Question

Find the equation of the line passing through the origin and the point of intersection of the lines $$5x + 7y = 3$$ and $$2x - 3y = 7$$.

[Hint: On solving 5x + 7y = 3 and 2x – 3y = 7, we get x = 2, y = –1. Now, find the equation of a line passing through the points A(2, –1) and O(0, 0).]

Answer 1

Given equations:

$$5x + 7y = 3 \quad ...(1)$$ 

$$2x - 3y = 7 \quad ...(2)$$

Multiply equation (1) by $$3$$ and equation (2) by $$7$$: 

$$15x + 21y = 9$$ 

$$14x - 21y = 49$$

Adding these two equations:

$$29x = 58$$

$$\implies x = 2$$

Substitute $$x = 2$$ in equation (1): 

$$5(2) + 7y = 3$$

$$10 + 7y = 3$$

$$7y = -7$$

$$\implies y = -1$$

So, the point of intersection is $$A(2, -1)$$.

The required line passes through the origin $$O(0, 0)$$ and $$A(2, -1)$$.

Slope of the line: $$m = \frac{-1 - 0}{2 - 0} = -\frac{1}{2}$$

Equation of the line passing through origin:

$$y = mx \implies y = -\frac{1}{2}x$$ 

$$2y = -x \implies x + 2y = 0$$

Thus, the required equation is $$x + 2y = 0$$.