green}{y_2})\]
I'm just having trouble understanding what they're even meaning. What point of intesection am I trying to find?
(4,8)(-2,-1) -1 - 8 -2 - 4 -9 -6 1.5
intersect point of equation y=1x+ 4 and PQ points it's same as `system of equations` questions where we have the 2 equations and we should find x and y values by using elimination/addition graphing matrices method
for this question the problem is they want us to find 2nd equation from PQ points
1.5 is right but better to keep it in fraction \[\huge\rm y=\frac{ 3 }{ 2 }x+b\] replaced m with slope which is 3/2 now we can pick one of the point to find y-intercept let's use point Q (-2,-1) substitute x and y for (-2,-1) solve for b \[-1= \frac{ 3 }{ 2 }(-2)+b\] b= ??
well pretty sure you know how to solve for b \[\rm -1=\frac{ 3 }{ \cancel{2} }(-\cancel{2})+b\] \[\rm -1=-3+b\] move to the left side \[\rm -1+3=b\] b=2 so 2nd equation (PQ) would be y=3/2x +2 `first equation` \[\large\rm y=1x+4\] `2nd equation` \[\large\rm y=\frac{ 3 }{ 2 }x+2\]
now we have two equations there are 4 ways to solve for x and y (in other words to find intersection point ) substitution graphing elimination/addition matrices ~~~ graphing one is easy if both equations are in standard form then u should use elimination and if one of the equations solved for either x or y variable then you can use substitution
for this question substitution would be easy :=)

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