Find the equation of the line passing through the point of intersectio

1.	Find the equation of the line passing through the point of intersection of the lines 4x – 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer» Given: 4x− 7y − 3 = 0 … (1) 2x − 3y + 1 = 0 … (2) To find: Equation of line passing through the point of intersection of lines. Concept Used: Point of intersection of two lines. Explanation: Solving (1) and (2) using cross - multiplication method: \(\frac{x}{-7-9}=\frac{y}{-6-4}=\frac{1}{-12+14}\) ⇒ x = - 8 , y = - 5 Thus, the point of intersection of the given lines is (- 8, - 5). Now, the equation of a line having equal intercept as a is \(\frac{x}{a}+\frac{y}{a}=1\) This line passes through ( - 8, - 5) \(-\frac{8}{a}-\frac{5}{a}=1\) ⇒ - 8 - 5 = a ⇒ a = - 13 Hence, the equation of the required line is \(\frac{x}{-13}=\frac{y}{-13}=1\) or x + y + 13 = 0

Find the equation of the line passing through the point of intersection of the lines 4x – 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer»

Given:

4x− 7y − 3 = 0 … (1)

2x − 3y + 1 = 0 … (2)

To find:

Equation of line passing through the point of intersection of lines.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2) using cross - multiplication method:

\(\frac{x}{-7-9}=\frac{y}{-6-4}=\frac{1}{-12+14}\)

⇒ x = - 8 , y = - 5

Thus, the point of intersection of the given lines is (- 8, - 5).

Now, the equation of a line having equal intercept as a is \(\frac{x}{a}+\frac{y}{a}=1\)

This line passes through ( - 8, - 5)

\(-\frac{8}{a}-\frac{5}{a}=1\)

⇒ - 8 - 5 = a

⇒ a = - 13

Hence, the equation of the required line is \(\frac{x}{-13}=\frac{y}{-13}=1\) or x + y + 13 = 0