Math Trailhead

We are given two points:

\(x_1 = -9\text{,}\) \(y_1 = 4\text{,}\) \(x_2 = 10\text{,}\) and \(y_2 = 10\text{.}\)

Start by finding the slope:

\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 4}{10 + 9} = \frac{{6}}{{19}} = {{\frac{6}{19}}}\)

Now we can use the point-slope formula to write the equation of this line:

\(y - y_1 = m(x - x_1)\)

\(m = {{\frac{6}{19}}}\)

\(x_1 = -9\)

\(y_1 = 4\)

\(y - 4 = {{\frac{6}{19}}}(x + 9)\)

To put the equation in slope-intercept form, distribute 6/19 and then add 4 to both sides.

\({y-4} = {{\frac{6}{19}}}x - {{\frac{6}{19}}} \cdot -9\)

\({y-4} = {\left({\frac{6}{19}}\right)x+{\frac{54}{19}}}\)

\(y = {{\frac{6}{19}}}x + {{\frac{130}{19}}}\)

Note: To multiply fractions, we multiply straight across. For example \(\frac{2}{5} \cdot 6 = \frac{2}{5} \cdot \frac{6}{1} = \frac{12}{5}\text{.}\)

To add or subtract fractions we need a least common denominator (LCD). For example, \(\frac{2}{5} - 3 = \frac{2}{5} - \frac{3}{1} = \frac{2}{5} - \frac{15}{5} = -\frac{13}{5}\text{.}\)

When trying to find the equation of a line with some information, start with the equation in slope-intercept form, that is, \(y = mx+b\text{.}\) Then substitute in any known information. In this case, we know the slope is \(m={-{\frac{3}{4}}}\text{,}\) or

\(\displaystyle{ y = {-{\frac{3}{4}}} x + b }\)

Next, substitute in the point \((3,3)\) or

Next, substract \({-{\frac{9}{4}}}\) from both sides:

and finally plug this value of b into the equation:

The graphed line passes through the points: \(A = (-3,13)\text{,}\) \(B = (-1,6)\text{,}\) \(C = (1,-1)\)

Select any pair of points on the line to compute the slope:

\(m = \frac{\Delta y}{\Delta x} = \frac{6-13}{-1 + 3} = {-{\frac{7}{2}}}\)

Use the point-slope form of a line:

\(y = m (x - x_A) + y_A\)

\(y = {-{\frac{7}{2}}\mathopen{}\left(x+1\right)+6}\)

The graphed line passes through the points: \(A = (-5,4)\text{,}\) \(B = (0,5)\text{,}\) \(C = (5,6)\)

Select any pair of points on the line to compute the slope:

\(m = \frac{\Delta y}{\Delta x} = \frac{5-4}{0 + 5} = {{\frac{1}{5}}}\)

Use the point-slope form of a line:

\(y = m (x - x_A) + y_A\)

\(y = {{\frac{1}{5}}x+5}\)

We’re parallel to the line \(y = {{\frac{1}{3}}x-5}\text{,}\) which has slope \(m = {{\frac{1}{3}}}\text{.}\)

Which means our slope must also be \(m = {{\frac{1}{3}}}\)

\(A = (5,2)\)

Use the point-slope form of a line:

\(y = m (x - x_A) + y_A\)

\(y = {{\frac{1}{3}}\mathopen{}\left(x-5\right)+2}\)

Use the point-slope form of a line:

\(y = m (x - x_A) + y_A\)

\(y = {{\frac{2}{3}}\mathopen{}\left(x+3\right)-6}\)

Compare the given equation with a generic point-slope equation:

\(\begin{aligned} y-y_{1} \amp = m(x-x_{1}) \\ {y-6} \amp = {5\mathopen{}\left(x-1\right)} \end{aligned}\)

We can see the slope is \(5\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(1,6\right)}\text{.}\)

Compare the given equation with a generic point-slope equation:

\(\begin{aligned} y \amp = m(x-x_{1})+y_1 \\ {y} \amp = {-5\mathopen{}\left(x+2\right)+5} \end{aligned}\)

We can see the slope is \(-5\text{,}\) and an apparent point on the line \((x_{1},y_{1})\) is \({\left(-2,5\right)}\text{.}\)

First, find the slope by using the slope formula:

\(\displaystyle m = \frac{y_A - y_B}{x_A - x_B}\)

\(\displaystyle m = \frac{6 + 6}{-8 + 1}\)

\(\displaystyle m = \frac{12}{-7}\)

\(\displaystyle m = {-{\frac{12}{7}}}\)

Then use the point-slope form of a line:

\(y = m (x - x_A) + y_A\)

\(y = {-{\frac{12}{7}}\mathopen{}\left(x+8\right)+6}\)

When trying to find the equation of a line with some information, start with the equation in slope-intercept form, that is, \(y = mx+b\text{.}\) Then substitute in any known information. We know neither the slope or the y-intercept in this case, but easily can find the slope from the two points:

Then substitute this into the slope-intercept form of the line:

Next, substitute in the point \((-4,0)\) or

Next, subtract \(3/2\) from both sides:

When an equation of a line is written in the form \(y=mx+b\text{,}\) it is said to be in slope-intercept form. In this form, \(m\) is the line’s slope, and \(b\) is the coordinate on the \(y\)-axis where the line intercepts the \(y\)-axis.

In this problem, the line’s equation is given as \(\displaystyle{ -{5}x+y= 6 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\)

\(\displaystyle{\begin{aligned} -{5}x+y \amp = 6 \\ -{5}x+y\mathbf{{}+{5}x} \amp = 6\mathbf{{}+{5}x} \\ y \amp = {5}x+6 \end{aligned} }\)

Now we can see the line’s slope is \({5}\text{,}\) and its \(y\)-intercept has coordinates \((0,6)\text{.}\)

For each of the lines in the answer, you should identify if the line is vertical, horizontal or oblique.

In this case, the line \({x+5y = 4}\) is horizontal. The answer is D.

The line \({3x+y = -4}\) in slope-intercept form, and the slope is -3 with y-intercept of -4. It appears that the line has this slope and y-intercept. The answer is A.

In this case, we need to find the equation of the line based on the graph. It’s generally easiest to find the equation in slope-intercept form, so we will need both the slope and the y-intercept. It hard to just look at the line to get either the slope or the y-intercept, but it appears that there are two points that are easy to see on the graph (since they fall on integer coordinates).

These points appear to be \((-4,-2)\) and \((1,2)\text{,}\) so we will find the line that passes through these points. First, find the slope from the two points:

\(\displaystyle{ m = \frac{-2-2}{-4-1} = {{\frac{4}{5}}} }\)

Then substitute this into the slope-intercept form of the line:

\(\displaystyle{ y = {{\frac{4}{5}}} x + b }\)

Next, substitute in the point \((-4,-2)\) or

\(\displaystyle{\begin{aligned} -2 \amp = ({{\frac{4}{5}}})(-4) + b \\ -2 \amp = + b \end{aligned} }\)

Next, substract \(\) from both sides:

\(\displaystyle{ {{\frac{6}{5}}} = b}\) and finally plug this value of b into the equation:

\(\displaystyle{ y = {\left({\frac{4}{5}}\right)x+\left({\frac{6}{5}}\right)} }\)

Solution

The best way to handle this problem, if the answer is not immediately clear to you,

is to add a positive quantity to \(n\) and see what happens to \(p\text{.}\)

Suppose \(m\) is positive and we repace \(n\) by \(n+m\text{.}\) Calling the resulting value

of \(p\) by the name \(p'\) we obtain the equation

\(p'=2(n+m)-10=2n-10+2m\text{.}\)

This tells us that when we increase \(n\) by \(m\) we increase \(p\) by \(2m\text{.}\)