The line closest to (-1, -7) ? (possible calculus prob?)?

Question

An equation of 5x+6y-3=0 is given and it wants me to find an x,y coordinate that is closest to -1, -7

I changed it over to slope form y = 1/2 - 5/6x
would I need to find the perpendicular and go from there?
and use y - y1 = m (x - x1) ?
and then re-input it back to the original formla?

Or is there a better way to do this?

Could anyone explain Madhukar's answer?
I see he received the perpendicular line by using (-1, -7) and inputting it into the perpendicular equation to get k.
now how does he go from k to the answers in fraction form?

Answer 1

You can get the point by calculating the slope.
Two points: (x, 1/2 - 5/6x) and (-1, -7).
( 1/2 - 5/6x + 7)/(x + 1) = 6/5
Multiply both sides by 30(x + 1),
15 - 25x + 210 = 36x + 36
61x = 189
x = 189/61
y = -127/61

Answer 2

Find the point on the line 5x + 6y - 3 = 0 that is closest to the point P(-1, -7).

The point in question will be on the line perpendicular to the given line that goes thru the point P.

To write the equation of the perpendicular line swap the x and y coefficients and change the sign of one of them (it doesn't matter which one).

Given line
5x + 6y - 3 = 0

Perpendicular line
6x - 5y - c = 0

Solve for the constant c by plugging in the point P(-1, -7).

6(-1) - 5(-7) + c = 0
-6 + 35 + c = 0
-29 + c = 0
c = 29

The equation of the perpendicular line is:
6x - 5y - 29 = 0

Now you have two equations and two unknowns.

5x + 6y - 3 = 0
6x - 5y - 29 = 0

Add five times the first equation to six times the second.

61x - 189 = 0
61x = 189
x = 189/61

Plug back into the first equation and solve for y.

5x + 6y - 3 = 0
5(189/61) + 6y - 3 = 0
945/61 + 6y = 3
6y = 3 - 945/61 = -762/61
y = -762/366 = -127/61

The closest point on the given line to P(-1, -7) is:

(x, y) = (189/61, -127/61)

Answer 3

The ah-ha 2d of this subject is composed of the 1st spinoff of the area formulation. The ah-ha 2d of the examine to the answer is composed of the slope of the perpendicular line. the area formulation is sqrt[(x sub a million - x sub 2)^2 + (y sub a million - y sub 2)] subsequently, we've the two ordered pairs (x, 3x + 9) and (0, 0) the hollow g(x) = sqrt[(x sub a million - x sub 2)^2 + (y sub a million - y sub 2)^2] = sqrt [(x - 0)^2 + (3x + 9 - 0)^2] to shrink, resolve for g ` (x) = 0 utilising the chain rule, we get g `(x) = a million/2 [2x + 2(3)(3x + 9)][(x - 0)^2 + (3x + 9 - 0)^2]^-0.5 yet g `(x) = 0 whilst the numerator is 0. So, resolve for a million/2 [2x + 2(3)(3x + 9)] = 0 a million/2 [2x + 2(3)(3x + 9)] = 0 x + 9x + 27 = 0 x = -2.7 Then plug the fee for x into the unique equation: y = 3(-2.7) + 9 y = 0.9 the element on the line y = 3x + 9 it is closest to the foundation is (-2.7, 0.9). to be taught the respond devoid of utilising calculus, use the perpendicular line defined via y = -a million/m x (we can try this with the aid of fact y = 3x + 9 is a 'uncomplicated curve' and the line perpendicular to y that incorporates the foundation might additionally diploma our required distance.) This area of intersection is at (3x + 9 = -x/3, 3x + 9) 3x + 9 = -x/3 9x + 27 = -x 10x = -27 x = -2.7 back, plug the fee of x into the unique equation & resolve for y: y = 3(-2.7) + 9 the element on the line y = 3x + 9 it is closest to the foundation is (-2.7, 0.9). Our recommendations utilising the two routes tournament. for this reason, our answer is genuine.

Answer 4

Eqn. of line perpendicular to 5x + 6y - 3 = 0 is of the form
6x - 5y + k = 0.
If it passes through the point (-1, -7), 6(-1) - 5(-7) + k = 0.
=> k = -29
Hence, the eqn. of perpendicular line through (-1, -7) is
6x - 5y - 29 = 0.
Solving this eqn. with the eqn. of the given line gives the closest point as
(189/61, -127/61)