Question

Find the equation of the line through the intersection of the lines 2x + 3y – 2 = 0 and x – 2y + 1 = 0 and having x-intercept equal to 3.

Answer 1

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

2x + 3y – 2 = 0 …(i)

x – 2y + 1 = 0 …(ii)

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

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

2x – 4y + 2 = 0 …(iii)

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

2x + 3y – 2 – 2x + 4y – 2 = 0

⇒ 7y – 4 = 0

⇒ 7y = 4

⇒ y = 4/7

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

Hence, the point of intersection P(x₁, y₁) is \((\frac{1}{7},\frac{4}{7})\)

Now, the equation of a line in intercept form is:

x/a + y/b = 1

where a and b are the intercepts on the axis.

Given that: a = 3

⇒ bx + 3y = 3b …(i)

If eq. (i) passes through the point \((\frac{1}{7},\frac{4}{7})\), we get

⇒ b + 12 = 21b

⇒ b – 21b = - 12

⇒ 20b = 12

⇒ b = 12/20 = 3/5

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

⇒ 3x + 15y = 9

⇒ x + 5y = 3

Hence, the required equation of line is x + 5y = 3