Question

What is the length of the segment, endpoints of which are intersections of parabolas y=x^2− 11/4 x− 7/4 and y=− 7/8 x^2+x+ 31/8 ?

Answer 1

The length of line segment is 5

Explanation:

we are given equation of parabolas as

`y=x^2-(11)/(4)x-(7)/(4)`

`y=-(7)/(8)x^2+x+(31)/(8)`

Firstly, we will find intersection points

we can set them equal

and then we can solve for x

`x^2-(11)/(4)x-(7)/(4)=-(7)/(8)x^2+x+(31)/(8)`

Multiply all sides by 8

`x^2\cdot \:8-(11)/(4)x\cdot \:8-(7)/(4)\cdot \:8=-(7)/(8)x^2\cdot \:8+x\cdot \:8+(31)/(8)\cdot \:8`

`8x^2-22x-14=-7x^2+8x+31`

`15x^2-30x-45=0`

now, we can factor it

`15(x^2-2x-3)=0`

`15(x-3)(x+1)=0`

`x=-1,x=3`

now, we can find y-values

At x=-1:

`y=(-1)^2-(11)/(4)(-1)-(7)/(4)`

y=2

so, we get point as

(-1,2)

At x=3:

`y=(3)^2-(11)/(4)(3)-(7)/(4)`

y=-1

so, we get point as

(3,-1)

now, we can find distance between these two points

(-1,2)

x1=-1 , y1=2

(3,-1)

x2=3 , y2=-1

now, we can find distance

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

now, we can plug values

`D=√((3+1)^2+(-1-2)^2)`

`D=5`

So,

The length of line segment is 5