Q. 183.7( 4 Votes )

Find the equation

Answer :

Given two lines are:


x – y + 1 = 0 …(i)


and 2x – 3y + 5 = 0 …(ii)


Now, point of intersection of these lines can be find out as:


Multiplying eq. (i) by 2, we get


2x – 2y + 2 = 0 …(iii)


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


2x – 2y + 2 – 2x + 3y – 5 = 0


y – 3 = 0


y = 3


On putting value of y in eq. (ii), we get


2x – 3(3) + 5 = 0


2x – 9 + 5 = 0


2x – 4 = 0


2x = 4


x = 2


So, the point of intersection of given two lines is:


(x, y) = (2, 3)


Let m be the slope of the required line


Equation of the line is


y – 3 = m(x – 2)


y – 3 = mx – 2m


mx – y – 2m + 3 = 0 …(i)


Since, the perpendicular distance from the point (3, 2) to the line is then





Squaring both the sides, we get



49(m2 + 1) = 25(m + 1)2


49m2 + 49 = 25(m2 + 1 + 2m)


49m2 + 49 = 25m2 + 25 + 50m


25m2 + 25 + 50m – 49m2 – 49 = 0


– 24m2 + 50m – 24 = 0


– 12m2 + 25m – 12 = 0


12m2 – 25m + 12 = 0


12m2 – 16m – 9m + 12 = 0


4m (3m – 4) – 3(3m – 4) = 0


(3m – 4)(4m – 3) = 0


3m – 4 = 0 or 4m – 3 = 0


3m = 4 or 4m = 3


or



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







4x – 3y + 1 = 0


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







3x – 4y + 6 = 0


hence, the required equation are 4x – 3y + 1 = 0 and 3x – 4y + 6 = 0


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