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3 years ago

Which equation contains the points (-1,-8) and (3,4)

Mathematics

1 answer:

katrin [286]3 years ago

8 0

Answer:

y = 3x –5

Step-by-step explanation:

for slope-intercept form, y = mx + b

find the slope, m

(-8 -4)/(-1-3) = m

-12/-4 =m

3 = m

Substitute for m and use either coordinate for y and x to calculate b

4 = 3(3) + b

4 = 9 + b. Subtract 9 from both sides

4 –9 = b

–5 = b. Rewrite the equation with the calculated values for m and b

y = 3x –5

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The total allowance is $18.5, and the amount spent daily is $3.2

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let T and x represent the total allowance and the amount spent daily respectively.

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Points A and B have coordinates (3,1,2) and (1,-5,4) respectively. Point C lies on line AB such that AC:BC=3:2. Find position ve

Sever21 [200]

Answer:

The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

Step-by-step explanation:

* Lets revise how to solve the problem

- If the endpoints of a segment are (x1 , y1 , z1) and (x2 , y2 , z2), and

point (x , y , z) divides the segment externally at ratio m1 :m2, then

$x=\frac{m_{1}x_{2}-m_{2}x_{1}}{m_{1} -m_{2}},y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}},z=\frac{m_{1}z_{2}-m_{2}z_{1}}{m_{1}-m_{2}}$

* Lets solve the problem

∵ AB is a segment where A = (3 , 1 , 2) and B = (1 , - 5 , 4)

∵ Point C lies on line AB such that AC : BC=3 : 2

∵ From the ratio AC = 3/2 AB

∴ C divides AB externally

- Lets use the rule above to find the coordinates of C

- Let Point A is (x1 , y1 , z1) , point B is (x2 , y2 , z2) and point C is (x , y , z)

and AC : AB is m1 : m2

∴ x1 = 3 , x2 = 1

∴ y1 = 1 , y2 = -5

∴ z1 = 2 , z2 = 4

∴ m1 = 3 , m2 = 2

- By using the rule above

∴ $x=\frac{3(1)-2(3)}{3-2}=\frac{3-6}{1}=\frac{-3}{1}=-3$

∴ $y=\frac{3(-5)-2(1)}{3-2}=\frac{x=-15-2}{1}=\frac{-17}{1}=-17$

∴ $z=\frac{3(4)-2(2)}{3-2}=\frac{12-4}{1}=\frac{8}{1}=8$

∴ The coordinates fo point c are (-3 , -17 , 8)

* The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

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