Question

For what value of the three straight lines 7x-11y+3=0, 4x+3y-9=0 and 13x+ay-48=0 pass through the same point.​

Answer 1

Given :  three straight lines 7x-11y+3=0, 4x+3y-9=0 and 13x+ay-48=0 pass through the same point.​

To find : Value of a

Solution:

7x-11y+3=0,

4x+3y-9=0

13x+ay-48=0

passes though  the same point.​  if

$\left|\begin{array}{ccc}7&-11&3\\4&3&-9\\13&a&-48\end{array}\right| =0$

=> 7 (-144 + 9a) - (-11)(-192 + 117) + 3(4a - 39) = 0

=> -1008 + 63a  -825 + 12a - 117 = 0

=> 75a  = 1950

=> a = 1950/75

=> a = 26

Value of a = 26

other way

7x-11y+3=0,      21x - 33y + 9 = 0

4x+3y-9=0      44x + 33y - 99 = 0

=> 65x = 90

=> x = 18/13

    y = 15/13

put values of x & y in   13x+ay-48=0  to find a

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