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The straight lines 1_1 and 1_2 pass through the origin and trisect the line segment of the line L: 9 x+5 y=45 between the axes. If m_{1} and m_{2} are the slopes of the lines 1_{1} and 1_{2}, then the point of intersection of the line y = (m1 + m2) x with L lies on.

Option: 1

6 x+y=10


Option: 2

6 x-y=15


Option: 3

y-2 x=5


Option: 4

y-x=5


Answers (1)

best_answer



\rightarrow \mathrm{P}_{\mathrm{x}}=\frac{2 \times 5+1 \times 0}{1+2}=\frac{10}{3}
\mathrm{P}_{\mathrm{y}}=\frac{0 \times 2+9 \times 1}{1+2}=3
\mathrm{P}:\left(\frac{10}{3}, 3\right)

Similarly \: \: \rightarrow Q_x=\frac{1 \times 5+2 \times 0}{1+2}=\frac{5}{3}

Q_y=\frac{1 \times 0+2 \times 9}{1+2}=6
Q:\left(\frac{5}{3}, 6\right)

\text { Now } \mathrm{m}_1=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}
\mathrm{~m}_2=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}

Now\: L_{1}:y\left ( m_{1}+m_{2} \right )x\Rightarrow y= \left ( \frac{9}{2} \right )x\Rightarrow 9x= 2y\: \dots\left ( 2 \right )
from (1) & (2)

9 x+5 y=45
9 \mathrm{x}-2 \mathrm{y}=0
-\quad+\quad-
7 \mathrm{y}=45 \Rightarrow \mathrm{y}=\frac{45}{7}
\Rightarrow x=\frac{10}{7} \\
which satisfy  y-x=5 Ans. 4
 

Posted by

Irshad Anwar

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