Question

Line JK passes through points J(-3,11) and K(1, -3) O 7x + 2y =-1 (g) 7x+2y = 1 O 14x+4y =-1 O 14x+4y = 1

Answer 1

Line JK passes through $J(-3,11)$ , $K(1,-3)$

General equation of line is $y-y_{1}=m(x-x_{1})$

Where 'm' is the slope

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$(x_{1},y_{1})=(-3,11),(x_{2},y_{2})=(1,-3)$

So, slope is,

$m=\frac{-3-11}{1-(-3)}=\frac{-14}{4}=\frac{-7}{2}$

Substitute m=-7/2 in $y-y_{1}=m(x-x_{1})$

$y-11=\frac{-7}{2}(x-(-3))$

On multiplying with 2 on both sides

$2(y-11)=-7(x+3)$

$2y-22=-7x-21$

Shifting the terms,

$7x+2y-22+21=0$

$7x+2y-1=0$

Adding 1 on both sides

$7x+2y=1$

Therefore,

The answer is $7x+2y=1$