Question

1. Given the points (4,-2) and (8,1), what is the equation of a line that connect the two points. Answer in slope-intercept form.

2. What is the slope-intercept form of a linear equation that passes through (-3,7) and is parallel to the equation y=2x+5

3.∠ABC is adjacent to ∠CBD. If the m∠ABC=4x+23, m∠CBD=6x+7, and m∠ABD=130°, what is the measure of angle ABC?

4.Point B lies between the points A and C. If AC=7x+11, AB=2x+29, and CB=4x+3, what is the length of segment CB?

5.What are the angle measures of the triangle? (Picture included below)

6.What is the area of the rectangle?(Picture included)

Answer 1

Answer:

Part 1) $y=\frac{3}{4}x-5$

Part 2) $y=2x+13$

Part 3) $m\angle ABC=63^o$

Part 4) $CB=87\ units$

Part 5) $m\angle A=68.75^o$,$m\angle B=42.5^o$,$m\angle C=68.75^o$

Part 6) $A=580\ units^2$

Step-by-step explanation:

Part 1) Given the points (4,-2) and (8,1), what is the equation of a line that connect the two points

step 1

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{1+2}{8-4}$

$m=\frac{3}{4}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$

$point\ (8,1)$

substitute

$y-1=\frac{3}{4}(x-8)$

step 3

Convert to slope intercept form

$y=mx+b$

isolate the variable y

$y-1=\frac{3}{4}x-6$

$y=\frac{3}{4}x-6+1$

$y=\frac{3}{4}x-5$

Part 2) What is the slope-intercept form of a linear equation that passes through (-3,7) and is parallel to the equation y=2x+5

we know that

If two line are parallel, then their slopes are the same

The slope of the given line is m=2

so

The slope of the line parallel to the given line is m=2

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=2\\point(-3,7)$

substitute

$7=2(-3)+b$

solve for b

$7=-6+b$

$b=7+6=13$

therefore

$y=2x+13$

Part 3) ∠ABC is adjacent to ∠CBD. If the m∠ABC=4x+23, m∠CBD=6x+7, and m∠ABD=130°, what is the measure of angle ABC?

we know that

$m\angle ABD=m\angle ABC+m\angle CBD$ ----> by addition angle postulate

substitute the given values

$130^o=(4x+23)^o+(6x+7)^o$

solve for x

$130^o=(10x+30)^o$

$10x=130-30$

$10x=100$

$x=10$

Find the measure of angle ABC

$m\angle ABC=(4x+23)^o$

substitute the value of x

$m\angle ABC=(4(10)+23)=63^o$

Part 4) Point B lies between the points A and C. If AC=7x+11, AB=2x+29, and CB=4x+3, what is the length of segment CB?

we know that

$AC=AB+CB$ -----> by addition segment postulate

Remember that

$CB=BC$

substitute the given values

$7x+11=(2x+29)+(4x+3)$

solve for x

$7x+11=(6x+32)$

$7x-6x=32-11$

$x=21$

Find the length of segment CB

$CB=4x+3$

substitute the value of x

$CB=4(21)+3$

$CB=87\ units$

Part 5) What are the angle measures of the triangle?

we know that

An isosceles triangle has two equal sides and two equal angles

The triangle ABC of the figure is an isosceles triangle

because

AB=BC

so

∠A=∠C ----> equation A

Remember that the sum of the interior angles of triangle must be equal to 180 degrees

so

∠A+∠B+∠C=180° ----> equation B

substitute equation A in equation B

2∠A+∠B=180°

substitute the given values

$2(2x+37)^o+(4x-21)^o=180^o$

solve for x

$4x+74+4x-21=180$

$8x+53=180$

$8x=180-53$

$8x=127$

$x=15.875$

Find the angle measures of the triangle

substitute the value of x

$m\angle A=m\angle C=2(15.875)+37=68.75^o$

$m\angle B=4(15.875)-21=42.5^o$

therefore

$m\angle A=68.75^o$

$m\angle B=42.5^o$

$m\angle C=68.75^o$

Part 6) What is the area of the rectangle?

we know that

A rectangle has opposite sides parallel and congruent

so

AB=CD

substitute the given values

$2x+3=3x-10$

solve for x

$3x-2x=3+10$

$x=13\ units$

Remember that the area of rectangle is equal to

$A=LW$

where

L is the length

W is the width

In this problem

$L=AB=2x+3$

$W=AD=x+7$

substitute the value of x

$L=AB=2(13)+3=29\ units$

$W=AD=13+7=20\ units$

The area is equal to

$A=(29)(20)=580\ units^2$