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

Answer:

The length of line segment is 5

Step-by-step explanation:

we are given equation of parabolas as

$y=x^2-\frac{11}{4}x-\frac{7}{4}$

$y=-\frac{7}{8}x^2+x+\frac{31}{8}$

Firstly, we will find intersection points

we can set them equal

and then we can solve for x

$x^2-\frac{11}{4}x-\frac{7}{4}=-\frac{7}{8}x^2+x+\frac{31}{8}$

Multiply all sides by 8

$x^2\cdot \:8-\frac{11}{4}x\cdot \:8-\frac{7}{4}\cdot \:8=-\frac{7}{8}x^2\cdot \:8+x\cdot \:8+\frac{31}{8}\cdot \:8$

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

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

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:

$y=(-1)^2-\frac{11}{4}(-1)-\frac{7}{4}$

y=2

so, we get point as

(-1,2)

At x=3:

$y=(3)^2-\frac{11}{4}(3)-\frac{7}{4}$

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

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, we can plug values

$D=\sqrt{(3+1)^2+(-1-2)^2}$

$D=5$

So,

The length of line segment is 5