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Find the equation of the line drawn through the point of intersection of the lines x – y = 1 and 2x – 3y + 1 = 0 and which is parallel to the line 3x + 4y = 12.

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Suppose the given two lines intersect at a point P(x1, y1). Then, (x1, y1) satisfies each of the given equations.

x – y = 1 …(i)

2x – 3y + 1 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 2, we get

2x – 2y = 2

or 2x – 2y – 2 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 3y + 1 – 2x + 2y + 2 = 0

⇒ - y + 3 = 0

⇒ y = 3

Putting the value of y in eq. (i), we get

x – 3 = 1

⇒ x = 1 + 3

⇒ x = 4

Hence, the point of intersection P(x1, y1) is (4, 3)

Now, we find the slope of the given equation 3x + 4y = 12

We know that the slope of an equation is

m = - a/b

⇒ m = -3/4

So, the slope of a line which is parallel to this line is also − 3/4

Then the equation of the line passing through the point (4, 3) having a slope − 3/4 is:

y – y1 = m (x – x1)

⇒ y - (3) = -3/4(x - 4)

⇒ y – 3 = - 3x + 12

⇒ 4y – 12 = -3x + 12

⇒3x + 4y – 12 – 12 = 0

⇒ 3x + 4y – 24 = 0

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