Straight Line Question 107
Question: The distance of the lines $ 2x-3y=4 $ from the point (1, 1) measured parallel to the line $ x+y=1 $ is[Orissa JEE 2002]
Options:
A) $ \sqrt{2} $
B) $ \frac{5}{\sqrt{2}} $
C) $ \frac{1}{\sqrt{2}} $
D)6
Correct Answer: A
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Answer:
Solution:
Þ $ \frac{x-1}{-1/\sqrt{2}}=\frac{y-1}{1/\sqrt{2}}=r $ Co-ordinates of any point on this line are $ ( 1-\frac{r}{\sqrt{2}},1+\frac{r}{\sqrt{2}} ) $ If this point lies on $ 2x-3y=4 $ , then $ 2( 1-\frac{r}{\sqrt{2}} )-3( 1+\frac{r}{\sqrt{2}} )=4 $
Þ $ r=\sqrt{2}. $
BETA
