The points (-6,r) and (10,-2) lie on a line with slope -(3)/(4). Find the missing coordinate r. r=

Identify Points and Slope: Identify the given points and the slope of the line. Points: (-6, r) and (10, -2) Slope: -(3/4) Use the slope formula to set up the equation. Slope = (y_2 - y_1)/(x_2 - x_1) -(3/4) = (-2 - r) / (10 - (-6))Set Up Equation: Simplify the right side of the equation. -(3/4) = (-2 - r) / (10 + 6) -(3/4) = (-2 - r) / 16Simplify Equation: Cross-multiply to solve for r. -(3/4) * 16 = -2 - r -3 * 4 = -2 - r -12 = -2 - rCross-Multiply to Solve: Solve for r. -12 = -2 - r -12 + 2 = -2 - r + 2 -10 = -r r = 10

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The points
(-6,r) and
(10,-2) lie on a line with slope
-(3)/(4). Find the missing coordinate
r.

r=

The points $(-6, r)$ and $(10,-2)$ lie on a line with slope $-\frac{3}{4}$ . Find the missing coordinate $r$ . $\newline$ $r=$

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Math Problems

Algebra 1

Find a missing coordinate using slope

Full solution

Q. The points

(-6, r)

and

(10,-2)

lie on a line with slope

-\frac{3}{4}

. Find the missing coordinate

r

\newline

r=

Identify Points and Slope: Identify the given points and the slope of the line. $\newline$ Points: $(-6, r)$ and $(10, -2)$ $\newline$ Slope: $-\left(\frac{3}{4}\right)$ $\newline$ Use the slope formula to set up the equation. $\newline$ Slope $= \frac{y_2 - y_1}{x_2 - x_1}$ $\newline$ $-\left(\frac{3}{4}\right) = \frac{-2 - r}{10 - (-6)}$
Set Up Equation: Simplify the right side of the equation. $\newline$ $-\frac{3}{4} = \frac{-2 - r}{10 + 6}$ $\newline$ $-\frac{3}{4} = \frac{-2 - r}{16}$
Simplify Equation: Cross-multiply to solve for $r$ . $-\frac{3}{4} \times 16 = -2 - r$ $-3 \times 4 = -2 - r$ $-12 = -2 - r$
Cross-Multiply to Solve: Solve for $r$ .
$-12 = -2 - r$
$-12 + 2 = -2 - r + 2$
$-10 = -r$
$r = 10$