Answer :
[tex]3x+ky=5[/tex]
Plugin the given point [tex](-1, 2)[/tex] and solve [tex]k[/tex]
[tex]3(-1)+k(2)=5\\-3+2k=5\\2k=8\\k=4[/tex]
Second part is bit tricky :
Plugging in the point [tex](1,0)[/tex] gives
[tex]3(1)+k(0)=5\\3+0=5\\3=5[/tex]
which is impossible. so there is no [tex]k[/tex] such that the line [tex]3x+ky=5[/tex] passes through [tex](1,0)[/tex]
Plugin the given point [tex](-1, 2)[/tex] and solve [tex]k[/tex]
[tex]3(-1)+k(2)=5\\-3+2k=5\\2k=8\\k=4[/tex]
Second part is bit tricky :
Plugging in the point [tex](1,0)[/tex] gives
[tex]3(1)+k(0)=5\\3+0=5\\3=5[/tex]
which is impossible. so there is no [tex]k[/tex] such that the line [tex]3x+ky=5[/tex] passes through [tex](1,0)[/tex]
3x+ky = 5
when we substitute (x,y) = (-1,2) in 3x+ky = 5
3(-1)+k(2) = 5
2k=8
k=4
when we substitute (x,y) = (1,0) in 3x+ky = 5
3(1)+k(0) = 5
3 = 5
therefore it is not possible to the line 3x+ky = 5 to pass into the point (1,0)
when we substitute (x,y) = (-1,2) in 3x+ky = 5
3(-1)+k(2) = 5
2k=8
k=4
when we substitute (x,y) = (1,0) in 3x+ky = 5
3(1)+k(0) = 5
3 = 5
therefore it is not possible to the line 3x+ky = 5 to pass into the point (1,0)