Question

Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*

Answer 1

A = (-1, 1) and B = (2, k) so:

`d(A,B)=√((x_B-x_A)^2+(y_B-y_A)^2)=√((2-(-1))^2+(k-1)^2)=\\\\=√(3^2+(k-1)^2)=√(9+k^2-2k+1)=√(k^2-2k+10)\\\\\\d(A,B)=5\\\\√(k^2-2k+10)=5\quad|(\ldots)^2\\\\\big|k^2-2k+10\big|=25\\\\k^2-2k+10=25\qquad\vee\qquad k^2-2k+10=-25\\\\k^2-2k-15=0\qquad\vee\qquad k^2-2k+35=0\\\\\\\Delta_1=(-2)^2-4\cdot1\cdot(-15)=64>0\qquad\text{two solutions}\\\\\Delta_2=(-2)^2-4\cdot1\cdot35=-136<0\qquad\text{no solutions}\\\\\\k_1=(2-√(64))/(2)=(2-8)/(2)=(-6)/(2)=\boxed{-3}`

`k_2=(2+√(64))/(2)=(2+8)/(2)=(10)/(2)=\boxed{5}`

Answer 2

k = -3

k =5

Explanation:

`d = √((x_2-x_1)^2+(y_2-y_1)^2)\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\`

`5=√(\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2)\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\`

`\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3`