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

Let AB lines segment of length $13$ units $A=(-6, 7)$

Let coordinate of B be 'm', $B=(-1, m)$

$AB=13$, $AB^2=169$

$(-6+1)^2+(7-m)^2=169$

$(-5)^2+(7-m)^2=169$

$25+49+m^2-14m=169$

$m^2-14m+74=169$

$m^2-14m-95=0$

$m^2-19m+5m-95=0$

$m(m-19)+5(m-19)=0$

$(m+5)(m-19)=0$

$m=-5$ and $m=19$

Thus, point B can be $(-1, -5)$ or $(-1, 19)$.