Question

the length of line segment is 13 units and coordinates of one end point are(- 6,7) if the coordinates of other end point is -1 find the abscissa of the other end point.

Answer 1

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Answer 2

Answer:

$x= -6-\sqrt{105}, \sqrt{105}-6$

Step-by-step explanation:

(x,y)

Where x is abscissa

Y is ordinate

We are given the coordinate of one end point are(- 6,7)

The ordinate of other point is -1

Let abscissa of that point be a

So, point = (a,-1)

We are also given the distance between these two points is 13 units

So, to find a we will use distance formula :

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

d = 13

$(x_1,y_1)=(- 6,7)$

$(x_2,y_2)=(a,-1)$

Substitute the values

$13 =\sqrt{(a-(-6))^2+(-1-7)^2}$

$13 =\sqrt{(a+6)^2+(-8)^2}$

Squaring both sides

$169=(a+6)^2+64$

$169-64=(a+6)^2+64$

$105=a^2+36+12a$

$a^2+36-105+12a=0$

$a^2+12a-69=0$

So, using discriminant rule:

$x= \frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

a= 1

b =12

c = -69

Substitute the values

$x= \frac{-12\pm\sqrt{(12^2-4(1)(-69))}}{2(1)}$

$x= -6-\sqrt{105}, \sqrt{105}-6$

Hence the abscissa of the other end point. is $x= -6-\sqrt{105}, \sqrt{105}-6$