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

Here intersection of the lines are given though which a line is passing. Solve this question

This is the Question Number 38, Exercise 12.2 of M.L Aggarwal.

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# Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and (i) parallel to the line x + 2y – 5 = 0 (ii) perpendicular to the x-axis.

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Given, line equations:

4x + 3y = 1 … (1)

5x + 4y = 2 … (2)

On solving the above equation to find the point of intersection, we have

Multiplying (1) by 4 and (2) by 3

16 x + 12y = 4

15x + 12y = 6

On subtracting, we get

x = -2

Putting the value of x in (1), we have

4(-2) + 3y = 1

-8 + 3y = 1

3y = 1 + 8 = 9

y = 9/3 = 3

Hence, the point of intersection is (-2, 3).

(i) Given line, x + 2y – 5 = 0

2y = -x + 5

y = -(1/2) x + 5/2

Slope (m) = -1/2

A line parallel to this line will have the same slope m = -1/2

So, the equation of line having slope m and passing through (-2, 3) will be

y – 3 = (-1/2) (x + 2)

2y – 6 = -x – 2

x + 2y = 4

(ii) As any line perpendicular to x-axis will be parallel to y-axis.

So, the equation of line will be

x = -2 ⇒ x + 2 = 0