Question

The vertices of an isosceles triangle are A(3, 6), B(7, 2), and C4, 3).

What is the equation of the triangle's line of symmetry?

Answer 1

If we graph the points

then we will get AC=CB

That means C is the point from where the axis of symmetry passes

Foot of perpendicular is midpoint of AB

Midpoint of AB

• ((3+7)/2,(6+2)/2)
• (10/2,8/2)
• (5,4)

Find equation of C and foot of perpendicular

Slope

• m=(4-3)/(5-4)=1

Equation

• y-3=1(x-4)
• y-3=x-4
• y=x-1

Answer 2

`y=x-1`

Explanation:

The line of symmetry of an isosceles triangle is perpendicular to its base.

Given vertices of the isosceles triangle:

• A = (3, 6)
• B = (7, 2)
• C = (4, 3)

By sketching the triangle with the given points, we can see that AB is its base. (See attached graph).

To find the equation for the line of symmetry, first find the slope of the line AB. To do this, use the coordinates of points A and B and the slope formula.

Define the points:

• 
  `\textsf{Let }(x_1,y_1)=(7, 2)`
• 
  `\textsf{Let }(x_2,y_2)=(3, 6)`

`\implies \textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(6-2)/(3-7)=-1`

As the line of symmetry is perpendicular to line AB, its slope is the negative reciprocal of the slope of line AB.

Therefore, the slope of the line of symmetry = 1.

The line of symmetry will pass through point C (4, 3).

Therefore, substitute the found slope and the coordinates of point C into the point-slope formula:

`\implies y-y_1=m(x-x_1)`

`\implies y-3=1(x-4)`

`\implies y-3=x-4`

`\implies y=x-1`

Therefore, the equation of the given isosceles triangle's axis of symmetry is:

`y = x - 1`