Question

How you could find the shortest distance from A(6, 5) to the line y = 5x – 10?

Answer 1

The distance between two points (a, b) and (c, d) is given by:

`d = √((a - c)^2 + (b - d)^2)`

So the distance between the point (6, 5) and the line y = 5x - 10 can be thought as the distance between the point (6, 5) and the point (x, 5x - 10)

Where:

(x, 5x - 10) denotes all the points in the line y = 5x – 10

That distance is given by:

`d = √((x - 6)^2 + (5x - 10 - 5)^2) = √((x - 6)^2 + (5x - 15)^2)`

Now we want to minimize this.

Because the distance is a positive quantity, we can try to minimize d^2 insted, so we have:

`d^2 = (√((x - 6)^2 + (5x - 15)^2))^2 = (x - 6)^2 + (5x - 15)^2}\\\\d^2 = x^2 - 2*x*6 + 36 + 25*x^2 - 2*15*x + (-15)^2\\\\d^2 = 26*x^2 - 42*x + 261`

Notice that this is a quadratic equation with a positive leading coefficient, which means that the arms of the graph will open upwards, then the minimum will be at the vertex of the parabola.

Remember that for a parabola:

y = a*x^2 + bx + c

the x-value of the vertex is:

x = -b/2a

Then for our parabola:

d^2 = 26*x^2 - 42*x + 261

The vertex is at:

x = -(-42)/(2*26) = 0.808

Then we just need to evaluate the distance equation in that value of x to get the shortest distance:

`d = √((0.808 - 6)^2 + (5*0.808 - 15)^2) = 12.129`

The shortest distance between the point A and the line is 12.129 units.