Respuesta :
Answer:
The length of the line segment is 5 unit.
Step-by-step explanation:
The given equations are
[tex]y=x^2-\frac{11}{4}x-\frac{7}{4}[/tex] .... (1)
[tex]y=-\frac{7}{8}x^2+x+\frac{31}{8}[/tex] .... (2)
Equate both equations.
[tex]x^2-\frac{11}{4}x-\frac{7}{4}=-\frac{7}{8}x^2+x+\frac{31}{8}[/tex]
[tex]\frac{1}{4}(4 x^2 - 11 x - 7) = \frac{1}{8}(-7 x^2 + 8 x + 31)[/tex]
[tex]2(4 x^2 - 11 x - 7) = -7 x^2 + 8 x + 31[/tex]
[tex]8 x^2 - 22 x - 14 = -7 x^2 + 8 x + 31[/tex]
[tex]15 x^2 - 30 x - 45 = 0[/tex]
[tex]15 x^2 - 45x+30x - 45 = 0[/tex]
[tex]15x(x-3)x+15(x-3)=0[/tex]
[tex]15(x+1)(x-3)x=0[/tex]
[tex]x=-1,3[/tex]
The value of y at x=-1.
[tex]y=(-1)^2-\frac{11}{4}(-1)-\frac{7}{4}=2[/tex]
The value of y at x=3.
[tex]y=(3)^2-\frac{11}{4}(3)-\frac{7}{4}=-1[/tex]
Therefore the intersection points of given parabolas are (-1,2) and (3,-1).
The length of line segment is
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]d=\sqrt{(3-(-1))^2+(-1-2)^2}[/tex]
[tex]d=\sqrt{16+9}[/tex]
[tex]d=5[/tex]
Therefore the length of the line segment is 5 unit.
