An Important Question of class 10 Based on Equation of a Straight Line Chapter of M.L Aggarwal for ICSE BOARD.

Here given that a straight line is passing through the origin and through the point of intersection of the given lines.

So, you have to find the equation of a straight line.

This is the Question Number 32, Exercise 12.1 of M.L Aggarwal.

# Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7

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Given line equations,

5x + 7y = 3 … (i)

2x – 3 y = 7 … (ii)

Now, performing multiplication of (i) by 3 and (ii) by 7, we get

15x + 21y = 9

14x – 21y = 49

On adding we get,

29x = 58

x = 58/29 = 2

Substituting the value of x in (i), we get

5(2) + 7y = 3

10 + 7y = 3

7y = 3 – 10

y = -7/7 = -1

Hence, the point of intersection of lines is (2, -1)

Now, the slope of the line joining the points (2, -1) and (0, 0) will be

m = y

_{2}– y_{1}/ x_{2}– x_{1}= (0 + 1)/ (0 – 2)

= -1/2

Equation of the line is given by:

y – y

_{1}= m (x – x_{1})y – 0 = -1/2 (x – 0)

2y = -x

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