Question

The orthocenter of ∆ABC where vertices are A(-1, 7) B(-7, 1) C(5, -5) is ​

Answer 1

Answer:

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Answer 2

ANSWER:

• The orthocentre of the triangle H = (- 3, 3)

GIVEN:

• A = (- 1, 7), B = (-7, 1), C = (5, - 5) .

TO FIND:

• The orthocentre of the triangle.

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{Slope = \dfrac{y_2 - y_1}{x_2-x_1}}}}}$

$\sf A = (- 1, 7), \ B = (-7, 1)$

$\sf(x_1, \ y_1) = (- 1, 7)$

$\sf(x_2, \ y_2) = (- 7, 1)$

$\sf Slope\ of\ AB = \dfrac{1 - 7}{-7+1}$

$\sf Slope\ of\ AB = \dfrac{ - 6}{-6}$

$\sf Slope\ of\ AB = 1$

$\sf Slope\ of\ AB\ \perp \ Slope\ of\ CF$

$\sf (Slope\ of\ AB) \times (Slope\ of\ CF) = - 1$

$\sf 1 \times (Slope\ of\ CF) = - 1$

$\sf Slope\ of\ CF= - 1$

$\boxed{ \bold{ \large{ \gray{y - y_1 = m(x - x_1)}}}}$

$\sf C = (x_1,\ y_1) =(5, -5)$

$\sf m = Slope\ of\ CF = - 1$

$\sf y + 5 = - 1(x - 5)$

$\sf - y - 5 = x - 5$

$\sf x + y - 5 + 5 = 0$

$\sf x + y = 0$

$\sf x = - y$

$\boxed{ \bold{ \large{ \gray{Slope = \dfrac{y_2 - y_1}{x_2-x_1}}}}}$

$\sf B = (-7, 1), \ C = (5, - 5)$

$\sf(x_1, \ y_1) = (- 7, 1)$

$\sf(x_2, \ y_2) = (5, - 5)$

$\sf Slope\ of\ BC = \dfrac{ - 5 - 1}{5 + 7}$

$\sf Slope\ of\ BC = \dfrac{ - 6}{12}$

$\sf Slope\ of\ BC = \dfrac{ - 1}{2}$

$\sf Slope\ of\ BC\ \perp \ Slope\ of\ AD$

$\sf (Slope\ of\ BC) \times(Slope\ of\ AD) = - 1$

$\sf \dfrac{ - 1}{2} \times(Slope\ of\ AD) = - 1$

$\sf Slope\ of\ AD= 2$

$\boxed{ \bold{ \large{ \gray{y - y_1 = m(x - x_1)}}}}$

$\sf C = (x_1,\ y_1) =( - 1, 7)$

$\sf m = 2$

$\sf y -7 = 2(x + 1)$

$\sf y -7 = 2x +2$

$\sf 2x- y +9 = 0$

$\sf We\ know\ that \ x = - y$

$\sf 2( - y) - y +9 = 0$

$\sf - 2y - y +9 = 0$

$\sf - 3y = - 9$

$\sf y = 3$

$\sf We\ know\ that \ x = - y$

$\sf x = - 3$

HENCE THE ORTHOCENTRE H = (- 3, 3).