Find the equation of the line drawn through the point of intersection of the lines x – 2y + 3 = 0 and 2x – 3y + 4 = 0 and passing through the point (4, -5).
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