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

Monday, July 28, 2014

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

For a right-angled triangle, the product of slopes of the two sides is -1.

The slope of a line passing through two point: (x1, y1) and (x2, y2) is given as:

slope = (y2 - y1)/(x2 - x1)

Hence, in this case:

$\displaystyle\frac{{{\left({x}+{3}\right)}-{\left({x}+{1}\right)}}}{{{\left({x}+{2}\right)}-{\left({x}\right)}}}\times\frac{{{\left({2}{x}+{4}\right)}-{\left({x}+{3}\right)}}}{{{\left({x}+{3}\right)}-{\left({x}+{2}\right)}}}$

$\displaystyle={\left(\frac{{2}}{{2}}\right)}\times\frac{{{x}+{1}}}{{1}}={x}+{1}$

For it to be a right angled triangle, x +1 = -1 or, x = -2.

We can check to make sure that the answer is right, by using Pythagoras Theorem, according to which the sum of the squares of two sides is equal to the square of the third side.

substituting x = -2 in the given points, we obtain coordinates as (-2, -1), (0, 1) and (1, 0)

Using the Pythagoras theorem:

$\displaystyle{\left({{\left({1}-{\left(-{1}\right)}\right)}}^{{2}}+{{\left({0}-{\left(-{2}\right)}\right)}}^{{2}}\right)}+{\left({{\left({0}-{1}\right)}}^{{2}}+{{\left({1}-{0}\right)}}^{{2}}\right)}={\left({{\left({0}-{\left(-{1}\right)}\right)}}^{{2}}+{{\left({1}-{\left(-{2}\right)}\right)}}^{{2}}\right)}$