Coordinate Geometry

Question

9. Find x so that the points (x,x+1),(x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

10. The points (a,10) and (–2,6) are 5 units apart. Find the value(s) of a



Anyone help with these questions because I don't understand. 

Note: There are no graphs for these questions

Answer 1

Davado,

The key to both of these problems is the Pythagorean theorem. a² + b² = c²

 

9)  First we evaluate the distance between the given coordinates in their general form:

 

     √{[x₂ - x₁]² + [y₂ - y₁]²} for each pairing

 

√{[(x + 2) - x]² + [(x + 3) - (x + 1)]²}

√{[2]² + [2]²}

√{8}   For this one, the distance between them is always the same, regardless of x.

 

√{[(x + 3) - x]² + [(2x + 4) - (x +1)]²}

√{[3]² + [x + 3]²}

√{9 + x² + 6x + 9}

√x² + 6x + 18}

 

√{[(x + 3) - (x + 2)]² + [(2x + 4) - (x + 3)]²}

√{[1]² + [x + 1]²}

√{1 + x² + 2x + 1}

√{x² + 2x + 2}

 

Just from looking at the three distances, it appears that the middle one would be the longest. So we set up Pythagorean theorem again with the middle one as c. Remember that a² means the sq.rt. is cancelled out.

 

x² + 2x + 2 + 8 = x² + 6x + 18

                   -8 = 4x

                    -2 = x

 

10) √[(-2 - a)² + (6 - 10)²] = 5

       [4 + 4a + a² + 16] = 25

                   a² + 4a - 5 = 0

                (a - 1)(a + 5) = 0

                                 a = 1 or -5