Question

the vertices of triangle OAB are O(0,0),A(3,4),B(4,3): then the length of the altitude from O is​

Answer 1

Answer:

O(0,0) ,A(3,4)B (4,3) Q . 1. Then the value OA is √(4^2+3^2) =5

2. Then the value of OB is √(3^2+4^2) =5

3. Then the value of AB is✓(1^2+1^2) =√2

OA is 5

OB is 5

AB is √2

AD^2 = (AB/2)^2 + OA^2 = (√49/√2) =7/√2

So , the length of median is 7/√2

Answer 2

Answer:

The length of altitude from O in ΔOAB is $\frac{7\sqrt{2}}{2}$

Step-by-step explanation:

We have a ΔOAB with vertices of the triangle as O(0,0), A(3,4), B(4,3). and now we have to calculate the length of altitude from O.

So altitude is the line from the vertex of the triangle which lies perpendicular to the opposite side of the vertex.

Step 1 of 3

For calculating the distance first it is necessary to know the endpoints of altitude.

Let the altitude be OC as shown in fig attached.

So we have the vertices of O. Now to get the vertices of C

Let the slope of line AB is M₁ which is given as

M₁ = $\frac{3-4}{4-3}$ = -1

And let the slope of altitude OC be M₂ which is given as

M₁M₂=-1

⇒M₂ = 1

Step 2 of 3

Then the equation of altitude OC is

y = M₂x +C

Then by putting the value of vertices of O we get

0 = 1*0 + C

so, C=0

The equation of line OC is

y = x                               ....(1)

Now the equation of line AB in the same way is

y = -x + 7            .....(2)

Step 3 of 3

Now we need intersection point of both lines AB and OC which can be calculated from eq(1) and eq(2)

x = 7/2 and y = 7/2

Then the distance between two points O and C is given as

$d = \sqrt{(x-y)^2+(a-b)^2$

Here x= 7/2, y = 7/2 and  a = 0, b = 0 then

$d=\sqrt{(\frac{7}{2}-0)^2+(\frac{7}{2}-0)^2 } =\frac{7\sqrt{2}}{2}$

Hence the length of altitude from O is $\frac{7\sqrt{2}}{2}$