# Determine the vector and parametric equations for the line passing through the points P(-1,-3) and Q(3,5).a) r=(3,5) + s(1,2), seR x=2t-1, y=4t-3, teR b) r=(1,-3) + s(4,8), seR ...

Determine the vector and parametric equations for the line passing through the points P(-1,-3) and Q(3,5).

a) r=(3,5) + s(1,2), seR

x=2t-1, y=4t-3, teR

b) r=(1,-3) + s(4,8), seR

x=4t-1, y=8t-3, teR

c) r=(1,0) + s(2,4), seR

x= t-1, y=2t-3, teR

d) all of the above

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A line is determined by 2 points. If both P and Q satisfies the equation, it is an equation of the line.

a) (-1,-3)=(3,5)+s(1,2) for s=-4 P belongs to the line

(3,5)=(3,5)+(1,2) for s=0 Q belongs to the line therefore

**It is a vector equation of the line (PQ)**

3=2t-1, 5=4t-3 for t=2 therefore the coordinates of Q satisfy the equation.

-1=2t-1, -3=4t-3 for t=0 therefore the coordinates of P satisfy the equation.

P and Q belongs to the line.** It is a parametric equation of the line (PQ)**

b)Use the same method. Do the points P and Q satisfy the equations?

(1,-3)+s(4,8)=(-1,-3) implies s=-1/2 and s=0 which is impossible. Therefore P doesn't belong to the given line.** **

**It is not a vector equation of the line (PQ)**

**b) and d) are not answers.**

**Remark:** in the question there is no negative sign in front of 1 in (1,-3)

What about c)?

(-1,-3)=(1,0)+s(2,4) implies that -1=1+2s and -3=4s

s=-3/ and s=-1 it is impossible. P doesn't belong to the given line. **It is not a vector equation of (PQ).**

**c) is not an answer.**

**The only correct answer is** a)