# Find intersection of these two lines the find area of triangle whose vertices are the  intersection point, s=0 on L1, and t=3 on L2. Given following: Let L1 be the line `vecr`=(2,9)+s(1,1). Let L2 be `vecv`=(-3,2)+t(1,2).

You need to write parametric equations of the lines such that:

`x_1 = 2 + s*1`

`y_1 = 9 + s*1`

x_2 = -3 + t*1

`y_2 = 2 + t*2`

You need to set the equations of `x_1`  and `x_2`  equal such that:

`2 + s*1 = -3 + t*1`

You need to set the equations of `y_1`  and `y_2`  equal such that:

`9 + s*1 = 2 + t*2`

You should solve for s and t the system of equations above such that:

`s-t = -3 - 2 `

`s - 2t = 2-9`

You should use substitution method, hence you need to write s in terms of t in the first equation such that:

`s = t - 5`

You need to substitute `t - 5`  for `s ` in the equation `s - 2t = -7`  such that:

`t - 5 - 2t = -7 =gt -t - 5 = -7 =gt -t = -7 + 5 =gt t = 2`

`s = 2 - 5 =gt s = -3`

You need to substitute 2 for t and -3 for s in parametric equations to find the point of intersection of lines such that:

`x = 2 + (-3) = -1`

`y = 9 - 3 = 6`

Hence, the line `L_1`  intercepts the line `L_2`  at the point `(-1,6).`

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