Question

the orthocentre of the triangle formed by the lines x+y+1=0,x-y-1=0,3x+4y+5=0 is

Answer 1

Answer on Question #51346 – Math – Analytic Geometry

Task

the orthocentre of the triangle formed by the lines $x + y + 1 = 0, x - y - 1 = 0, 3x + 4y + 5 = 0$ is

Solution

Introduce equations of sides of the triangle

AB: $x + y + 1 = 0$ (1)

BC: $x - y - 1 = 0$ (2)

AC: $3x + 4y + 5 = 0$ (3)

(2)

Equation of AB is $y = -x - 1$, its slope is $k1 = -1$.

Equation of BC is $y = x - 1$, its slope is $k2 = 1$.

Note that $k1 * k2 = -1$, it means that AB and BC are perpendicular, hence, triangle ABC is right. In right triangle ABC, the orthocenter is the polygon vertex B of the right angle.

Solving (1) and (2):

$$\left\{ \begin{array}{l} x + y + 1 = 0 \\ x - y - 1 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} (x + y + 1) + (x - y - 1) = 0 \\ x - 1 = y \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2x = 0 \\ y = x - 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 0 \\ y = -1 \end{array} \right. \Rightarrow B(0; -1)$$

Answer: the orthocenter of the triangle is $(0; -1)$.

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