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${:[AB=sqrt((4-(-1)^(2)+(-3-9)^(2):})],[=sqrt147],[=12.12]:} {:[CD=sqrt((-1-6)^(2)+(5-(-4)^(2):})],[=sqrt38],[=6.16]:} The diagram shows the points A(0,-1),B(p,p) and C(p,q). (a) Given that the length of AB is 5 units, find the value of p. (b) Given that angle BAC= angle BCA, find the value of q.$

$\begin{aligned} A B & =\sqrt{\left(4-(-1)^{2}+(-3-9)^{2}\right.} \\ & =\sqrt{147} \\ & =12.12 \end{aligned}$ $\newline$ $\begin{aligned} C D & =\sqrt{(-1-6)^{2}+\left(5-(-4)^{2}\right.} \\ & =\sqrt{38} \\ & =6.16 \end{aligned}$ $\newline$ The diagram shows the points $A(0,-1), B(p, p)$ and $C(p, q)$ . $\newline$ (a) Given that the length of $A B$ is $5$ units, find the value of $p$ . $\newline$ (b) Given that angle $B A C=$ angle $B C A$ , find the value of $q$ .

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Math Problems

Algebra 1

Transformations of quadratic functions

Full solution

\begin{aligned} A B & =\sqrt{\left(4-(-1)^{2}+(-3-9)^{2}\right.} \\ & =\sqrt{147} \\ & =12.12 \end{aligned}

\newline

\begin{aligned} C D & =\sqrt{(-1-6)^{2}+\left(5-(-4)^{2}\right.} \\ & =\sqrt{38} \\ & =6.16 \end{aligned}

\newline

The diagram shows the points

A(0,-1), B(p, p)

and

C(p, q)

\newline

(a) Given that the length of

A B

5

units, find the value of

p

\newline

(b) Given that angle

B A C=

angle

B C A

, find the value of

q

Calculate AB length: Calculate the length of AB using the coordinates of A and B. $\newline$ $AB = \sqrt{(p - 0)^2 + (p - (-1))^2}$ $\newline$ Set $AB$ equal to $5$ units and solve for $p$ . $\newline$ $5 = \sqrt{p^2 + (p + 1)^2}$
Set AB to $5$ units: Square both sides to remove the square root. $25 = p^2 + p^2 + 2p + 1$ Combine like terms. $25 = 2p^2 + 2p + 1$
Square both sides: Subtract $25$ from both sides to set the equation to zero. $\newline$ $0 = 2p^2 + 2p - 24$
Subtract $25$ : Divide the entire equation by $2$ to simplify. $\newline$ $0 = p^2 + p - 12$
Divide by $2$ : Factor the quadratic equation. $\newline$ $(p + 4)(p - 3) = 0$
Factor quadratic equation: Set each factor equal to zero and solve for $p$ . $p + 4 = 0$ or $p - 3 = 0$ $p = -4$ or $p = 3$
Solve for $p$ : Since $p$ is a coordinate, it cannot be negative in this context. $p = 3$
Use angle BAC: Use the fact that angle $BAC = angle BCA$ to find $q$ . $\newline$ Since triangle $ABC$ is isosceles, $AC = BC$ . $\newline$ Use the coordinates of $A, B,$ and $C$ to write the expressions for $AC$ and $BC$ . $\newline$ $AC = \sqrt{(p - 0)^2 + (q - (-1))^2}$ $\newline$ $BC = \sqrt{(p - p)^2 + (q - p)^2}$
Substitute $p = 3$ : Substitute $p = 3$ into the expressions for AC and BC. $\newline$ $AC = \sqrt{(3 - 0)^2 + (q - (-1))^2}$ $\newline$ $BC = \sqrt{(3 - 3)^2 + (q - 3)^2}$
Simplify AC and BC: Simplify the expressions for AC and BC. $\newline$ AC = $\sqrt{9 + (q + 1)^2}$ $\newline$ BC = $\sqrt{0 + (q - 3)^2}$
Set $AC = BC$ : Set $AC$ equal to $BC$ and solve for $q$ . $\newline$ $\sqrt{9 + (q + 1)^2} = \sqrt{(q - 3)^2}$ $\newline$ Square both sides to remove the square roots. $\newline$ $9 + (q + 1)^2 = (q - 3)^2$
Expand squared terms: Expand the squared terms. $9 + q^2 + 2q + 1 = q^2 - 6q + 9$
Combine like terms: Combine like terms and solve for $q$ . $10 + 2q = -6q + 9$ $8q = -1$ $q = -\frac{1}{8}$