Geometry

Mathematics

1 answer:

juin [17]4 years ago

6 0

Answer:
y-6 = (7/6)(x-2)

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Explanation:

The median DT goes from point D, which is given, to point T which we have to find. The point T is the midpoint of segment EF.

Use the midpoint formula to find the midpoint. Add up the x coordinates of E and F to get
-1+5 = 4
Then divide that by 2
4/2 = 2
The x coordinate of point T is x = 2

Repeat for the y coordinates
8+4 = 12
12/2 = 6
The y coordinate of point T is y = 6

So the point T is (x,y) = (2,6)

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We know point D is (-4,-1)
We just found point T to be (2,6)

Using the slope formula, the slope is..
m = (y2-y1)/(x2-x1)
m = (6-(-1))/(2-(-4))
m = (6+1)/(2+4)
m = 7/6
The slope is m = 7/6

Use either the coordinates of D or T. I'll use point T. So (x1,y1) = (2,6) meaning x1 = 2 and y1 = 6

We'll plug m = 7/6, x1 = 2 and y1 = 6 into the point slope equation
y-y1 = m(x-x1)
y-6 = (7/6)(x-2)

which is the final answer

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NOT SURE THO :)

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Would be very happy if you helped.Don’t spam guys

Reptile [31]

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$\\ \sf\longmapsto sin\Theta=\dfrac{40}{41}$

Now

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$\\ \sf\longmapsto cos\Theta=\sqrt{1-\dfrac{1600}{1682}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{1681-1600}{1681}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{81}{1681}}$

$\\ \sf\longmapsto cos\Theta=\dfrac{9}{41}$

We know

$\boxed{\sf tan\Theta=\dfrac{Sin\theta}{Cos\Theta}}$

$\\ \sf\longmapsto \dfrac{tan\Theta}{1-tan^2\Theta}$

$\\ \sf\longmapsto \dfrac{\dfrac{sin\Theta}{cos\Theta}}{1-\dfrac{sin^2\Theta}{cos^2\Theta}}$

$\\ \sf\longmapsto \dfrac{\dfrac{\left(\dfrac{40}{41}\right)}{\left(\dfrac{9}{41}\right)}}{1-\dfrac{\left(\dfrac{40}{41}\right)^2}{\left(\dfrac{9}{41}\right)^2}}$