Answer :

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

2x – 3y = 0 …(i)


4x – 5y = 2 …(ii)


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


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


4x – 6y = 0 …(iii)


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


4x – 5y – 4x + 6y = 2 – 0


y = 2


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


2x – 3(2) = 0


2x – 6 = 0


2x = 6


x = 3


Hence, the point of intersection P(x1, y1) is (3, 2)



Now, we know that, when two lines are perpendicular, then the product of their slope is equal to -1


m1 × m2 = -1


Slope of the given line × Slope of the perpendicular line = -1



The slope of the perpendicular line = 2


So, the slope of a line which is perpendicular to the given line is 2


Then the equation of the line passing through the point (3, 2) having slope 2 is:


y – y1 = m (x – x1)


y – 2 = 2(x – 3)


y – 2 = 2x – 6


2x – y – 6 + 2 = 0


2x – y – 4 = 0



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