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.


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




So, the slope of a line which is parallel to this line is also


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


y – y1 = m (x – x1)



y – 3 = - 3x + 12


4y – 12 = -3x + 12


3x + 4y – 12 – 12=0


3x + 4y – 24 = 0



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