Question

Point B has coordinates ​(4,1​). The​ x-coordinate of point A is -8. The distance between point A and point B is 13 units. What are the possible coordinates of point​ A?

Answer 1

Answer:

The possible coordinates of point A are (-8,-4) or (-8,6)

Step-by-step explanation:

The distance between two points is given by the formula

$d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} }$

For point A $(x_{1} , y_{1})$; $x_{1}= -8$ and $y_{1}$ is unknown

For point B $(x_{2} , y_{2})$; (4,1) i.e $x_{2}=4$ and $y_{2}=1$

and d = 13

Putting the values into the equation,

$13= \sqrt{[(4-(-8)]^{2} + (1-y_{1})^{2} }\\13= \sqrt{[(4+8)]^{2} + (1-y_{1})^{2} }\\13= \sqrt{12^{2} + (1-y_{1})^{2} }\\13^{2} = 12^{2} + (1-y_{1})^{2} \\169 = 144 +(1-y_{1})^{2}\\169-144 = (1-y_{1})^{2}\\25 = 1 -2y_{1} +y_{1}^{2} \\y_{1}^{2} -2y_{1}+1-25 =0\\y_{1}^{2} -2y_{1}-24 =0\\y_{1}^{2} -6y_{1} + 4y_{1} -24 =0 \\y_{1}(y_{1}-6) +4(y_{1} -6) = 0\\(y_{1}+4)(y_{1}-6) =0\\(y_{1}+4)=0 or (y_{1}-6) =0$

$y_{1}+4= 0$ or $y_{1}-6 =0$

$y_{1} =-4$ or $y_{1} = 6$

Hence, the possible coordinates of point A are (-8,-4) or (-8,6)