Question

HELP, WILL MARK BRAINLIEST+100 POINTS. PLEEEEES Square EFGH is drawn on a coordinate plane. Diagonal FH is on the line y − 3 = negative one third(x + 9). What is the slope of the diagonal GE? negative one third one third −3 3 ____________________________________________ 2. Find the equation of a line that is parallel to line g that contains (P, Q). coordinate plane with line g that passes through the points negative 3 comma 2 and 0 comma 5 3x − y = 3P − Q 3x + y = Q − 3P x − y = P − Q x + y = Q − P ___________________________________________ 3. Rectangle TUVW is on a coordinate plane at T (a, b), U (a + 2, b + 2), V (a + 5, b − 1), and W (a + 3, b − 3). What is the slope of the line that is parallel to the line that contains side WV? −2 2 −1 1 ___________________________________________ 4. What is the y-intercept of the line perpendicular to the line y = four thirdsx + 1 that includes the point (4, 1)? 2 negative thirteen thirds 4 negative nineteen thirds

I PUT ALL THE PICTURES OF THE QUESTIONS BELOW IF THAT HELPS

Answer 1

1) Slope of diagonal GE: 3

2) Equation of the line: $x-y=P-Q$

3) Slope of the line: 1

4) y-intercept: 4

Step-by-step explanation:

1)

Here the diagonal FH is given by the line with equation

$y-3=-\frac{1}{3}(x+9)$

This line is written in the form

$y-y_0 = m(x-x_0)$

where m is the slope of the line and $(x_0,y_0)$ are the coordinates of a point on the line.

Therefore, for line FH we have:

$m=-\frac{1}{3}$ (slope)

$(x_0,y_0)=(-9,3)$

Diagonal GE is perpendicular to diagonal FH (the diagonals in a square are perpendicular): this means that its slope must be equal to the inverse reciprocal of the slope of FH. THerefore, the slope of diagonal GE is

$m_{GE}=-\frac{1}{m_{FH}}=-\frac{1}{-1/3}=3$

2)

Two lines are said to be parallel when they have same slope.

Here, we see that the line given in the plot passes through the points

$(x_1,y_1)=(0,5)\\(x_2,y_2)=(-3,2)$

Therefore the slope of the line shown is

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{0-(-3)}=1$

This means that also the line that we want to find must have slope of 1:

$m=1$

Now we can use the following form to write the equation of this line:

$y-y_0 = m(x-x_0)$

where:

$m=1$

$(x_0, y_0)=(P,Q)$ is the point that must be on the line

Therefore,

$y-Q=1(x-P)\\y-Q=x-P\\x-y=P-Q$

3)

In order to solve this problem, we have to find the slope of the side WV.

The coordinates of the two points are:

$V(a+5,b-1)\\W(a+3,b-3)$

Therefore, the slope of side WV is given by:

$m=\frac{y_W-y_V}{x_W-x_V}=\frac{(b-3)-(b-1)}{(a+3)-(a+5)}=\frac{-2}{-2}=1$

And we said that two lines are parallel if they have same slope: therefore, the slope of the line parallel to the line that contains side VW is also 1.

4)

The line given in this problem has equation:

$y=\frac{4}{3}x+1$

where

$m=\frac{4}{3}$ is the slope of the line

$q=+1$ is the y-intercept of the line

Here we want to find the line perpendicular to the given line and passing through the point (4,1).

A line perpendicular to the given one must have a slope which is equal to the negative of the reciprocal of the original one, therefore:

$m'=-\frac{1}{m}=-\frac{1}{4/3}=-\frac{3}{4}$

Now we can use the form

$y-y_0 = m'(x-x_0)$

and the coordinates of the point are

$(x_0,y_0)=(4,1)$

Substituting,

$y-1=-\frac{3}{4}(x-4)\\y-1=-\frac{3}{4}x+3\\y=-\frac{3}{4}x+4$

We have written the line in the form $y=mx+q$, with $q=4$: so the y-intercept is 4.

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