Question

Find the equation of straight line passing through the point of intersection of the lines 3x + 4y - 11 = 0 and 2x - 3y + 4 = 0 and parallel to the line 2x + 3y = 5.

Answer 1

Given ,

The equation of straight line passing through the point of intersection of the lines 3x + 4y - 11 = 0 and 2x - 3y + 4 = 0 and parallel to the line 2x + 3y = 5

Let ,

3x + 4y = 11 --- (i)

2x - 3y = -4 --- (ii)

Multiply eq (i) by 2 and eq (ii) by 3 , we get

6x + 8y = 22 --- (iii)

6x - 9y = -12 --- (iv)

Subtract eq (iv) from eq (iii) , we get

6x + 8y - (6x - 9y) = 22 - (-12)

8y + 9y = 34

17y = 34

y = 34/17

y = 2

Put y = 2 in eq (i) , we get

3x + 4(2) = 113x + 8 = 113x = 3x = 3/3x = 1

Therefore , the point of intersection of the lines 3x + 4y - 11 = 0 and 2x - 3y + 4 = 0 is (1 , 2)

We know that , the slope of the line passing passing through (x, y) and (xo , yo) is

$\boxed { \tt{Slope \: (m) = \frac{ y - y_{o}}{x- x_{o}}}}$

Thus ,

Slope (m) = (2 - y)/(1 - x)

Now , Slope of 2x + 3y - 5 = 0 will be

Slope (m) = -A/B = -2/3

We know that , if two lines are parrallel to each other, then

$\boxed{ \tt{m_{1} = m_{2}}}$

Thus ,

-2/3 = (2 - y)/(1 - x)

-2 + 2x = 6 - 3y

2x - 3y - 8 = 0

Therefore , the required equation of the line is 2x - 3y - 8 = 0

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Answer 2

Given ,

The equation of straight line passing through the point of intersection of the lines 3x + 4y - 11 = 0 and 2x - 3y + 4 = 0 and parallel to the line 2x + 3y = 5

Let ,

3x + 4y = 11 --- (i)

2x - 3y = -4 --- (ii)

Multiply eq (i) by 2 and eq (ii) by 3 , we get

6x + 8y = 22 --- (iii)

6x - 9y = -12 --- (iv)

Subtract eq (iv) from eq (iii) , we get

6x + 8y - (6x - 9y) = 22 - (-12)

8y + 9y = 34

17y = 34

y = 34/17

y = 2

Put y = 2 in eq (i) , we get

3x + 4(2) = 113x + 8 = 113x = 3x = 3/3x = 1

Therefore , the point of intersection of the lines 3x + 4y - 11 = 0 and 2x - 3y + 4 = 0 is (1 , 2)

We know that , the slope of the line passing passing through (x, y) and (xo , yo) is

$\boxed { \tt{Slope \: (m) = \frac{ y - y_{o}}{x- x_{o}}}}$

Thus ,

Slope (m) = (2 - y)/(1 - x)

Now , Slope of 2x + 3y - 5 = 0 will be

Slope (m) = -A/B = -2/3

We know that , if two lines are parrallel to each other, then

$\boxed{ \tt{m_{1} = m_{2}}}$

Thus ,

-2/3 = (2 - y)/(1 - x)

-2 + 2x = 6 - 3y

2x - 3y - 8 = 0

Therefore , the required equation of the line is 2x - 3y - 8 = 0

________________ Keep Smiling ☺️