Question

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).

Answer 1

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)

Slope of line joining two points = \(\frac{y_2-y_1}{x_2-x_1}\)

∴ m_AB = \(\frac{-5-2}{4-1}=\frac{-7}{3}\)

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