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If the line 2x + y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k equals

Option 1)
$\frac{29}{5}$

Option 2)
$5$

Option 3)
$6$

Option 4)
$\frac{11}{5}$

Answers (2)

best_answer

As learnt in

Selection formula -

$x= \frac{mx_{2}+nx_{1}}{m+n}$

$y= \frac{my_{2}+ny_{1}}{m+n}$

- wherein

If P(x,y) divides the line joining A(x₁,y₁) and B(x₂,y₂) in ration $m:n$

$: A(1,1);B(2,4)$

$p(x_{1}y_{1})\: divides\: line \: segment\: \: AB\: in\: the\: ratio\: 3:2$

$x_{1}= \frac{3(2)+2(1)}{5}= \frac{8}{5}$ $y_{1}= \frac{3(4)+2(1)}{5}= \frac{14}{5}$

$2x+y=k\: \: passes\: through\: P(x_{1},y_{1})$

$\therefore 2\times \frac{8}{5}+\frac{14}{5}= k\Rightarrow k=6$

Option 1)

$\frac{29}{5}$

This option is incorrect.

Option 2)

$5$

This option is incorrect.

Option 3)

$6$

Let that point be P $(x_1,y_1)$ .

Then applying Section Formula -

If $(x,y)$ P(x , y) divides the line joining A $(x_{1}-y_{1}}})$ & B $(x_{2}-y_{2}}})$ in the ratio m : n, then ;

$x=\frac{mx_{2}+nx_{1}}{m+n}$ $y=\frac{my_{2}+ny_{1}}{m+n}$

$x_1=\dfrac{3*2+2*1}{5}= \dfrac{8}{5}$

$y_1=\dfrac{3*4+2*1}{5}= \dfrac{14}{5}$

The point P lies on given line. So,

$k=2x+y=2*\dfrac{8}{5}+\dfrac{14}{5}=\dfrac{30}{5}=6$

This option is correct.

Option 4)

$\frac{11}{5}$

This option is incorrect.

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