Isosceles triangle JKL has a perimeter of $32$ units and the given vertices. $\newline$ - J $(-3,-9)$ $\newline$ - $K(-3,6)$ $\newline$ - L $(x,-1.5)$ $\newline$ What is a possible $x$ -coordinate for point $L$ ?

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Write a quadratic function from its x-intercepts and another point

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Q. Isosceles triangle JKL has a perimeter of

32

units and the given vertices.

\newline

- J

(-3,-9)

\newline

K(-3,6)

\newline

- L

(x,-1.5)

\newline

What is a possible

x

-coordinate for point

L

Calculate JK Length: Since $JKL$ is an isosceles triangle, two sides are equal. $JK$ is vertical, so its length is the difference in $y$ -coordinates between $J$ and $K$ . $\newline$ Length of $JK = 6 - (-9) = 15$ units.
Find Sum of Other Sides: The perimeter of the triangle is $32$ units. If we subtract the length of $JK$ from the perimeter, we get the sum of the lengths of the other two sides, $JL$ and $KL$ . $\newline$ Perimeter - $JK$ = $32 - 15 = 17$ units.
Determine JL and KL Length: In an isosceles triangle, JL and KL are equal in length. So, each side is half of $17$ units. $\newline$ Length of JL = Length of KL = $\frac{17}{2} = 8.5$ units.
Express JL Length in Terms of $x$ : Now we need to find the $x$ -coordinate for point L. Since J and K have the same $x$ -coordinate, the $x$ -coordinate of L will determine the length of JL and KL. $\newline$ We can use the distance formula to express the length of JL in terms of $x$ : $\newline$ Length of JL = $\sqrt{(x - (-3))^2 + ((-1.5) - (-9))^2} = 8.5$ units.
Simplify Equation: Simplify the equation: $\newline$ $\sqrt{(x + 3)^2 + 7.5^2} = 8.5$ $\newline$ Square both sides to remove the square root: $\newline$ $(x + 3)^2 + 7.5^2 = 8.5^2$
Calculate Squares: Calculate the squares: $\newline$ $(x + 3)^2 + 56.25 = 72.25$ $\newline$ $(x + 3)^2 = 72.25 - 56.25$ $\newline$ $(x + 3)^2 = 16$
Take Square Root: Take the square root of both sides: $\newline$ $x + 3 = \pm\sqrt{16}$ $\newline$ $x + 3 = \pm4$
Solve for x: Solve for x: $\newline$ $x = -3 \pm 4$ $\newline$ So, $x$ can be $-3 + 4$ or $-3 - 4$ . $\newline$ $x = 1$ or $x = -7$