Question

Find the co-ordinates of the orthocentre of the triangle, the equations of whose sides are x + y = 1,2x + 3y = 6, 4x − y + 4 = 0, without finding the co−ordinates of its vertices.

Answer 1

Answer on Question #56865 – Math – Analytic Geometry

Find the co-ordinates of the orthocentre of the triangle, the equations of whose sides are $x + y = 1,2x + 3y = 6$, $4x - y + 4 = 0$, without finding the co-ordinates of its vertices.

Solution

Rewrite equations $x + y = 1,2x + 3y = 6$, $4x - y + 4 = 0$ as $y = -x + 1$, $y = -\frac{2}{3}x + 2$, $y = 4x + 4$ respectively.

Given:

$$a: y = -x + 1$$ $$b: y = -\frac{2}{3}x + 2$$ $$c: y = 4x + 4$$

Three lines $y = m_1x + b_1, y = m_2x + b_2, y = m_3x + b_3$ intersect at a single point $(x^*; y^*)$.

Given $m_1, b_1, m_2, b_2, m_3$. Find $b_3$.

To find the point of intersection of straight lines

$$y = m_1x + b_1, y = m_2x + b_2,$$

equate the right-hand sides of the previous equalities:

$$m_1x + b_1 = m_2x + b_2,$$ $$(m_1 - m_2)x = b_2 - b_1$$ $$\Rightarrow x = -\frac{b_1 - b_2}{m_1 - m_2}.$$

Substitute for $x$ into $y = m_1x + b_1$:

$$y = m_1x + b_1,$$ $$y = m_1 \cdot \left(-\frac{b_1 - b_2}{m_1 - m_2}\right) + b_1,$$ $$y = \frac{-m_1b_1 + m_1b_2 + b_1m_1 - b_1m_2}{m_1 - m_2},$$ $$y = \frac{m_1b_2 - b_1m_2}{m_1 - m_2},$$ $$y = \frac{b_1m_2 - b_2m_1}{m_2 - m_1}.$$

Thus, the point of intersection of straight lines (1) is

$$(x^*; y^*) = \left(-\frac{b_1 - b_2}{m_1 - m_2}; \frac{b_1m_2 - b_2m_1}{m_2 - m_1}\right).$$

The line $y = m_3x + b_3$ also goes through point (2), therefore $y^* = m_3x^* + b_3$, hence

$$b _ {3} = y ^ {*} - m _ {3} x ^ {*} = \frac {b _ {1} m _ {2} - b _ {2} m _ {1}}{m _ {2} - m _ {1}} - m _ {3} \cdot \left(- \frac {b _ {1} - b _ {2}}{m _ {1} - m _ {2}}\right)$$ $$b _ {3} = \frac {b _ {1} \left(m _ {2} - m _ {3}\right) - b _ {2} \left(m _ {1} - m _ {3}\right)}{m _ {2} - m _ {1}}.$$

When two straight lines are perpendicular, the product of their slopes is $(-1)$ .

If equation of side $a$ is $y = -x + 1$ , then its slope is $k_{1} = -1$ , hence the slope of perpendicular straight line is $k_{2} = \frac{-1}{k_{1}} = \frac{-1}{-1} = 1$ and equation of the height drawn to the side $a$ is

$$h _ {a}: y = x + b _ {a}$$

If equation of side $b$ is $y = -\frac{2}{3} x + 2$ , then its slope is $k_{3} = -\frac{2}{3}$ , hence the slope of perpendicular straight line is $k_{4} = \frac{-1}{k_{3}} = \frac{-1}{-2/3} = \frac{3}{2}$ and equation of the height drawn to the side $b$ is

$$h _ {b}: y = \frac {3}{2} x + b _ {b}$$

Straight lines $h_a, b, c$ intersect at a single point. Using formula (3), put

$m_{1} = -\frac{2}{3}, b_{1} = 2, m_{2} = 4, b_{2} = 4, m_{3} = 1$ and we can find $b_{a}$ :

$$b _ {a} = \frac {2 (4 - 1) - 4 (- \frac {2}{3} - 1)}{4 + \frac {2}{3}} = \frac {1 9}{7}.$$

Straight lines $h_b, a, c$ intersect at a single point. Using formula (3) put

$m_{1} = -1, b_{1} = 1, m_{2} = 4, b_{2} = 4, m_{3} = \frac{3}{2}$ and we can find $b_{b}$ :

$$b _ {b} = \frac {1 \left(4 - \frac {3}{2}\right) - 4 \left(- 1 - \frac {3}{2}\right)}{4 + 1} = 2. 5.$$

From (4), (5), (6), (7) it follows that

$$h _ {a}: y = x + \frac {1 9}{7},$$ $$h _ {b}: y = \frac {3}{2} x + 2. 5$$

Orthocentre: is the point of intersection of heights $h_a, h_b$ . Using (8) and (9) obtain the following system of equations:

$$\left\{ \begin{array}{l} y = x + \frac {1 9}{7} \\ y = \frac {3}{2} x + 2. 5 \end{array} \right. \to \frac {3}{2} x + 2. 5 = x + \frac {1 9}{7} \to \left(\frac {3}{2} - 1\right) x = \frac {1 9}{7} - 2. 5 \to$$ $$\rightarrow x = \frac {3}{7}; y = x + \frac {1 9}{7} = \frac {3}{7} + \frac {1 9}{7} = 3 \frac {1}{7}$$

Thus, $\left(\frac{3}{7}; 3\frac{1}{7}\right)$ is orthocenter.

Answer: $\left(\frac{3}{7}, 3\frac{1}{7}\right)$ .

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