Find the point on the line 3 x + 3 y + 4 =0 which is closest to the point ( -3, 5 ).

Question

anonymous · Answer

We have to find the orthogonal line, how can we find this line?

$$3x+3y+4=0$$

$$y=-x-\frac{4}{3}~~~~~~(Line~1)$$

$$y-y_o=m*(x-x_o)$$

$$m=1$$

$$y-5=x-(-3)$$

$$y=x+8~~~~~~(Line~2)$$

now we have to find where this two lines will be the same value

$$-x-\frac{4}{3}=x+8$$

$$-\frac{4}{3}-8=2x$$

$$\boxed{\boxed{x=-\frac{14}{3}}}$$

now we have to find the Y, to do this verification we can use both equation, I'll choose the first one

$$y=-x-\frac{4}{3}$$

$$y=-\left(-\frac{14}{3}\right)-\frac{4}{3}$$

$$y=\frac{14}{3}-\frac{4}{3}$$

$$\boxed{\boxed{y=\frac{10}{3}}}$$

$$\therefore~P=\left(-\frac{14}{3},\frac{10}{3}\right)$$