Question

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Solve the following problems, do not forget that every point must be accompanied by its respective procedure.


a. A (1,1) and B (2,1)

b. C (0.5) and D (5.0)

c. E (3,2) and F (4,5)

d. G (0, -3) and H (-6.0)

e. P (-3, 4) and Q (3, 4)


2. Find the equation of the line


a. It has slope 4 and passes through the point (−3, −2).

b. It passes through the point P (2, −3) and is parallel to the line of equation y = −x + 7.

c. It has slope 3 and goes through the point (-2, 7)

d. It has slope -4/3 and passes through the point (-1, 7).

e. It has Slope 3/2 and cuts to the vertical axis Y at point - 2.


PLEASE HELP!!!

Answer 1

                    Question a

Find the equation of the line. It has slope 4 and passes through the point (−3, −2).

As the equation of a line in slope-intercept form  is

$y=mx+b$

where $m$ is the slope and $b$ being the y-intercept.

here   $m=4$

⇒  $y=4x+b$  ← is the partial equation

to find $b$ substitute $(-3, -2)$ into the partial equation.

$-2=4\left(-3\right)+b$

$\mathrm{Switch\:sides}$

$4\left(-3\right)+b=-2$

$-12+b=-2$

$b=10$

So,

⇒ $y=4x+10$

Therefore, $y=4x+10$ is the equation of line which has slope 4 and passes through the point (−3, −2).

                       Question b

Find the equation of the line. It passes through the point P (2, −3) and is parallel to the line of equation y = −x + 7.

As the equation of a line in slope-intercept form  is

$y=mx+b$

as   $y=-x+7$

here   $m=-1$

As the point is $(-2, 7)$

Using the point slope formula

$y-y_1=m\left(x-x_1\right)$

$y-7=m\left(x-\left(-2\right)\right)$

$y-7=4\left(x-\left(-2\right)\right)$

$y-7=4\left(x+2\right)$

$y-7+7=4\left(x+2\right)+7$

$y=4x+15$

Therefore,  $y=4x+10$ is the equation of line which passes through the point P (2, −3) and is parallel to the line of equation y = −x + 7.

                       Question c

Find the equation of the line. It has slope 3 and goes through the point (-2, 7).

As the equation of a line in slope-intercept form  is

$y=mx+b$

where $m$ is the slope and $b$ being the y-intercept.

here   $m=3$

⇒  $y=3x+b$  ← is the partial equation

to find $b$ substitute $\left(-2,\:7\right)$ into the partial equation.

$7=3\left(-2\right)+b$

$\mathrm{Switch\:sides}$

$3\left(-2\right)+b=7$

$-6+b=7$

$6+b+6=7+6$

$b=13$

So,

⇒   $y=3x+13$

Therefore, $y=3x+13$  is the equation of line which has slope 3 and passes through the point  $\left(-2,\:7\right)$.

                         

                          Question d

Find the equation of the line. It has slope -4/3 and passes through the point (-1, 7).

As the equation of a line in slope-intercept form  is

$y=mx+b$

where $m$ is the slope and $b$ being the y-intercept.

here   $m=\frac{-4}{3}$

⇒  $y=\left(\frac{-4}{3}\right)x+b$  ← is the partial equation

to find $b$ substitute $\left(-1,\:7\right)$ into the partial equation.

$7=\left(\frac{-4}{3}\right)\left(-1\right)+b$

$\mathrm{Switch\:sides}$

$\left(\frac{-4}{3}\right)\left(-1\right)+b=7$

$\frac{4}{3}\:+b=7\:\:\:$

$b=7\:-\frac{4}{3}$

$b=\frac{17}{3}$

So,

⇒   $y=\left(\frac{-4}{3}\right)x+\frac{17}{3}$  is the equation of line which has slope   $m=\frac{-4}{3}$ and passes through the point (-1, 7).

                          Question e

Find the equation of the line. It has Slope 3/2 and cuts to the vertical axis Y at point - 2.

As

$m\:=\frac{3}{2}\:$

and

at  $x=0$, the value of $y=-2$

So

the y-intercept will be $c = -2$

Let the equation be

$y=mx+c$

so

$\:y=\left(\frac{3}{2}\right)x+\left(-2\right)$

$y=\left(\frac{3}{2}\right)x-2$

Therefore, $y=\left(\frac{3}{2}\right)x-2$  the equation of the line which It has Slope 3/2 and cuts to the vertical axis Y at point - 2.

Keywords: equation of a line , slop , y-intercept, slop-intercept form

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Answer 2

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