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ICSE Class X Math 2013 Paper

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Question : 28 of 46
Marks: +1, -0
In △ABC,A(3,5),B(7,8)\triangle A B C, A(3,5), B(7,8) and C(1,−10)C(1,-10). Find the equation of the median through AA.
Solution:  
Here DD is mid point of BCB C. (Given )
∴\therefore The co-ordinate of D=(7+12,8−102)D = \left( \frac{7+1}{2}, \frac{8-10}{2} \right)
=(4,−1)= (4,-1)
Now equation of median ADAD,
y−y1=y2−y1x2−x1(x−x1)y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)
Here, x1=3,y1=5,x2=4,y2=−1x_1 = 3, y_1 = 5, x_2 = 4, y_2 = -1
y−5=−1−54−3(x−3)y - 5 = \frac{-1-5}{4-3}(x - 3)
y−5=−61(x−3)y - 5 = \frac{-6}{1}(x - 3)
y−5=−6x+18y - 5 = -6x + 18
y=−6x+18+5y = -6x + 18 + 5
y=−6x+23y = -6x + 23
6x+y−23=06x + y - 23 = 0
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