If the line 2x - y - 4 = 0 divides the line s | vedclass

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(D) Let the ratio be $m:n = k:1$. The point $(a, b)$ divides the segment joining $(2, -1)$ and $(1, -4)$ in the ratio $k:1$. Using the section formula

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Let the ratio be $m:n = k:1$. The point $(a, b)$ divides the segment joining $(2, -1)$ and $(1, -4)$ in the ratio $k:1$. Using the section formula,$a = \frac{k(1) + 1(2)}{k+1} = \frac{k+2}{k+1}$ and $b = \frac{k(-4) + 1(-1)}{k+1} = \frac{-4k-1}{k+1}$. Since $(a, b)$ lies on the line $2x - y - 4 = 0$,we have $2(\frac{k+2}{k+1}) - (\frac{-4k-1}{k+1}) - 4 = 0$. Multiplying by $(k+1)$,we get $2k + 4 + 4k + 1 - 4(k+1) = 0$,which simplifies to $6k + 5 - 4k - 4 = 0$,so $2k + 1 = 0$,giving $k = -\frac{1}{2}$. Thus,$\frac{m}{n} = -\frac{1}{2}$. Substituting $k = -\frac{1}{2}$ into the coordinates: $a = \frac{-0.5+2}{-0.5+1} = \frac{1.5}{0.5} = 3$ and $b = \frac{-4(-0.5)-1}{-0.5+1} = \frac{2-1}{0.5} = 2$. Finally,$4(a - b(\frac{m}{n})^2) = 4(3 - 2(-\frac{1}{2})^2) = 4(3 - 2(\frac{1}{4})) = 4(3 - 0.5) = 4(2.5) = 10$.

If the line $2x - y - 4 = 0$ divides the line segment joining the points $(2, -1)$ and $(1, -4)$ at the point $(a, b)$ in the ratio $m:n$,then $4(a - b(\frac{m}{n})^2) = $

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