हिंदी

A(1, 4), B(3, 2) and C(7, 5) are the vertices of a $$\triangle \mathrm{ABC}$$. Find: i. the co-ordinates of the centroid G of $$\triangle \mathrm{ABC}$$, ii. the equation of a line through G

Advertisements
Advertisements

प्रश्न

A(1, 4), B(3, 2) and C(7, 5) are the vertices of a $$\triangle \mathrm{ABC}$$. 

Find:

  1. the co-ordinates of the centroid G of $$\triangle \mathrm{ABC}$$, 
  2. the equation of a line through G and parallel to AB.
योग
Advertisements

उत्तर

i. Centroid $$\mathrm{G}(x, y)$$ of $$\triangle \mathrm{ABC}$$ is:

$$x = \frac{1 + 3 + 7}{3} = \frac{11}{3}$$ 

$$y = \frac{4 + 2 + 5}{3} = \frac{11}{3}$$

So, $$\mathrm{G} = \left(\frac{11}{3}, \frac{11}{3}\right)$$.

ii. Slope of line $$\mathrm{AB}$$: 

$$m = \frac{2 - 4}{3 - 1}$$

$$= \frac{-2}{2}$$

$$= -1$$

Since the required line is parallel to $$\mathrm{AB}$$, its slope is also $$m = -1$$. 

Passing through $$\mathrm{G}\left(\frac{11}{3}, \frac{11}{3}\right)$$: 

$$y - \frac{11}{3} = -1\left(x - \frac{11}{3}\right)$$

$$y - \frac{11}{3} = -x + \frac{11}{3}$$

$$x + y = \frac{22}{3}$$

$$3x + 3y - 22 = 0$$

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Equation of a Straight Line - EXERCISE 14B [पृष्ठ २३२]

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 14 Equation of a Straight Line
EXERCISE 14B | Q 26. | पृष्ठ २३२
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×