Question

What is the slope-intercept equation that represents a midsegment of the triangle below?

*Hint* There are three possible answers.

Answer 1

1)
`y=-6x-6.5`

2)
`y=0.6x+0.1`

3)
`y=-0.5x+1.75`

Step-by-step explanation:

Since we know that the segment joining the mid-points of two sides of a triangle is known as mid-segment of triangle. A triangle has three mid-segments.

1) Let us find midpoints of side ED and DF using midpoint formula.

`\text{Midpoint}=((x_1+x_2)/(2),(y_1+y_2)/(2))`

`\text{Midpoint of segment DE}=((-4+1)/(2),(1+4)/(2))`

`\text{Midpoint of segment DE}=((-3)/(2),(5)/(2))=(-1.5,2.5)`

`\text{Midpoint of segment DF}=((-4+2)/(2),(1-2)/(2))`

`\text{Midpoint of segment DF}=((-2)/(2),(-1)/(2))=(-1,-0.5)`

Now let us find slope of line joining points (-1.5,2.5) and (-1,-0.5) using slope formula.

`m=(y_2-y_1)/(x_2-x_1)`

`m=(-0.5-2.5)/(-1--1.5)`

`m=(-3)/(-1+1.5)`

`m=(-3)/(0.5)=-6`

Now let us substitute our values in slope intercept form of the line to find y-intercept.

`y=mx+b`, where m= slope of the line, b= y-intercept.

`-0.5=-6(-1)+b`

`-0.5=6+b`

`-6.5=b`

Upon substituting m=-6 and b= -6.5 in slope intercept form of line, we will get,

`y=-6x-6.5`

Therefore, 1st equation that represents the mid-segment parallel to side EF of the given triangle will be
`y=-6x-6.5`.

2) Let us find midpoint of segment EF.

`\text{Midpoint of segment EF}=((1+2)/(2),(4-2)/(2))`

`\text{Midpoint of segment EF}=((3)/(2),(2)/(2))=(1.5,1)`

Now let us find the slope of the line passing through points (1.5,1) and (-1,-0.5).

`m=(1--0.5)/(1.5--1)`

`m=(1.5)/(2.5)=0.6`

Now let us find y-intercept of line parallel to segment DE.

`-0.5=0.6(-1)+b`

`-0.5=-0.6+b`

`0.1=b`

Upon substituting m=0.6 and b=0.1 in slope intercept form of line we will get,

`y=0.6x+0.1`

Therefore, second equation that represents the mid-segment parallel to side DE of the given triangle will be
`y=0.6x+0.1`.

3) Now let us find the slope of the mid-segment joining mid-points of segment DE and EF that are (-1.5,2.5) and (1.5,1).

`m=(1-2.5)/(1.5--1.5)`

`m=(-1.5)/(3)=-0.5`

Now let us find y-intercept of our line.

`y=mx+b`

`1=-0.5(1.5)+b`

`1=-0.75+b`

`1.75=b`

`y=-0.5x+1.75`

Therefore, the third equation that represents the mid-segment parallel to side DF of the given triangle will be
`y=-0.5x+1.75`.