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4. find the coordinates of the orthocenter of

abc

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Guided practice

4. find the coordinates of the orthocenter of

abc

(5,1)

Step-by-step explanation:

The orthocenter will be the intersection of the altitudes.

The altitudes are found by drawing from a vertex of the triangle to the opposite side such that the line segment you draw is perpendicular the line segment it intersects with.

The most obvious equation is the one from from vertex B to side AY which y=1 (the purple one).

So we already know the y-coordinate has to be 1.

If we are able to find one of the other equations we can find the corresponding x of the intersections there.

Let's look at the green one; the line drawn from vertex Y to side AB.

So let's find the equation for this line.

We know it has point (3,-2) and that is is perpendicular to the line AB.

So we know that perpendicular lines have opposite reciprocal slopes so we will need to start with finding the slope of AB to obtain the slope of our altitude formed by vertex Y and side AB.

So A is at (3,5) and B is (9,1)

To find the slope, I'm going to line up the points vertically and subtract, then put 2nd difference over 1st difference.

(9 , 1)

-(3 , 5)

-----------

6 -4

The slope of AB is -4/6 or -2/3 after reducing (divided top and bottom by 2 to obtain this reduction).

The opposite reciprocal of -2/3 is 3/2.

So the slope of the equation that is in green is 3/2 and it goes through point (3,-2).

I'm going to use point-slope form to find the line.

[tex]y-y_1=m(x-x_1)[/tex] is called point-slope form because [tex](x_1,y_1)[/tex] is a point on the line and [tex]m[/tex] is the slope.

Let's plug in the information we have:

[tex]y-(-2)=\frac{3}{2}(x-3)[/tex]

[tex]y+2=\frac{3}{2}(x-3)[/tex]

The system of equations we want to solve is:

[tex]y+2=\frac{3}{2}(x-3)[/tex]

[tex]y=1[/tex]

I'm going to substitute 1 for y in the first equation since y=1:

[tex]y+2=\frac{3}{2}(x-3)[/tex]

[tex]1+2=\frac{3}{2}(x-3)[/tex]

[tex]3=\frac{3}{2}(x-3)[/tex]

If you don't like the fraction multiply both sides by 2 to obtain:

[tex]6=3(x-3)[/tex]

Divide both sides by 3:

[tex]\frac{6}{3}=\frac{3(x-3)}{3}[/tex]

Simplify:

[tex]2=x-3[/tex]

Add 3 on both sides:

[tex]x=5[/tex]

So the orthocenter is at (x,y)=(5,1)

I drew the drawing below just to show you what this looks like.

I actually drew this drawing before doing the math. I tried to draw the line segments from a vertex to an opposite side so that the angle formed from my line segment I drew to the opposite side was perpendicular (meaning forming 90 degree angles).

[tex]Find the coordinates of the orthocenter of δyab that has vertices at y(3, –2), a(3, 5), and b(9, 1).[/tex]

First you should plot the points and connect them so you can see the triangle.

Then you need to find the slope of each line segment. Then you need to get the slope that would be perpendicular to each individual slope.

Then you got to the OPPOSITE vertex of each line segment, and use the perpendicular slope from that vertex to draw a line, then you will see that all 3 of them intersect at (5,-1)

Firstly, we need to draw graph

we can see that

this is right angled triangle

and we know that

Orthocenter is the point of intersection of the altitudes

and Each leg in a right triangle forms an altitude

So, in a right-angled triangle, the orthocenter lies at the vertex containing the right angle

Hence, point B will be orthocenter

so,

Orthocenter is (4,0)........Answer

[tex]Find the coordinates of the orthocenter of triangle abc with vertices a(0,0), b(4,0), and c(4,2) a.[/tex]

Orthocenter of a triangle with one vertex at the origin D(0,0), E(0,7), F(6,3)

The orthocenter is the meet of the altitudes.

DE is the line x=0. That's the y axis. The altitude is the perpendicular through F(6,3), so is the horizontal line y=3.

DF is the line

6y=3x

y=(1/2)x

Perpendicular through E(0,7) is

y - 7 = (-2)(x - 0)

y = -2x + 7

That's the other altitude. Intersecting with y=3

3 = -2x + 7

-4 = -2x

x = 2

y = 3

orthocenter is (2,3)

See the attached

Step-by-step explanation:

When in doubt, draw a diagram.

The orthocenter of this acute triangle will be within its bounds. That should tell you right away that the y-coordinate of it will not be 8, but must be between 2 and 6.

The line perpendicular to BC through A must have a y-intercept greater than the y-coordinate of A, so cannot be 5. Whatever it is, the y-coordinate of the orthocenter will be less, so again, your answer fails the reasonableness test.

The perpendicular line to BC through A is ...

... y = (-1/2)(x -2) +6 = -x/2 +7 . . . . . . looks like you had a sign error in (-1/2)(-2)

The intersection of that line and x=6 is ...

... y = -6/2 +7 = 4

[tex]Ineed on finding the orthocenter of a triangle, i got the x right, x=6 but i got the y value wrong,[/tex]

By elimination, we can remove B and D because they lie outside of the triangle.

We know that the orthocenter is where the altitudes of a triangle intersect.

The side AB is horizontal, so you know that the altitude of AB is vertical. The altitude of AB also connects to vertex C.

Since the orthocenter must lie on the altitude of AB, and the altitude is vertical, the orthocenter must lie directly above vertex C. This means that the x-coordinate of the orthocenter is the same as the x-coordinate of vertex C.

The only choices where the x-coordinate is the same as the one for vertex C is C and D. But we eliminated choice D, meaning the answer is choice C.

( -1,-5 )

Step-by-step explanation:

We have the co-ordinates A( -3,3 ), B( -1,7 ) and C( 3,3 ).

We will find the orthocenter using below steps:

1. First, we find the equations of AB and BC.

The general form of a line is y=mx+b where m is the slope and b is the y-intercept.

Using the formula of slope given by [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1} }[/tex], we will find the slope of AB and BC.

Now, slope of AB is [tex]m=\frac{7-3}{-1+3}[/tex] i.e. [tex]m=\frac{4}{2}[/tex] i.e. [tex]m=2[/tex].

Putting this 'm' in the general form and using the point B( -1,1 ), we get the y-intercept as,

y = mx + b i.e. 1 = 2 × (-1) + b i.e. b = 3.

So, the equation of AB is y = 2x + 3.

Also, slope of BC is [tex]m=\frac{3-7}{3+1}[/tex] i.e. [tex]m=\frac{-4}{4}[/tex] i.e. [tex]m=-1[/tex].

Putting this 'm' in the general form and using the point B( -1,1 ), we get the y-intercept as,

y = mx + b i.e. 1 = (-1) × (-1) + b i.e. b = 0.

So, the equation of BC is y = -x.

2. We will find the slope of line perpendicular to AB and BC.

When two lines are perpendicular, then the product of their slopes is -1.

So, slope of line perpendicular to AB is [tex]m \times 2 = -1[/tex] i.e. [tex]m=\frac{-1}{2}[/tex]

So, slope of line perpendicular to BC is [tex]m \times (-1) = -1[/tex] i.e. m = 1.

3. We will now find the equations of line perpendicular to AB and BC.

Using the slope of line perpendicular to AB i.e. [tex]m=\frac{-1}{2}[/tex] and the point opposite to AB i.e. C( 3,3 ), we get,

y = mx+b i.e. [tex]3=\frac{-1}{2} \times 3 + b[/tex] i.e. [tex]b=\frac{9}{2}[/tex]

So, the equation of line perpendicular to AB is [tex]y=\frac{-x}{2} +\frac{9}{2}[/tex]

Again, using the slope of line perpendicular to BC i.e. m = 1 and the point opposite to BC i.e. A( -3,3 ), we get,

y = mx + b i.e. 3 = 1 × -3 + b i.e. b = 6.

So, the equation of line perpendicular to BC is y = x+6

4. Finally, we will solve the obtained equations to find the value of ( x,y ).

As, we have y = x+6 and [tex]y=\frac{-x}{2}+\frac{9}{2}[/tex]

This gives, [tex]y=\frac{-x}{2}+\frac{9}{2}[/tex] → [tex]x+6=\frac{-x}{2} +\frac{9}{2}[/tex] → 2x+12 = -x+9 → 3x = -3 → x = -1.

So, y = x+6 → y = -1+6 → y=5.

Hence, the orthocenter of the ΔABC is ( -1,5 ).

[tex](0,0.75) \:or\:(0,\frac{3}{4})[/tex]

Step-by-step explanation:

Hi there!

1) Firstly, connect the points to draw a triangle.

2) From each vertex either with a pair of square or with a software trace a perpendicular line to the opposite side.

3) The concurrent point, i.e. the intersection point of these three altitudes is the orthocenter. Orthocenter means the the right center.

In equilateral triangles the Orthocenter coincides with the Centroid.

4) Finally, the coordinates of the Orthocenter found is (0,0.75)

[tex]Find the coordinates of the orthocenter of abc. a(-1,0) b(0,4) c(3,0)[/tex]

So (5,1) is the orthocenter.

Step-by-step explanation:

So we have to find the slopes of all three lines in burgundy (the line segments of the triangle). We also need to find the equations for the altitudes with respect from all sides of the triangle (we are looking for perpendicular lines).

The vertical line there is just going to be x=a number so that line is x=3 because all the points on that line are of the form (3,y). x=3 says we don't care what y is but x will always be 3. So the line for AY is x=3.

So the altitude of the triangle with respect to that side (that line segment) would be a line that is perpendicular to is which would be a horizontal line y=1. I got y=1 because it goes through vertex B(9,1) and y=1 is perpendicular to x=3.

So we now need to find the equations of the other 2 lines.

One line has points A(3,5) and B(9,1).

To find the slope, you may use [tex]\frac{y_2-y_1}{x_2-x_1}[/tex].

Or you could just line up the points vertically and subtract then put 2nd difference over 1st difference.

Like this:

(9 , 1)

-(3 ,5)

----------

6 -4

So the slope is -4/6=-2/3.

So a line that is perpendicular will have opposite reciprocal slope. That means we are looking for a line with 3/2 as the slope. We want this line from segment AB going to opposite point Y so this line contains point (3,-2).

Point slope-form is

[tex]y-y_1=m(x-x_1)[/tex] where [tex](x_1,y_1)[/tex] is a point on the line and [tex]m[/tex] is the slope.

So the line is:

[tex]y-(-2)=\frac{3}{2}(x-3)[/tex]

[tex]y+2=\fac{3}{2}x-\frac{9}{2}[/tex]

Subtract 2 on both sides:

[tex]y=\frac{3}{2}x-\frac{9}{2}-2[/tex]

Simplify:

[tex]y=\frac{3}{2}x-\frac{13}{2}[/tex].

Let's find the the third line but two lines is plenty, really. The othorcenter is where that perpendicular lines will intersect.

Now time for the third line.

BY has points (9,1) and (3,-2).

The slope can be found by lining up the points vertically and subtracting, then put 2nd difference over 1st difference:

(9 ,1)

-(3,-2)

---------

6 3

So the slope is 3/6=1/2.

A perpendicular line will have opposite reciprocal slope. So the perpendicular line will have a slope of -2.

We want this line segment to go through A(3,5).

We are going to use point-slope form:

[tex]y-5=-2(x-3)[/tex]

Add 5 on both sides:

[tex]y=-2(x-3)+5[/tex]

Distribute:

[tex]y=-2x+6+5[/tex]

Combine like terms:

[tex]y=-2x+11[/tex]

So the equation of the 3rd altitude line is y=-2x+11.

So the equations we want to find the intersection to is:

y=(3/2)x-(13/2)

y=1

y=-2x+11

I like the bottom two equations so I'm going to start there and then use my third line to check some of my work.

y=1

y=-2x+11

Replacing 2nd y with 1 since y=1:

1=-2x+11

Subtract 11 on both sides:

1-11=-2x

Simplify:

-10=-2x

Divide both sides by -2:

5=x

The point of intersection between y=1 and y=-2x+11 is (5,1).

Let's see if (5,1) is on that third line.

y=(3/2)x-(13/2)

1=(3/2)(5)-(13/2)

1=(15/2)-(13/2)

1=(2/2)

1=1

So (5,1) is the intersection of all three lines.

So (5,1) is the orthocenter.

[tex]Find the coordinates of the orthocenter of δyab that has vertices at y(3, –2), a(3, 5), and b(9, 1).[/tex]

For convenience, I'll name the three points: A(1, 3), B(1,12), C(8,10)

The orthocenter is where the three altitudes of the triangle meet. That is, where the perpendicular lines from each vertex to the opposite side meet.

the goal in this case is to find out where the perpendicular line from C to AB, the perpendicular line from B to AC, and the perpendicular line form A to BC meet.

It is enough to find two perpendicular lines.

step one: find the equation of two perpendicular lines

find the slope of AB: m=(12-3)/(1-1), notice the denominator is 0, so the slope doesn't exist, that means AB is a vertical line parallel to the y axis, so the perpendicular line has a slope of o, a horizontal line. this horizontal line goes through C(8,10), so the equation is y=10

use the same method, slope of AC: (10-3)/(8-1)=1, so the perpendicular line from B to AC has a slope of -1. y=-x+b

Now use the point B(1,12) to find b: 12=-1+b =>b=13

so the equation for this line is y=-x+13

Step two: solve the system of equations: y=10

y=-x+13

we get x=3, y=10, which are the coordinates of the orthocenter.

I've also found the equation of the third line, which is y=(7/2)x-1/2, the point (3,10) is also on this line. So it works!