Answer:
The length of line segment is 5
Step-by-step explanation:
we are given equation of parabolas as
[tex]y=x^2-\frac{11}{4}x-\frac{7}{4}[/tex]
[tex]y=-\frac{7}{8}x^2+x+\frac{31}{8}[/tex]
Firstly, we will find intersection points
we can set them equal
and then we can solve for x
[tex]x^2-\frac{11}{4}x-\frac{7}{4}=-\frac{7}{8}x^2+x+\frac{31}{8}[/tex]
Multiply all sides by 8
[tex]x^2\cdot \:8-\frac{11}{4}x\cdot \:8-\frac{7}{4}\cdot \:8=-\frac{7}{8}x^2\cdot \:8+x\cdot \:8+\frac{31}{8}\cdot \:8[/tex]
[tex]8x^2-22x-14=-7x^2+8x+31[/tex]
[tex]15x^2-30x-45=0[/tex]
now, we can factor it
[tex]15(x^2-2x-3)=0[/tex]
[tex]15(x-3)(x+1)=0[/tex]
[tex]x=-1,x=3[/tex]
now, we can find y-values
At x=-1:
[tex]y=(-1)^2-\frac{11}{4}(-1)-\frac{7}{4}[/tex]
y=2
so, we get point as
(-1,2)
At x=3:
[tex]y=(3)^2-\frac{11}{4}(3)-\frac{7}{4}[/tex]
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
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
now, we can plug values
[tex]D=\sqrt{(3+1)^2+(-1-2)^2}[/tex]
[tex]D=5[/tex]
So,
The length of line segment is 5