Answer :

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

x – 2y + 3 = 0 …(i)


2x – 3y + 4 = 0 …(ii)


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


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


2x – 4y + 6 = 0 …(iii)


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


2x – 3y + 4 – 2x + 4y – 6 = 0


y – 2 = 0


y = 2


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


x – 2(2) + 3 = 0


x – 4 + 3 = 0


x – 1 = 0


x = 1


Hence, the point of intersection P(x1, y1) is (1, 2)



Let AB is the line drawn from the point of intersection (1, 2) and passing through the point (4, -5)


Firstly, we find the slope of the line joining the points (1, 2) and (4, -5)




Now, we have to find the equation of line passing through point (4, -5)


Equation of line: y – y1 = m(x – x1)




3y + 15 = -7x + 28


7x + 3y + 15 – 28 = 0


7x + 3y – 13 = 0


Hence, the equation of line passing through the point (4, -5) is 7x + 3y – 13 = 0



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