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Q : 15      The perpendicular from the origin to the line  y=mx+c  meets it at the point  (-1,2). Find the values of m and c

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We can say that line passing through point (0,0) \ and \ (-1,2)  is perpendicular to line y=mx+c
Now,
The slope of the line  passing through the point (0,0) \ and \ (-1,2) is , m = \frac{2-0}{-1-0}= -2
lines are perpendicular
Therefore,
m = -\frac{1}{m'} = \frac{1}{2}               - (i)
Now, the point (-1,2) also lies on the line y=mx+c
Therefore,
2=\frac{1}{2}.(-1)+C\\ C = \frac{5}{2} \ \ \ \ \ \ \ \ \ \ \ -(ii)
Therefore, the value of m and C is \frac{1}{2} \ and \ \frac{5}{2}   respectively

Posted by

Gautam harsolia

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