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astraxan [27]
3 years ago
13

Which equation represents the line passing through the points (4, −1) and (1, 0)? Select one: A. x + 3y = 1 B. 3x + y = 1 C. x −

3y = −1 D. −x + 3y = 1
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

Gradient of a line =( y2 - y1)/x2 - x1

From the question, x1=4,y1=-1,x2=1,y2=0.

Therefore, substitute for these values in the above formula

m = 0-(-1)/1-4

= 0+1/-3

= -1/3.

Therefore, y-y1/x-x1 = -1/3

y - y1 = -1/3(x - x1)

y - (-1) = -(x-x1)/3

y+1 = - (x - 4)/3

Multiply through by 3

3y+3= -x+4

3y +x=4-3

x+3y= 1

Therefore the answer is A.

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Ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication
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Answer:

x=1\,,\,y=-1

Step-by-step explanation:

Given: ax+by=a-b\,,\,bx-ay=a+b

To solve: the given linear equations

Solution:

Consider the equations:

A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0

By method of cross multiplication:

\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}

For equations: ax+by=a-b\,,\,bx-ay=a+b

ax+by-(a-b)=0\\bx-ay-(a+b)=0

Take A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)

So,

\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1

6 0
3 years ago
Just #14 plz thanks!!!
ziro4ka [17]
You can do this! You’re a smart person :) don’t ever give up
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Indicate the method you would use to prove the two 's . If no method applies, enter "none".SSS
balu736 [363]
I hope this helps you




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