1. The base of an isosceles triangle JKL has the endpoints J and K. Point K is located at (2,1) and the top of the triangle, point L, is located at (-1,5). Use the distance formula to find the location of point J.
If we have to use the distance formula, then we have
LK = sqrt(9+16) = 5
so, JL also is 5.
The question is, where is J?
Note that since LK = 5, it is the hypotenuse of a right triangle, with legs parallel to the axes. Thus, the base of the triangle is parallel to the x-axis.
So, since the base extends to the left of the line x = -1 the same distance it extends to the right, namely 3,
J = (-4,1)
This problem would have been harder if the triangle had been oriented differently.
To find the location of point J, we can use the fact that an isosceles triangle has two sides of equal length. In this case, the two equal sides are JK and KL.
First, let's find the length of side KL using the distance formula:
Distance of KL = √[(x2 - x1)² + (y2 - y1)²]
Substituting the coordinates of points K and L into the formula, we have:
Distance of KL = √[(-1 - 2)² + (5 - 1)²]
= √[(-3)² + (4)²]
= √[9 + 16]
= √25
= 5
Now, since JK and KL are equal sides, the length of JK is also 5.
Let's assume the coordinates of point J are (x, y). Using the distance formula again, we can find the length of JK:
Distance of JK = √[(x - 2)² + (y - 1)²]
Since JK = KL = 5, we can equate the two distances:
√[(x - 2)² + (y - 1)²] = 5
Squaring both sides of the equation, we have:
(x - 2)² + (y - 1)² = 25
Expanding the equation, we get:
x² - 4x + 4 + y² - 2y + 1 = 25
Simplifying further, we have:
x² - 4x + y² - 2y = 20
To find the location of point J, we need to find the values of x and y that satisfy this equation.