# What is the coordinate of the midpoint of line segment TR with T(8, -3) and R (-6,8) What is an equation for the line parallel to y= 3x+ 4 that contains (3,10)

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Midpoint formula = `((x(T)+x(R))/2,(y(T)+y(R))/2)`

Midpoint of TR is `((8+(-6))/2,(-3+8)/2)=(1,5/2)`

The line parallel to y=3x+4 has a slope of 3.

We need a line through (3,10) which has a slope of 3

We use the point slope formula

`y-y_1=m(x-x_1)`

In our case m=3 and `x_1` = 3 and `y_1` = 10

We get

y - 10 = 3(x - 3)

Simplifying to

y = 3x + 1

So a) `(1,5/2)` and b) `y=3x + 1`