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Solution


Q.

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point (2,-43) divides the third side BC internally in the ratio 2 : 1, the equation of the side BC is          [2024]
1 x + 3y + 2 = 0  
2 x – 3y – 6 = 0  
3 x + 6y + 6 = 0  
4 x – 6y – 10 = 0  

Ans.

(1)

Let D(2, -43) be the point on BC which divides BC in ratio 2 : 1.

Let coordinates of B be (x1, 14 - 4x1) and coordinates of C be (x2, 3x2 - 52)

Now, 2x2 + x13 = 2, 2(3x2 - 52) + 14 - 4x13 = -43

⇒   2x2 + x1 = 6, 3x2 - 4x1 = -4 + 5 - 14

⇒   2x2 + x1 = 6, 3x2 - 4x1 = -13

On solving these two equations, we get, x1 = 4, x2 = 1

So B(4, –2) and C(1, –1) are the required points.

Equation of BC : y+1=-13(x-1)⇒3y+x+2=0