1. find a point

Find the point on the line 4x+4y+3=0 which is closet to P (1, 2)?

So I re wrote my line as y=-x-(3/4)
Then i take d^2=(x-1)^2+(-x-(3/4)-2)^2=2x^2+(7x/2)+(137/16)
Once I take the dirivative i get 4x+(7/2) soove for x i get -(7/8) and then y = 1/8
But its wrong. What did i do wrong?

2. Re: find a point

M670

The slope intercept form of the equation of the given line L is : y = -1x-(3/4)
the closest point on the line L from the point P(1,2) is the point that the perpendicular line to the line L , passing from the point P, intecects the line L.
the equation of the line that passes from the point P and is perpendicular to L is y = x+1

to find the coordinates of the point of intercection of these two lines solve the system of the two simultaneous equations
y = -1x-3/4
y=x+1

the answer is x=-7/8 and y= 1/8 therefore the point is Q(-7/8 , 1/8)

Pls revise the equation of the line in the plane .

3. Re: find a point

Originally Posted by MINOANMAN
M670

The slope intercept form of the equation of the given line L is : y = -1x-(3/4)
the closest point on the line L from the point P(1,2) is the point that the perpendicular line to the line L , passing from the point P, intecects the line L.
the equation of the line that passes from the point P and is perpendicular to L is y = x+1

to find the coordinates of the point of intercection of these two lines solve the system of the two simultaneous equations
y = -1x-3/4
y=x+1

the answer is x=-7/8 and y= 1/8 therefore the point is Q(-7/8 , 1/8)

Pls revise the equation of the line in the plane .

Point Q = (-7/8,1/8) is incorrect when I submitted that...

4. Re: find a point

Originally Posted by M670
Find the point on the line 4x+4y+3=0 which is closet to P (1, 2)?
So I re wrote my line as y=-x-(3/4)
Then i take d^2=(x-1)^2+(-x-(3/4)-2)^2=2x^2+(7x/2)+(137/16)
Once I take the dirivative i get 4x+(7/2) soove for x i get -(7/8) and then y = 1/8
But its wrong. What did i do wrong?

No calculus needed. Where does the line $\displaystyle 4x-4y+4=0$ intersect the given line?

5. Re: find a point

The question is asking me to find the point closest to (1,2) on the line$\displaystyle y=-x-\frac{3}{4}$
Then I get $\displaystyle d^2=(x-1)^2+(-x-\frac{3}{4}-2)^2$ which is = $\displaystyle 2x^2+\frac{7x}{2}+\frac{137}{16}$
Then once I get $\displaystyle d^2'=4x+\frac{7}{2}$ I set it equal to 0 and solve for x
Thus getting $\displaystyle x=-\frac{7}{8} and y = \frac{1}{8}$

Shoot I think I found my mistake my original formula is $\displaystyle -4x+4y+3=0$ which would make it $\displaystyle 4y=4x-3$then I should have $\displaystyle y=x-\frac{3}{4}$

6. Re: find a point

All this time it was the Minus sign that messed me up........

7. Re: find a point

OK MAN
Next time give us correct information....