Triangle ABC has vertices A(3,1), B(5,7) and C(1, y). Find all y so that angle C is a right angle?

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Answer 1 · 2016-09-06T19:57:46

The two possible values of $y$ $y$ are $3$ $3$ and $5$ $5$ .

For this problem, we need to consider AC be perpendicular to BC.

Since the lines are perpendicular, by the slope formula we have:

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = - \frac{x_{2} - x_{1}}{y_{2} - y_{1}}$ $(y_2 - y_1)/(x_2 - x_1) = -(x_2 - x_1)/(y_2 - y_1)$

$\frac{y - 7}{1 - 5} = - \frac{1 - 3}{y - 1}$ $(y - 7)/(1 - 5) = -(1 - 3)/(y - 1)$

$(y - 7) (y - 1) = 2 (- 4)$ $(y - 7)(y - 1) = 2(-4)$

$y^{2} - 7 y - y + 7 = - 8$ $y^2 - 7y - y + 7 = -8$

$y^{2} - 8 y + 15 = 0$ $y^2 - 8y + 15 = 0$

$(y - 3) (y - 5) = 0$ $(y - 3)(y - 5) = 0$

$y = 3 and 5$ $y = 3 and 5$

Hopefully this helps!