Question

The equation of straight line through the intersection of the lines x – 2y = 1 and x + 3y = 2 and parallel to 3x + 4y = 0 is

Options

• 3x + 4y + 5 = 0

• 3x + 4y – 10 = 0

• 3x + 4y – 5 = 0

• 3x + 4y + 6 = 0

Answer 1

3x + 4y – 5 = 0

Explanation:

The line is passing through the intersection of lines x – 2y = 1 and x + 3y = 2. This line is parallel to the line 3x + 4y = 0.

Let the equation of the line by y = mx + c

Since this is parallel to the line 3x + 4y = 0

∴ m = `- 3/4`.

Putting this value in y = mx + c

We get, y = `- 3/4 x + c`  ......(i)

Since this passing through the point which is intersection of x – 2y = 1 and x + 3y = 2, we solve for the values of x and y

x – 2y = 1
x + 3y = 2
–    –     –
– 5y = – 1

⇒ y = `1/5`

Putting value of y in the equation of any of the two lines, say x – 2y = 1

`x - 2 xx 1/5` = 1 ⇒ x = `1 + 2/5 = 7/5`

So, point of intersection is `(7/5, 1/5)` and putting these values in equation (1) we get

`1/5 = - 3/4 xx 7/5 + c`

⇒ c = `1/5 + 21/20 = (4 + 21)/20`

⇒ c = `25/20 = 5/4`

Hence equation of the line is y = `(-3)/4 x + 5/4`

or 4y = – 3x + 5 or 3x + 4y – 5 = 0