Math Trailhead

We are given two points:

\(x_1 = -4\text{,}\) \(y_1 = -5\text{,}\) \(x_2 = -7\text{,}\) and \(y_2 = 8\text{.}\)

Start by finding the slope:

\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 + 5}{-7 + 4} = \frac{{13}}{{-3}} = {-{\frac{13}{3}}}\)

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

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

\(m = {-{\frac{13}{3}}}\)

\(x_1 = -4\)

\(y_1 = -5\)

\(y + 5 = {-{\frac{13}{3}}}(x + 4)\)

To put the equation in slope-intercept form, distribute -13/3 and then add -5 to both sides.

\({y-\left(-5\right)} = {-{\frac{13}{3}}}x - {-{\frac{13}{3}}} \cdot -4\)

\({y-\left(-5\right)} = {-\left({\frac{13}{3}}\right)x-{\frac{52}{3}}}\)

\(y = {-{\frac{13}{3}}}x + {-{\frac{67}{3}}}\)

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 \(m={-{\frac{1}{5}}}\text{,}\) or

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

Next, substitute in the point \((2,1)\) or

Next, substract \({-{\frac{2}{5}}}\) from both sides:

and finally plug this value of b into the equation:

The graphed line passes through the points: \(A = (-9,-3)\text{,}\) \(B = (-4,-1)\text{,}\) \(C = (1,1)\)

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

\(m = \frac{\Delta y}{\Delta x} = \frac{-1 + 3}{-4 + 9} = {{\frac{2}{5}}}\)

Use the point-slope form of a line:

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

\(y = {{\frac{2}{5}}\mathopen{}\left(x+4\right)-1}\)

The graphed line passes through the points: \(A = (-8,3)\text{,}\) \(B = (0,6)\text{,}\) \(C = (8,9)\)

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

\(m = \frac{\Delta y}{\Delta x} = \frac{6-3}{0 + 8} = {{\frac{3}{8}}}\)

Use the point-slope form of a line:

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

\(y = {{\frac{3}{8}}x+6}\)

We’re parallel to the line \(y = {{\frac{4}{5}}x-5}\text{,}\) which has slope \(m = {{\frac{4}{5}}}\text{.}\)

Which means our slope must also be \(m = {{\frac{4}{5}}}\)

\(A = (-2,8)\)

Use the point-slope form of a line:

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

\(y = {{\frac{4}{5}}\mathopen{}\left(x+2\right)+8}\)

Use the point-slope form of a line:

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

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

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

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

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

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

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

We can see the slope is \(-2\text{,}\) 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:

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

\(\displaystyle m = \frac{-8 + 2}{-3 + 8}\)

\(\displaystyle m = \frac{-6}{5}\)

\(\displaystyle m = {-{\frac{6}{5}}}\)

Then use the point-slope form of a line:

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

\(y = {-{\frac{6}{5}}\mathopen{}\left(x+3\right)-8}\)

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 \((-1,2)\) or

Next, subtract \(1\) 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 \(\displaystyle{ -{7}x+y= 10 }\text{.}\) It would be helpful to algebraically rearrange this into slope-intercept form: \(y= mx+b\text{.}\)

\(\displaystyle{\begin{aligned} -{7}x+y \amp = 10 \\ -{7}x+y\mathbf{{}+{7}x} \amp = 10\mathbf{{}+{7}x} \\ y \amp = {7}x+10 \end{aligned} }\)

Now we can see the line’s slope is \({7}\text{,}\) and its \(y\)-intercept has coordinates \((0,10)\text{.}\)

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

In this case, the line \({2x+6y = 6}\) is horizontal. The answer is A.

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

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 \((-1,1)\) and \((-3,1)\text{,}\) so we will find the line that passes through these points. First, find the slope from the two points:

\(\displaystyle{ m = \frac{1-1}{-1 + 3} = {0} }\)

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

\(\displaystyle{ y = {0} x + b }\)

Next, substitute in the point \((-1,1)\) or

\(\displaystyle{\begin{aligned} 1 \amp = ({0})(-1) + b \\ 1 \amp = + b \end{aligned} }\)

Next, substract \(\) from both sides:

\(\displaystyle{ {1} = b}\) and finally plug this value of b into the equation:

\(\displaystyle{ y = {1} }\)

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 \(2m\text{.}\)