Question

9. Find the equation of the straight line parallel to the line 3x + 4y=7 and passing through the point
of intersection of the lines x - 2y - 3 = 0 and x + 3y-6=0.​

Answer 1

Hey

First of all find the point

x - 2y - 3 = 0 ......(1)

x + 3y - 6 = 0 ..... (2)

(2) - (1)

3y + 2y -6 + 3 = 0

5y - 3 = 0

y = 3/5

sub y in (2) { u can sub in any equation }

x + 3(3/5) -6 = 0

x + 9/5 - 6 = 0

x + 9 - 30/5 = 0

x - 21/5 = 0

x = 21/5

Now the point is (21/5 , 3/5)

slope = -a / b

m = -3/4

The equation of straight line parallel to line 3x + 4y = 7 and passing through point (21/5 , 3/5) is

Refer to the attachment from here

Answer 2

The equation of the required line is 15x+20y = 75.

Given,

The equation of three lines 3x + 4y=7, x - 2y - 3 = 0,and x + 3y-6=0.​

To Find,

The equation of the line parallel to 3x + 4y=7 and passing through the intersection of x - 2y - 3 = 0,and x + 3y-6=0.​

Solution,

First we will find the intersection point of the lines.

x - 2y - 3 = 0 ......(1)

x + 3y - 6 = 0 ..... (2)

(2) - (1)

3y + 2y -6 + 3 = 0

5y - 3 = 0

y = 3/5

Now, substituting the value of y in (2)

x + 3(3/5) -6 = 0

x + 9/5 - 6 = 0

x + 9 - 30/5 = 0

x - 21/5 = 0

x = 21/5

Now the point is (21/5 , 3/5)

The slope of line slope 3x + 4y=7 = -a / b = -3/4

Now, the equation of the required line will be

(y-3/5) = -3/4(x-21/5)

5y-3 = -3/4(5x-21)

20y-12 = -15x+63

15x+20y = 75

Hence, the equation of the required line is 15x+20y = 75.

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