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# PLZ HELP!!!

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Triangle ABC has vertices A(0, 0), B(0, 3) and C(5, 0). A point P inside the triangle is √10 units from point A and √13 units from point B. How many units is P from point C? Express your answer in simplest radical form.

OOPS!!!

Jul 12, 2018
edited by CluelesssPersonnn  Jul 12, 2018

#1
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Triangle ABC has vertices A(0, 0), B(0, 3) and C(5, 0).
A point P inside the triangle is v10 units from point A and v13 units from point B.
How many units is P from point C?

$$\text{Circle at A:}$$

$$\begin{array}{|rcll|} \hline (x-x_A)^2 +(y-y_A)^2 &=& r_A^2 \quad & | \quad x_A = 0 \\ && \quad & | \quad y_A = 0 \\ && \quad & | \quad r_A = \sqrt{10} \\ x^2 +y^2 &=& 10 & (1) \\ \hline \end{array}$$

$$\text{Circle at B:}$$

$$\begin{array}{|rcll|} \hline (x-x_B)^2 +(y-y_B)^2 &=& r_B^2 \quad & | \quad x_B = 0 \\ && \quad & | \quad y_B = 3 \\ && \quad & | \quad r_B = \sqrt{13} \\ x^2 +(y-3)^2 &=& 13 & (2) \\ \hline \end{array}$$

$$\text{Point P at (x_P,y_P):}$$

$$\begin{array}{|lrcll|} \hline & x_P^2 +y_P^2 &=& 10 \qquad (3) \quad & | \quad x^2 +y^2 = 10\\ & x_P^2 +(y_P-3)^2 &=& 13 \qquad (4) \quad & | \quad x^2 +(y-3)^2 = 13 \\ \hline (3)-(4): & x_P^2 +y_P^2 - ( x_P^2 +(y_P-3)^2 ) &=& 10-13 \\ & x_P^2 +y_P^2 - x_P^2 - (y_P-3)^2 &=& -3 \\ & y_P^2 - (y_P-3)^2 &=& -3 \\ & y_P^2 - y_P^2 + 6y_P -9 &=& -3 \\ & 6y_P -9 &=& -3 \\ & 6y_P &=& 6 \\ & \mathbf{ y_P } & \mathbf{=}& \mathbf{1} \\\\ \hline & x_P^2 +y_P^2 &=& 10 \quad & | \quad y_P=1 \\ & x_P^2 + 1&=& 10 \\ & x_P^2 &=& 9 \\ & \mathbf{ x_P } & \mathbf{=}& \mathbf{3} \\ \hline \end{array}$$

$$\text{Distance P and C :}$$

$$\begin{array}{|rcll|} \hline P(3,1) \\ C(5, 0) \\ \text{Distance} &=& \sqrt{(3-5)^2+(1-0)^2} \\ &=& \sqrt{(-2)^2+1^2} \\ &=& \sqrt{4+1} \\ & \mathbf{=}& \mathbf{\sqrt{5} } \\ \hline \end{array}$$

Jul 12, 2018

#1
+22343
+2

Triangle ABC has vertices A(0, 0), B(0, 3) and C(5, 0).
A point P inside the triangle is v10 units from point A and v13 units from point B.
How many units is P from point C?

$$\text{Circle at A:}$$

$$\begin{array}{|rcll|} \hline (x-x_A)^2 +(y-y_A)^2 &=& r_A^2 \quad & | \quad x_A = 0 \\ && \quad & | \quad y_A = 0 \\ && \quad & | \quad r_A = \sqrt{10} \\ x^2 +y^2 &=& 10 & (1) \\ \hline \end{array}$$

$$\text{Circle at B:}$$

$$\begin{array}{|rcll|} \hline (x-x_B)^2 +(y-y_B)^2 &=& r_B^2 \quad & | \quad x_B = 0 \\ && \quad & | \quad y_B = 3 \\ && \quad & | \quad r_B = \sqrt{13} \\ x^2 +(y-3)^2 &=& 13 & (2) \\ \hline \end{array}$$

$$\text{Point P at (x_P,y_P):}$$

$$\begin{array}{|lrcll|} \hline & x_P^2 +y_P^2 &=& 10 \qquad (3) \quad & | \quad x^2 +y^2 = 10\\ & x_P^2 +(y_P-3)^2 &=& 13 \qquad (4) \quad & | \quad x^2 +(y-3)^2 = 13 \\ \hline (3)-(4): & x_P^2 +y_P^2 - ( x_P^2 +(y_P-3)^2 ) &=& 10-13 \\ & x_P^2 +y_P^2 - x_P^2 - (y_P-3)^2 &=& -3 \\ & y_P^2 - (y_P-3)^2 &=& -3 \\ & y_P^2 - y_P^2 + 6y_P -9 &=& -3 \\ & 6y_P -9 &=& -3 \\ & 6y_P &=& 6 \\ & \mathbf{ y_P } & \mathbf{=}& \mathbf{1} \\\\ \hline & x_P^2 +y_P^2 &=& 10 \quad & | \quad y_P=1 \\ & x_P^2 + 1&=& 10 \\ & x_P^2 &=& 9 \\ & \mathbf{ x_P } & \mathbf{=}& \mathbf{3} \\ \hline \end{array}$$

$$\text{Distance P and C :}$$

$$\begin{array}{|rcll|} \hline P(3,1) \\ C(5, 0) \\ \text{Distance} &=& \sqrt{(3-5)^2+(1-0)^2} \\ &=& \sqrt{(-2)^2+1^2} \\ &=& \sqrt{4+1} \\ & \mathbf{=}& \mathbf{\sqrt{5} } \\ \hline \end{array}$$

heureka Jul 12, 2018