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2 years ago

Pls answer the question in the pic I'll give u brainly

Mathematics

2 answers:

yuradex [85]2 years ago

8 0

Answer:9 and 21

Step-by-step explanation:I used a calculator

atroni [7]2 years ago

4 0

It’s probably 9 and 21

•Hope this helps!•

~have a wonder day/night~

•diva•

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Step-by-step explanation:

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3 years ago

I need help with this geometry question asap!

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Answer:

(a) Theorem 9

Step-by-step explanation:

Any of the given theorems can be used to prove lines are parallel. We need to find the one that is applicable to the given geometry.

<h3>Analysis</h3>

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2 years ago

Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)

Formula used :

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = -1

Equation of AB with slope -1 and the coordinates (-1,0) is,

(y – 0) = -1(x – (-1))

y+x=-1………………(1)

Similarly, for AC

Mid point of AC = $(\frac{2+2}{2},\frac{3-3}{2})=(2,0)$

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of AC with slope 0 and the coordinates (2,0) is,

(y – 0) = 0(x – 2)

y=0 ………………(2)

By solving equation (1) and (2),

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of CF = -1

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF

(y – 0) = -1(x – (-1))

y+x=-1………………(3)

Similarly, mid point of BC = $(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB = $\frac{-3+3}{-4-2}=0$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))

y+3=0

y=-3………………(4)

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)

7 0

3 years ago