Question

# If 7x – y + 3 = 0; x + y – 3 = 0 are tow sides of an isosceles triangle and the third side passes through (1, 0) then the equation of the third side is

- x + 3y + 1 = 0
- x – 3y – 1 = 0
- x – 3y + 1 = 0
- x + 3y – 1 = 0

Hint:

### Any triangle with any two sides that have the same length is known as an isosceles triangle. An isosceles triangle has two equal-sized angles that are opposite from one other and have equal sides. Here we have given that if 7x – y + 3 = 0; x + y – 3 = 0 are two sides of an isosceles triangle and the third side passes through (1, 0) then what is the equation of the third side.

## The correct answer is: x – 3y – 1 = 0

### The equation of a line has the following typical forms:

- Form of a slope-intercept
- Normal form
- Intercept form

The first-degree line's general equation in two variables is written as: Ax + By +C = 0.

The equation of any line passing through the given point $(1,0)$ will be:

$y-0=m(x−1)$

Since it makes equal angle theta with the given lines 7x – y + 3 = 0 and x + y – 3 = 0, therefore

So from this we can say that the two possible equations of third side are:

y - 0 = -3(x − 1)

y = 3 - x

or

y - 0 = -1/3(x − 1)

3y = 1 - x

x + 3y - 1 = 0

So from this we can say that the two possible equations of third side are:

y - 0 = -3(x − 1)

y = 3 - x

or

y - 0 = -1/3(x − 1)

3y = 1 - x

x + 3y - 1 = 0

So here we used the concept of the triangles, and the equations of the lines to solve this problem. We know that the equation of a straight line in slope-intercept form is given as y = mx+c, so we used that here. So the equation of the third side is x + 3y - 1 = 0.