हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Given values:

5x + 7y = 3   ...(Equation 1)

2x − 3y = 7   ...(Equation 2)

Multiply Equation 1 by 3 and Equation 2 by 7 to eliminate y:

15x + 21y = 9   ...(Equation 3)

14x − 21y = 49   ...(Equation 4)

Now, adding Equations 3 and 4:

29x = 58

x = `58/29`

∴ x = 2

Substitute x = 2 back into Equation 1:

5(2) + 7y = 3

10 + 7y = 3

7y = 3 − 10

7y = −7

∴ y = −1

The required line passes through the origin (0, 0) and the intersection point (2, −1):

`m = (y_2 - y_1)/(x_2 - x_1)`

`m = (-1 - 0)/(2 - 0)`

∴ `m = -1/2`

Let’s solve the equation:

y = mx

`y = -1/2x`

2y = −x

∴ x + 2y = 0

Hence, the equation of the line passing through the origin and the point of intersection is x + 2y = 0.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Equation of a line - Exercise 12A [पृष्ठ २४५]

APPEARS IN

नूतन Mathematics [English] Class 10 ICSE
अध्याय 12 Equation of a line
Exercise 12A | Q 19. | पृष्ठ २४५
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×