P is the point (4,2) Q is the point (12,5) Work out the equation of the line PQ?

Question

the mid point is (8,5) while the length of the line is (8.5)

Answer 1

Slope formula

m = y₂- y₁/ x₂- x₁

Ordered Pairs

(4, 2)(12, 5)

m = 5 - 2 / 12 - 4

m = 3/8

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Slope intercept fork

y = mx + b

2 = 3/8(4) + b

2 = 12/8 + b

2 = 3/2 + b

2 - 3/2 = 3/2 + b - 3/2

2 / 1 - 3/2 = b

4/2 - 3/2 = b

1/2 = b

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The equation

y = 3/8x + 1/2

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Answer 2

Slope of line
5-2 / 12-4 = 3 / 8
hence eq of line is
y-2 = 3 / 8 ( x-4)
8y - 16 = 3x - 12
3x - 8y = 4

Answer 3

gradient of the line = change in y / change in x

= (5-2) / (12-4)

=3/8

Using the point P (4,2) and the formula

(y -y1)=gr(x-x1)

(y -2)=3/8(x-4)

8(y -2)=3(x-4)

8y-16=3x -12

3x -8y +4 =0

or 8y=3x + 4

y=3/8x +1/2

Answer 4

Given two points, the equation of a line is:
(y-y1)/(x-x1) =(y2-y1)/x2-x1)
So (y-2)/(x-4) = (5-2)/(12-4)
(y-2)/(x-4) = 3/8
3x-12 =8y-16
3x-8y +4=0
The midpoint is not (8,5). It is (8,3.5).
The length of the line is sqrt(73) = approx 8.544.

Answer 5

rise/run
(5-2)/(12-4)

y=mx+b
m= 3/8
2= (3/8)*4 + b

b=.5

y=(3/8)x+(1/2)