Q : 14     Find the coordinates of the foot of perpendicular from the point  (-1,3)  to the line 3x-4y-16=0.

Answers (1)
G Gautam harsolia

Let suppose the foot of perpendicular is (x_1,y_1)
We can say that line passing through the point (x_1,y_1) \ and \ (-1,3)  is perpendicular to the line 3x-4y-16=0
Now,
The slope of the line 3x-4y-16=0 is , m' = \frac{3}{4}
And
The slope of the line  passing through the point (x_1,y_1) \ and \ (-1,3)is, m = \frac{y-3}{x+1}
lines are perpendicular
Therefore,
m = -\frac{1}{m'}\\ \frac{y_1-3}{_1+1} = -\frac{4}{3}\\ 3(y_1-3)=-4(x_1+1)\\ 4x_1+3y_1=5 \ \ \ \ \ \ \ \ \ -(i)
Now, the point (x_1,y_1) also lies on the line 3x-4y-16=0
Therefore,
3x_1-4y_1=16 \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii)
we will get
x_1 = \frac{68}{25} \ and \ y_1 =-\frac{49}{25}
Therefore, (x_1,y_1) = \left ( \frac{68}{25},-\frac{49}{25} \right )

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