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Points J (3, 1) and K (-1, -3) are two vertices of an...

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Points J (3, 1) and K (-1, -3) are two vertices of an...

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Points J (3, 1) and K ( -1, -3) are two vertices of an isosceles triangle. If L is the third vertex and has a y-coordinate of 6, what is the x-coordinate of L?

A. -3
B. -4
C. -5
D. -6
E. -7

The OA is D.

I don't have clear this PS question.

Can I determine the x-coordinate value of L using the expression?
$$d\left(P_1,P_2\right)=\sqrt{\left(X_2-X_1\right)^2+\left(Y_2-Y_1\right)^2}$$
distance between 2 points in the coordinate plane. Where P1 and P2 can be, P1 (J or K) and P2 (L)

I appreciate if any expert explain it for me. Thank you so much.

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GMAT/MBA Expert

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Hello AAPL.

Let's take a look at your question.

You are right. We have to use the formula $$d\left(P_1,P_2\right)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}.$$ We have that J=(3,1), K=(-1,-3) and L=(x,6).

Then $$d\left(J,\ L\right)=\sqrt{\left(x-3\right)^2+\left(6-1\right)^2}=\sqrt{\left(x-3\right)^2+25}$$ $$d\left(K\ ,\ L\right)=\sqrt{\left(x+1\right)^2+\left(6+3\right)^2}=\sqrt{\left(x+1\right)^2+81}$$ and $$d\left(J,\ K\right)=\sqrt{\left(-1-3\right)^2+\left(-3-1\right)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}.$$ Now, since the triangle is isosceles, we have that $$d\left(J,\ L\right)=d\left(K,\ L\right)\ \Leftrightarrow\ \ \ \sqrt{\left(x-3\right)^2+25}=\sqrt{\left(x+1\right)^2+81}$$ $$\Leftrightarrow\ \left(x-3\right)^2+25\ =\ \left(x+1\right)^2+81\ $$ $$\Leftrightarrow x^2-6x+9+25\ =\ x^2+2x+1+81$$ $$\Leftrightarrow\ -8x=48$$ $$\Leftrightarrow\ x=-6.$$ This is why the correct answer is the option D=-6.

I hope this explanation may help you.

I'm available if you'd like a follow-up.

Regards.

_________________
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