Question

the vertices of triangle ABC are A(-5,7) B(-4,-5)and c(4,5) find the slope of the attitude of the triangle​

Answer 1

Answer:

• Slope of altitude from A = -0.8
• Slope of altitude from B = 4.5
• Slope of altitude from C = 0.08333...

Step-by-step explanation:

There are three altitudes: one from A onto BC, one from B onto CA, and one from C onto AB.

The altitude from A onto BC is the segment from A to BC that is perpendicular to BC.  Since slopes of perpendicular lines multiply to give -1, the slope of the altitude from A is -1/(slope of BC).

    Slope of BC = (5 - -5)/(4 - -4) = (5+5)/(4+4) = 10/8

⇒  slope of altitude from A = -8/10 = -0.8

Similarly:

    slope of CA = (7 - 5)/(-5 - 4) = 2/-9

⇒  slope of altitude from B = 9/2 = 4.5

and

    slope of AB = (-5 - 7)/(-4 - -5) = -12/1 = -12

⇒  slope of altitude from C = 1/12 = 0.08333...

Answer 2

Answer:

SOLUTION :-

Given that,

The points A ( -5 , 7 ), B ( -4 , -5 ) and C ( 4 , 5 ) are the vertices of ΔABC .

$\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]}$

Here,

$\bullet\sf \ x_1=-5 \ , \ y_1=7 \\\\ \bullet \ x_2= -4 \ , \ y_2= -5 \\\\\bullet \ x_3= 4 \ , \ y_3= 5$

Area of triangle ABC,

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -5(-5-5)+-4(5-7)+4(7-(-5) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ -5(-5-5)+-4(5-7)+4(7+5) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ (-5\times -10)+(-4\times -2 )+(4\times 12) \ ] \\\\\longrightarrow \dfrac{1}{2}\times [ \ 50+8 +48 \ ] \\\\\longrightarrow \dfrac{1}{2}\times 106 \\\\\longrightarrow\bf 53 \ sq.units$

Area of ΔABC = 53 sq.units