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3 years ago

Which equation is parallel to the line LaTeX: y=\frac{1}{2}x+3y = 1 2 x + 3and passes through the point (10, -5)?

Mathematics

1 answer:

sukhopar [10]3 years ago

8 0

Answer:

$Equation\ of\ line:\ y=\frac{1}{2}x-10$

Step-by-step explanation:

$Let\ the\ required\ equation\ is\ y=mx+c\\\\where\ m\ is\ the\ slope\ of\ the\ equation\ and\ c\ is\ y-intercept\\\\It\ is\ parallel\ to\ the\ equation\ y=\frac{1}{2}x+3\\\\Hence\ slope\ of\ these\ two\ lines\ will\ be\ same.\\\\Slope\ of\ y=\frac{1}{2}x+3\ is\ \frac{1}{2}\\\\Hence\ slope\ of\ y=mx+c\ is\ \frac{1}{2}\Rightarrow m=\frac{1}{2}\\\\Equation:y=\frac{1}{2}x+c\\\\Line\ passes\ through\ (10,-5).\ Hence\ this\ point\ satisfies\ the\ equation\ of\ line.\\\\-5=\frac{1}{2}\times 10+c$

$-5=-5+c\\\\c=-10$

$Equation\ of\ line:\ y=\frac{1}{2}x-10$

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The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

#initialize a function named gcd which takes in two parameters

if a>b:

#checks if a is greater than b

return gcd (b, a)

#if true interchange the Parameters and Recall the function

elif a == 0:

return b

elif a == 1:

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elif((a%2 == 0)and(b%2==0)):

#even numbers leave no remainder when divided by 2, checks if a and b are even

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

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return gcd(a, b/2)

else :

return gcd(a, b-a)

#since it's a recursive function, it recalls the function with new parameters until a certain condition is satisfied

print(gcd(124, 244))

print()

#leaves a space after the first output

print(gcd(4424, 2111))

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Paraphin [41]

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