Question

find the value of k for which the pair of equation 3 X + K Y + 2 Y + 5 = 20, 6 x minus 8 Y + 7 is equal to zero represent parallel lines​

Answer 1

k=6

Step-by-step explanation:

here the equations are 3X +KY+2Y+5=20   (1)

                                         6X-8Y +7 =0             (2)

by rewriting the equations in standard form ax+by+c=0

we get

3X +(k+2) Y+5-20=0  or  3X+(k+2)Y-15=0

           and                        6X -   8Y   +7 =0        

now for parallel line we have condition

$\frac{a_{1} }{a_{2} } =\frac{b_{1} }{b_{2} } \neq \frac{ c_{1} }{c_{2} }$   for two lines $a_{1}X +b_{1} Y+C_{1} =0 and a_{2}X +b_{2} Y+C_{2} =0$

hence

$\frac{3}{6} = \frac{k+2}{-8} \neq \frac{-15}{7}$

equating 1st two and doing cross multiplication we get

3×-8 = 6(k+2)

-24=6k+12

6k=12+24

6k=36

k=36/6

k=6

 

                               

Answer 2

Answer:

Step-by-step explanation: