Question

The point a(0,5) B(4,2) and C(0,2) form the vertices of a right triangle. What is the equation of the line that form the hypotenuse

Answer 1

The equation of the hypotenuse line is

4y + 3x = 20

The equation of a line is represented by y = mx + b where m is the slope and b is the y intercept.

The hypotenuse of a right triangle is the longest side of the triangle.

From the coordinate given , we can see that the longest line will pass through point A and B

A ( 0,5) , B(4,2)

slope of the line = 5-2/0-4

= -3/4

The equation of a line is obtained by the equation;

y - y₁ = m(x - x₁ )

y - 2 = -3/4 ( x - 4)

y = -3/4x + 5

multiply through by 4

4y = -3x + 20

4y + 3x = 20

Therefore the equation of the hypotenuse is 4y + 3x = 20

Answer 2

The required equation of the hypotenuse is 3x + 4y - 20 = 0

Solution:

Given, the points A(0,5), B(4,2), and C(0,2) form the vertices of a right triangle in the coordinate plane.

We are to find the equation of the line that forms the hypotenuse.

Assume the points in the co –ordinate plane, A and C are points on y – axis and B, C are points on a line perpendicular to y – axis. Which means AC and BC are legs of right angle triangle.

Thus, we note that the side AB is the hypotenuse of the triangle ABC.

Slope of a line passing through the points (a, b) and (c, d) is given by:

`m=(d-b)/(c-a)`

`\text { So, the slope of } \mathrm{AB} \text { is } \mathrm{m}=(2-5)/(4-0)=(-3)/(4)`

Since the line AB passes through the point A(0, 5), so its equation in point slope form will be

`\begin{array}{l}{y-y_(1)=m\left(x-x_(1)\right)} \\\\ {y-5=(-3)/(4)(x-0)} \\\\ {4(y-5)=-3 x}\end{array}`

On rearranging terms to get standard form of equation,

3x + 4y -20 = 0

Hence, the required equation of the hypotenuse is found