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  • A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Q : 11          A line perpendicular to the line segment joining the points  (1,0)  and  (2,3)  divides it in the ratio 1:n. Find the equation of the line. 

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Co-ordinates of point which divide line segment joining the points  (1,0)  and  (2,3)  in the ratio 1:n is
\left ( \frac{n(1)+1(2)}{1+n},\frac{n.(0)+1.(3)}{1+n} \right )= \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )
Let the slope of the perpendicular line is m
And Slope of  line segment joining the points  (1,0)  and  (2,3) is
m'= \frac{3-0}{2-1}= 3
Now, slope of perpendicular line is
m = -\frac{1}{m'}= -\frac{1}{3}
Now, equation of line passing through point (x_1,y_1) and with slope m is
(y-y_1)= m(x-x_1)
equation of line passing through point \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right ) and with slope  -\frac{1}{3} is
(y- \frac{3}{1+n})= -\frac{1}{3}(x- (\frac{n+2}{1+n}))\\ 3y(1+n)-9=-x(1+n)+n+2\\ x(1+n)+3y(1+n)=n+11
Therefore, equation of line  is  x(1+n)+3y(1+n)=n+11

Posted by

Gautam harsolia

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