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NeTakaya
3 years ago
12

Which point is on the line y = -2x + 3?

Mathematics
2 answers:
Musya8 [376]3 years ago
8 0

(1, 1)

You would do your 'y' first which is three, then you would go down two and over one to get (1, 1)

Yeah, I know that is confusing, if you want me to explain better lemme know

Hope this helps :)

luda_lava [24]3 years ago
3 0

m = (y2 - y1) / (x2 - x1)                                                                                             now, if 2 lines are perpendicular, then the product of their slopes is -1          what is the slope of x -y = 5 ?    

to find out, rewrite it in slope - intercept form : y = mx + n, with m = slope and n = intercept    

so x - y = 5 <=> y = x - 5  

hence, its slope is m = 1  

then, what's the slope of ANY line perpedicular to it ? let M be the slope of that line  

as we said above: m * M = -1  

with m = 1 => M = -1  

coming back to the definition of the slope, given 2 points on it   (y2 - y1) / (x2 - x1) = M   we know y2 = 3 and x2 = 2 AND M = -1  

replace them, and what do u get u think ? the eqn of the line passing through (2,3) and is perpendicular to x - y = 5  

hence, (3 - y)/(2 - x) = -1 => y - 3 = 2 - x => y = -x + 5  

once we know its equation, how do we find the distance between that point on it (2,3) and x - y = 5 ?  

1stly, we find the intersection point of the new line and the one given; to do this, we need to solve the linear system of the their equations  

x - y = 5  

x + y = 5  

add 1st to 2nd  

2x = 10 => x = 5 => y = 0  

hence, the intersection between the given line and the perpendicular to it passing through (2,3) is (5,0)

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MDI<br> Solve the following system<br> -2x + 2y = -2<br> 2x + 3y = 12
s344n2d4d5 [400]

Answer:

x = 2.25, y = 2.5

Step-by-step explanation:

-2x + 2y = -2 .............equ (1)

2x + 3y = 12 .............equ (2)

equ (1) + equ (2)

-2x + 2y = -2

+ 2x + 2y = 12

=> 0 + 4y = 10

=> 4y/4 = 10/4

=> y = 2½

Substitute for y in equ (2)

i. e, 2x + 3y = 12

=> 2x + 3(10/4) =12

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x = 2.25, y = 2.5

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3 years ago
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