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  • Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Q : 15     Find the distance of the line  \small 4x+7y+5=0  from the point  \small (1,2)  along the line \small 2x-y=0. 

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point \small (1,2) lies on line 2x-y =0
Now, point of intersection of lines  2x-y =0    and    \small 4x+7y+5=0  is \left ( -\frac{5}{18},-\frac{5}{9} \right )
Now, we know that the distance between two point is given by 
d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|
d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|
d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |
d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}
Therefore, the distance of the line  \small 4x+7y+5=0  from the point  \small (1,2)  along the line \small 2x-y=0  is   \frac{23\sqrt5}{18} \ units

Posted by

Gautam harsolia

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