Question

Find the coordinates of the orthocenter of ΔYAB that has vertices at Y(3, –2), A(3, 5), and B(9, 1).

Answer 1

Answer:

(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.

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

Let's plug in the information we have:

$y-(-2)=\frac{3}{2}(x-3)$

$y+2=\frac{3}{2}(x-3)$

The system of equations we want to solve is:

$y+2=\frac{3}{2}(x-3)$

$y=1$

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

$y+2=\frac{3}{2}(x-3)$

$1+2=\frac{3}{2}(x-3)$

$3=\frac{3}{2}(x-3)$

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

$6=3(x-3)$

Divide both sides by 3:

$\frac{6}{3}=\frac{3(x-3)}{3}$

Simplify:

$2=x-3$

Add 3 on both sides:

$x=5$

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).