Question

all of rhombus (abcd) passes through the point ((6, 6)) and is perpendicular to the graph of (y = \frac{3}{4}x - 11). (overline{cd}) is parallel to (overline{ab}) and passes through the point ((-6, 10)). select the equation in slope - intercept form of the line that includes (overline{cd}).
a (y = \frac{4}{3}x + 2)
b (y = -\frac{4}{3}x + 2)
c (y = -\frac{3}{4}x + 2)
d (y = \frac{3}{4}x + 2)

Explanation:

Step1: Find slope of AD (perpendicular to \( y = \frac{3}{4}x - 11 \))

The slope of a line perpendicular to a line with slope \( m \) is \( -\frac{1}{m} \). For \( y = \frac{3}{4}x - 11 \), slope \( m = \frac{3}{4} \), so slope of AD is \( -\frac{4}{3} \).

Step2: Find slope of CD (parallel to AB, and AB is parallel to CD? Wait, no: CD is parallel to AB, but AD is perpendicular to the given line. Wait, the problem says CD is parallel to AB? Wait, no, re-reading: \( \overline{AD} \) passes through (6,6), is perpendicular to \( y = \frac{3}{4}x - 11 \). \( \overline{CD} \) is parallel to \( \overline{AB} \)? Wait, no, maybe typo? Wait, the question is about the line including \( \overline{CD} \), which is parallel to \( \overline{AB} \)? Wait, no, maybe \( \overline{CD} \) is parallel to \( \overline{AB} \), but \( \overline{AD} \) is perpendicular to the given line. Wait, no, the key is: the line for CD is parallel to AB? Wait, no, the problem says "the line that includes \( \overline{CD} \)" passes through (-6,10) and is... Wait, no, let's re-express:

Wait, the problem: \( \overline{AD} \) passes through (6,6), is perpendicular to \( y = \frac{3}{4}x - 11 \). Then \( \overline{CD} \) is parallel to \( \overline{AB} \)? Wait, no, maybe \( \overline{CD} \) is parallel to \( \overline{AB} \), but actually, the line for CD: we need to find its equation. Wait, maybe the line CD is parallel to AB, but AB is parallel to CD? Wait, no, the problem says "the line that includes \( \overline{CD} \) passes through (-6,10) and is... Wait, no, the options are in slope-intercept. Let's check the slope of CD: since CD is parallel to AB, but AD is perpendicular to \( y = \frac{3}{4}x - 11 \), so slope of AD is \( -\frac{4}{3} \). Then AB is parallel to CD? Wait, no, maybe the line CD has slope equal to AB, but AB is parallel to CD. Wait, no, the problem says "the line that includes \( \overline{CD} \) passes through (-6,10) and is... Wait, no, let's check the options. The options have slopes: A: \( \frac{4}{3} \), B: \( -\frac{4}{3} \), C: \( -\frac{3}{4} \), D: \( \frac{3}{4} \). Wait, no, earlier step: slope of AD is \( -\frac{4}{3} \), but CD is parallel to AB? Wait, no, maybe I messed up. Wait, the line for CD: let's see, the problem says "the line that includes \( \overline{CD} \) passes through (-6,10)". Wait, no, re-reading: "Select the equation in slope-intercept form of the line that includes \( \overline{CD} \). \( \overline{CD} \) is parallel to \( \overline{AB} \) and passes through the point (-6,10). \( \overline{AD} \) passes through (6,6) and is perpendicular to \( y = \frac{3}{4}x - 11 \)."

Ah! So \( \overline{AD} \) is perpendicular to \( y = \frac{3}{4}x - 11 \), so slope of AD is \( -\frac{4}{3} \). Then \( \overline{AB} \) is parallel to \( \overline{CD} \), but \( \overline{AD} \) is perpendicular to \( \overline{AB} \) (since it's a rhombus, adjacent sides are perpendicular? Wait, no, rhombus has adjacent sides equal, but not necessarily perpendicular (that's a square). Wait, no, maybe the problem has a typo, but let's proceed.

Wait, the line for CD: it passes through (-6,10), and we need to find its slope. Since CD is parallel to AB, and AD is perpendicular to the given line (slope \( \frac{3}{4} \)), so slope of AD is \( -\frac{4}{3} \). Then AB is perpendicular to AD (since it's a rhombus, adjacent sides are perpendicular? Wait, no, rhombus with perpendicular sides is a square. Maybe the problem means that AB is parallel to CD, and AD is perpendicular to the given line, so AB has slope equal to the given line? No, the given line is \( y = \frac{3}{4}x - 11 \), slope \( \frac{3}{4} \). Then AD is perpendicular, slope \( -\frac{4}{3} \). Then AB is parallel to CD, so AB has slope... Wait, no, the line for CD: let's check the options. The options are:

A. \( y = \frac{4}{3}x + 2 \)

B. \( y = -\frac{4}{3}x + 2 \)

C. \( y = -\frac{3}{4}x + 2 \)

D. \( y = \frac{3}{4}x + 2 \)

Wait, the line CD passes through (-6,10). Let's plug x = -6 into each option:

A: \( y = \frac{4}{3}(-6) + 2 = -8 + 2 = -6 eq 10 \)

B: \( y = -\frac{4}{3}(-6) + 2 = 8 + 2 = 10 \). Hey, that works!

C: \( y = -\frac{3}{4}(-6) + 2 = \frac{18}{4} + 2 = 4.5 + 2 = 6.5 eq 10 \)

D: \( y = \frac{3}{4}(-6) + 2 = -4.5 + 2 = -2.5 eq 10 \)

So the correct option is B, since when x = -6, y = 10, which matches the point (-6,10) that CD passes through. Also, the slope of CD: since CD is parallel to AB, and AD is perpendicular to the line with slope \( \frac{3}{4} \), so slope of AD is \( -\frac{4}{3} \), and AB is perpendicular to AD (since it's a rhombus, adjacent sides are perpendicular? Wait, no, but in the options, B has slope \( -\frac{4}{3} \), which when x=-6, y=10: \( 10 = -\frac{4}{3}(-6) + 2 = 8 + 2 = 10 \), which is correct.

Answer:

B. \( y = -\frac{4}{3}x + 2 \)