Question

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.
a. (-3,1)
b. (-2,1)
c. (-4,1)
d. (-4,5)

Explanation:

Step1: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

Step2: Calculate the distance for each option

• Option A: For \(J(-3,1)\) and \(K(2,1)\), \(d_{JK}=\sqrt{(2-(-3))^2+(1 - 1)^2}=\sqrt{(2 + 3)^2+0}=\sqrt{25}=5\). For \(J(-3,1)\) and \(L(-1,5)\), \(d_{JL}=\sqrt{(-1-(-3))^2+(5 - 1)^2}=\sqrt{( - 1+3)^2+4^2}=\sqrt{4 + 16}=\sqrt{20}\). For \(K(2,1)\) and \(L(-1,5)\), \(d_{KL}=\sqrt{(-1 - 2)^2+(5 - 1)^2}=\sqrt{(-3)^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5\). Since \(d_{JK}=d_{KL}\), this forms an isosceles triangle.
• Option B: For \(J(-2,1)\) and \(K(2,1)\), \(d_{JK}=\sqrt{(2-(-2))^2+(1 - 1)^2}=\sqrt{16}=4\). For \(J(-2,1)\) and \(L(-1,5)\), \(d_{JL}=\sqrt{(-1-(-2))^2+(5 - 1)^2}=\sqrt{1 + 16}=\sqrt{17}\). For \(K(2,1)\) and \(L(-1,5)\), \(d_{KL}=\sqrt{(-1 - 2)^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25}=5\). Since \(d_{JK}

eq d_{KL}
eq d_{JL}\), this does not form an isosceles triangle.

• Option C: For \(J(-4,1)\) and \(K(2,1)\), \(d_{JK}=\sqrt{(2-(-4))^2+(1 - 1)^2}=\sqrt{36}=6\). For \(J(-4,1)\) and \(L(-1,5)\), \(d_{JL}=\sqrt{(-1-(-4))^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25}=5\). For \(K(2,1)\) and \(L(-1,5)\), \(d_{KL}=\sqrt{(-1 - 2)^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25}=5\). Since \(d_{JL}=d_{KL}\), but we are looking for the base \(JK\) (as per the problem statement about the base of the isosceles triangle \(JKL\)), this is not the correct case for the base - vertex relationship.
• Option D: For \(J(-4,5)\) and \(K(2,1)\), \(d_{JK}=\sqrt{(2-(-4))^2+(1 - 5)^2}=\sqrt{36 + 16}=\sqrt{52}\). For \(J(-4,5)\) and \(L(-1,5)\), \(d_{JL}=\sqrt{(-1-(-4))^2+(5 - 5)^2}=\sqrt{9}=3\). For \(K(2,1)\) and \(L(-1,5)\), \(d_{KL}=\sqrt{(-1 - 2)^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25}=5\). Since \(d_{JK}

eq d_{KL}
eq d_{JL}\), this does not form an isosceles triangle.

Answer:

A. (-3,1)