Question

The vertices of a triangle are a(4,10) b(9,-4)and c (-1,-2)find the eq of

1 side bc
2 the median bisecting bc​

Answer 1

Answer:

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Answer 2

(a) The equation of line bc is y=-0.2(x)-2.2.

(b) The equation of median is x=4.

Step-by-step explanation:

If a line passes through two points $(x_1,y_1)$ and $(x_2,y_2)$, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The vertices of a triangle are a(4,10) b(9,-4)and c (-1,-2).

(a)

We need to find the equation of side bc.

The line bc passes through the points b(9,-4)and c (-1,-2). So, the equation of line is

$y-(-4)=\frac{-2-(-4)}{-1-9}(x-9)$

$y+4=\frac{2}{-10}(x-9)$

$y+4=-0.2(x-9)$

$y+4=-0.2(x)-0.2(-9)$

$y+4=-0.2(x)+1.8$

Subtract 4 from both sides.

$y=-0.2(x)+1.8-4$

$y=-0.2(x)-2.2$

The equation of line bc is y=-0.2(x)-2.2.

(b)

We need to find the equation of median bisecting bc​.

The midpoint of bc is

$Midpoint=m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})= (\frac{9-1}{2},\frac{-4-2}{2})=(4,-3)$

Median bisecting bc passes through the point m(4,-3) and a(4,10). So, the equation of median is

$y-(-3)=\frac{10-(-3)}{4-4}(x-4)$

$y+3=\frac{13}{0}(x-4)$

$0(y+3)=13(x-4)$

$0=(x-4)$

$x=4$

Therefore, the equation of median is x=4.

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