Question

transform the following equations into the form L1+kL2=0 and find the point of concurrence of the family of straight lines represented by the equation i)(2+5k)x-3(1+2k)y+2-k=0​

Answer 1

SOLUTION

TO DETERMINE

Transform the following equations into the form

$\sf{L_1 + kL_2 = 0 \: }$

and find the point of concurrence of the family of straight lines represented by the equation

(2+5k)x-3(1+2k)y+2-k=0

EVALUATION

Here the family of straight lines represented by the equation

$\sf{(2 + 5k)x - 3(1 + 2k)y + 2 - k = 0 \: }$

Which can be rewritten as

$\implies \sf{2x + 5kx- 3y - 6ky + 2 - k = 0}$

$\implies \sf{(2x - 3y + 2)+k(5x - 6y - 1) = 0 \: }$

Which of the form

$\sf{L_1 + kL_2 = 0 \: }$

Where

$\sf{L_1 : \: 2x - 3y + 2 = 0} \: \: .....(1)$

$\sf{L_2 : \: 5x - 6y - 1 = 0} \: \: \:. ...(2)$

Now the point of concurrence of the family of straight lines represented by the equation is the point of intersection of the the family of straight lines represented by the equation

Now 2× Equation (1) - Equation (2) gives

$\sf{ - x + 5 = 0 \: }$

$\implies \sf{x = 5 \: }$

From Equation (1)

$\sf{3y = 12 \: }$

$\implies \sf{y = 4}$

Hence the required point of concurrence of the family of straight lines represented by the equation is ( 5, 4 )

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

Draw the graph of the linear equation

3x + 4y = 6.

At what points, the graph cuts X and Y-axis

https://brainly.in/question/18185770