Question

Find all of the points on the line y = 2x + 5
which are 4 units from the point (−1, 7).

Answer 1

if the point is \((a,b)\) the you know it looks like \((a, 2a+5)\) and so the square of the distance between that point and \((-1,7)\) will be
\[(a+1)^2+(2a+5-7)^2=16\] or
\[(a+1)^2+(2a-2)^2=16\] solve the quadratic equation for \(a\)

Answer 2

You know the formula for the distance between two points.

\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Use your known point as one of the points.

\[\sqrt{(x+1)^2 + (2x+5-7)^2} = 4\]\[(x+1)^2 + (2x-2)^2 = 16\]\[x^2+2x+2 + 4x^2 - 8x + 4 = 16\]\[5x^2 -6x -10 = 0\]Solving that for \(x\) will give you the x values of the points, and the y-values you'll find via \(y = 2x+5\)

Answer 3

\[(a+1)^2+(2a-2)^2=16\]
\[a^2+2a+1+4a^2-8a+4=16\]
\[5a^2-6a-11=0\] solve for \(a\)

Answer 4

Oops, yes, how'd I do that? Thanks @satellite73 !

Answer 5

Thank you guys :)