Mathematics
Can someone help me with these 2 equations?

Question Description

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Final Answer

Formulas:

d=√((x1-x2)^2+(y1-y2)^2)   distance between 2 points

m=((x1+x2)/2, (y1+y2)/2)  midpoint

13) y= 3/2x + 4       A(-6,-5)

√13 = √((-6-x)^2 + (-5-y)^2)    raising to the second power to eliminate √

13 =(6+x)^2 + (5+y)^2            

(6+x)^2 = 4       and     (5+y)^2 = 9    because 13 = 4+9 (you have more possibilities but they don't work when                                                                                           you check in the equation) 

6+x=2               and       5+y=3         because you take √

x= 2-6=-4          and       y=3-5=-2

Substituting x=-4 in the linear equation

y = 3/2 (-4) + 4 = -12/2 + 4 = -6 + 4 = -2   The solution is correct

A(-6,-5)  

B(-4,-2)

M= ((-6-4)/2, (-5-2/2)) = (-10/2, -7/2) = (-5,-3.5)

M (-5,-3.5) or M(-5,-3 1/2)

14) y = -4/3 x +2       A(-3,6)   

5 = √((-3-x)^2)+(6-y)^2)    raising too the second power to eliminate √

25 = (x+3)^2 + (6-y)^2

(x+3)^2 = 9        and       (6-y)^2=16     because 9 + 16 = 25  (you have more possibilities but they don't work                                                                                                        when you check in the equation)

(x+3)=3              and       (6-y)=4          because you take √

x=0                    and        y=5-4=2

Substituting x=0 in the linear equation:

y= -4/3(0)+2 = 2   the solution is correct.

A(-3,6)

B(0,2)

M=((-3+0)/2, (6+2)/2) = (-3/2,4)

M(-1.5,4) or M(-1 1/2,4)

IleanaP (262)
New York University

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