How To Find The Gradient Of A Line Segment

8.three Gradient of a line (EMA6B)

Gradient

The gradient of a line is determined by the ratio of vertical alter to horizontal change.

Slope (\(thou\)) describes the gradient or steepness of the line joining two points. In the effigy below, line \(OQ\) is the least steep and line \(OT\) is the steepest.

To derive the formula for gradient, we consider any right-angled triangle formed from \(A \left(x_{1};{y}_{1}\right)\) and \(B \left({x}_{2};{y}_{2}\correct)\) with hypotenuse \(AB\) as shown in the diagram below. The gradient is adamant by the ratio of the length of the vertical side of the triangle to the length of the horizontal side of the triangle. The length of the vertical side of the triangle is the difference in \(y\)-values of points \(A\) and \(B\). The length of the horizontal side of the triangle is the deviation in \(x\)-values of points \(A\) and \(B\).

Therefore, slope is determined using the following formula:

Slope \(\left(m\correct)=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{one}}\) or \(\dfrac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}\)

Think to be consistent: \(m\ne \dfrac{{y}_{1}-{y}_{two}}{{10}_{two}-{x}_{one}}\)

To learn more near determining the gradient you lot can watch the following video.

Video: 2GC6

Worked example 3: Gradient between two points

Find the slope of the line between points \(E \left(two;5\right)\) and \(F\left(-3;9\right)\).

Draw a sketch

Assign values to \(\left({ten}_{i};{y}_{ane}\right)\) and \(\left({x}_{2};{y}_{2}\right)\)

Allow the coordinates of \(Eastward\) be \(\left({x}_{one};{y}_{1}\right)\) and the coordinates of \(F\) be \(\left({ten}_{2};{y}_{2}\correct)\).

\[{x}_{i} = 2 \quad {y}_{1} = v \quad {ten}_{2} = -3 \quad {y}_{2} = ix\]

Write down the formula for slope

\[thou = \frac{{y}_{2}-{y}_{ane}}{{x}_{2}-{x}_{1}}\]

Substitute known values

\brainstorm{align*} m_{EF} & = \frac{9 - 5}{-3 - ii} \\ & = \frac{four}{-5} \end{align*}

Write the last reply

The gradient of \(EF=-\frac{four}{5}\)

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Worked example 4: Gradient betwixt two points

Given \(G\left(7;-ix\right)\) and \(H\left(x;0\right)\), with \({m}_{GH}=3\), find \(ten\).

Depict a sketch

Assign values to \(\left({ten}_{1};{y}_{one}\correct)\) and \(\left({10}_{2};{y}_{2}\right)\)

Let the coordinates of \(G\) be \(\left({10}_{1};{y}_{one}\correct)\) and the coordinates of \(H\) be \(\left({x}_{2};{y}_{2}\right)\)

\[{10}_{i} = 7 \quad {y}_{1} = -9 \quad {ten}_{2} = x \quad {y}_{ii} = 0\]

Write down the formula for gradient

\[m=\frac{{y}_{ii}-{y}_{i}}{{x}_{2}-{x}_{1}}\]

Substitute values and solve for x

\begin{align*} 3 & = \frac{0 - \left(-nine\correct)}{x - seven} \\ 3\left(10 - 7\right)& = 9 \\ x - seven & = \frac{9}{3} \\ x - vii & = 3 \\ x & = three + 7 \\ & = 10 \finish{align*}

Write the terminal answer

The coordinates of \(H\) are \(\left(10;0\right)\).

Therefore \(x = x\).

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Textbook Exercise 8.iii

\(A(vii;x)\) and \(B(-four;1)\)

Let the coordinates of \(A\) be \(\left({10}_{1};{y}_{1}\right)\) and the coordinates of \(B\) be \(\left({x}_{2};{y}_{2}\right)\)

\[{10}_{ane} = 7 \quad {y}_{1} = 10 \quad {ten}_{2} = -4 \quad {y}_{2} = 1\] \brainstorm{align*} m_{AB} &= \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{ane - 10}{-4 - 7}\\ & = \frac{nine}{11} \end{align*}

\(A(-five;-9)\) and \(B(3;two)\)

Let the coordinates of \(A\) be \(\left({x}_{1};{y}_{1}\right)\) and the coordinates of \(B\) exist \(\left({x}_{2};{y}_{ii}\right)\)

\[{x}_{one} = -five \quad {y}_{1} = -9 \quad {x}_{two} = iii \quad {y}_{2} = ii\] \begin{align*} m_{AB} & = \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{2 - (-9)}{3 - (-5)}\\ & = \frac{xi}{8} \stop{marshal*}

\(A(x-three;y)\) and \(B(x;y+4)\)

Let the coordinates of \(A\) be \(\left({x}_{one};{y}_{i}\correct)\) and the coordinates of \(B\) be \(\left({ten}_{2};{y}_{2}\right)\)

\[{x}_{one} = ten - 3 \quad {y}_{1} = y \quad {ten}_{2} = x \quad {y}_{2} = y + 4\] \brainstorm{align*} m_{AB} &= \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{y + 4 - y}{ten - (x - 3)}\\ & = \frac{4}{three} \terminate{align*}

You are given the post-obit diagram:

Calculate the gradient (\(chiliad\)) of line \(AB\).

Let the coordinates of \(A\) exist \(\left({x}_{one};{y}_{1}\correct)\) and the coordinates of \(B\) be \(\left({x}_{2};{y}_{2}\right)\)

\[{x}_{1} = -1 \quad {y}_{1} = 2 \quad {x}_{ii} = ii \quad {y}_{two} = \text{0,5}\] \brainstorm{align*} 1000 & = \frac{y_B - y_A}{x_B - x_A} \\ & = \frac{(\text{0,5}) - (\text{two})}{(\text{2}) - (-\text{1})} \\ & = \frac{-\text{i,5}}{3} \\ & = -\text{0,5} \stop{align*}

Therefore the gradient \(thou\) of the line \(AB\) is \(-\text{0,v}\).

You are given the following diagram:

Calculate the gradient (\(m\)) of line \(AB\).

Allow the coordinates of \(A\) be \(\left({10}_{1};{y}_{ane}\right)\) and the coordinates of \(B\) be \(\left({ten}_{two};{y}_{2}\correct)\)

\[{10}_{ane} = -2 \quad {y}_{one} = -\text{ane,five} \quad {x}_{two} = ane \quad {y}_{2} = 3\] \begin{align*} thousand & = \frac{y_B - y_A}{x_B - x_A} \\ & = \frac{(\text{3}) - (-\text{1,5})}{(\text{ane}) - (-\text{2})} \\ & = \frac{\text{4,5}}{\text{three}} \\ & = \text{1,5} \finish{align*}

Therefore the slope \(k\) of the line \(AB\) is \(\text{i,five}\).

\(C(xvi;ii)\) and \(D(8;p)\).

Let the coordinates of \(C\) exist \(\left({x}_{ane};{y}_{ane}\right)\) and the coordinates of \(D\) be \(\left({x}_{2};{y}_{2}\correct)\)

\[{10}_{1} = sixteen \quad {y}_{1} = two \quad {x}_{2} = 8 \quad {y}_{2} = p\] \begin{align*} m_{CD} &= \frac{y_2 - y_1}{x_2 - x_1} \\ \frac{2}{three} &= \frac{p - 2}{8 - xvi} \\ \frac{2}{3} \times (-8)&= p - ii \\ \frac{-16}{iii} + 2 & = p \\ \frac{-16 + vi}{iii} &= p \\ \frac{-x}{iii} &= p \cease{align*}

\(C(iii;2p)\) and \(D(9;xiv)\).

Allow the coordinates of \(C\) be \(\left({x}_{1};{y}_{1}\right)\) and the coordinates of \(D\) be \(\left({x}_{2};{y}_{2}\right)\)

\[{x}_{i} = 3 \quad {y}_{1} = 2p \quad {10}_{2} = 9 \quad {y}_{2} = 14\] \begin{align*} m_{CD} &= \frac{y_2 - y_1}{x_2 - x_1}\\ \frac{2}{three} &= \frac{14 - 2p}{ix - 3}\\ \frac{2}{3} \times (vi)&= 14 - 2p\\ 4 & = 14 - 2p \\ 2p & = 14 - 4 \\ p & = \frac{ten}{2} \\ & = five \finish{align*}

You are given the post-obit diagram:

You are also told that line \(AB\) has a slope (\(m\)) of \(\text{two}\).

Calculate the missing co-ordinate of point \(B\).

Let the coordinates of \(A\) be \(\left({x}_{1};{y}_{ane}\right)\) and the coordinates of \(B\) exist \(\left({ten}_{two};{y}_{two}\right)\)

\[{x}_{one} = -i \quad {y}_{1} = 0 \quad {10}_{2} = one \quad {y}_{ii} = y\] \begin{align*} m_{AB} &= \frac{y_2 - y_1}{x_2 - x_1}\\ 2 &= \frac{y - 0}{1 - (-1)}\\ 2 & = \frac{y}{ii} \\ 4 &= y \stop{align*}

You are given the following diagram:

You are also told that line \(AB\) has a gradient (\(m\)) of \(-\text{1,v}\).

Calculate the missing co-ordinate of point \(B\).

Let the coordinates of \(A\) be \(\left({x}_{ane};{y}_{1}\right)\) and the coordinates of \(B\) exist \(\left({x}_{two};{y}_{two}\right)\)

\[{x}_{one} = -one \quad {y}_{one} = 3 \quad {x}_{2} = ten \quad {y}_{ii} = -3\] \begin{align*} m_{AB} &= \frac{y_2 - y_1}{x_2 - x_1}\\ -\text{1,5} &= \frac{-3 - three}{ten - (-i)}\\ -\text{one,5} & = \frac{-half-dozen}{10 + 1} \\ -\text{one,5}(10 + one) & = -6 \\ -\text{1,5}x - \text{1,5} & = -six \\ \text{ane,5}x & = \text{4,5} \\ x &= 3 \end{align*}

Straight lines (EMA6C)

Straight line

A directly line is a set up of points with a constant gradient between whatsoever 2 of the points.

Consider the diagram below with points \(A\left(x;y\right)\), \(B\left({ten}_{ii};{y}_{two}\right)\) and \(C\left({x}_{1};{y}_{1}\right)\).

We have \({m}_{AB} = {grand}_{BC} = {thou}_{Air-conditioning}\) and \(one thousand = \dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{one}} = \dfrac{{y}_{1} - {y}_{2}}{{x}_{1} - {10}_{two}}\)

The general formula for finding the equation of a directly line is \(\dfrac{y - {y}_{i}}{x - {10}_{1}} = \dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{i}}\) where \(\left(x;y\right)\) is any point on the line.

This formula can also be written as \(y - {y}_{one} = one thousand\left(x - {10}_{1}\right)\).

The standard grade of the direct line equation is \(y = mx + c\) where \(thou\) is the gradient and \(c\) is the \(y\)-intercept.

The worked example below shows ane method of determining the equation of a straight line. The equation of a direct line tin can as well be determined by start finding the gradient and then substituting 1 of the given points into \(y = mx + c\).

Worked instance 5: Finding the equation of a direct line

Detect the equation of the straight line through \(P\left(-1;-5\right)\) and \(Q\left(5;4\right)\).

Draw a sketch

Assign values

Let the coordinates of \(P\) exist \(\left({ten}_{one};{y}_{1}\right)\) and the coordinates of \(Q\) be \(\left({x}_{2};{y}_{two}\right)\).

\[{x}_{1} = -i \quad {y}_{1} = -5 \quad {x}_{2} = 5 \quad {y}_{ii} = 4\]

Write down the full general formula of the line

\[\frac{y - {y}_{i}}{x - {x}_{one}} = \frac{{y}_{2} - {y}_{1}}{{ten}_{two} - {x}_{i}}\]

Substitute values and make \(y\) the subject area of the equation

\brainstorm{marshal*} \frac{y - \left(-5\right)}{x - \left(-1\right)} & = \frac{4 - \left(-v\right)}{five - \left(-1\right)} \\ \frac{y + 5}{x + 1} & = \frac{3}{2} \\ 2\left(y + five\right) & = iii\left(x + ane\right) \\ 2y + 10 & = 3x + iii \\ 2y & = 3x - vii \\ y & = \frac{three}{2}x - \frac{7}{2} \stop{marshal*}

Write the terminal respond

The equation of the straight line is \(y=\frac{3}{ii}x-\frac{7}{2}\).

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Parallel and perpendicular lines (EMA6D)

Two lines that run parallel to each other are ever the same altitude apart and take equal gradients.

If two lines intersect perpendicularly, then the product of their gradients is equal to \(-\text{1}\).

If line \(WX \perp YZ\), then \({m}_{WX}\times {m}_{YZ} = -i\). Perpendicular lines have gradients that are the negative inverses of each other.

The following video shows some examples of calculating the slope of a line and determining if 2 lines are perpendicular or parallel.

Video: 2GCJ

Worked instance vi: Parallel lines

Prove that line \(AB\) with \(A\left(0;two\right)\) and \(B\left(2;6\right)\) is parallel to line \(CD\) with equation \(2x-y=2\).

Depict a sketch

(Be careful - some lines may look parallel just are non!)

Write down the formula for gradient

\[m = \frac{{y}_{2} - {y}_{1}}{{x}_{two} - {x}_{1}}\]

Substitute values to find the gradient for line \(AB\)

\begin{align*} {grand}_{AB} & = \frac{vi - two}{ii - 0} \\ & = \frac{four}{two} \\ & = 2 \cease{align*}

Check that the equation of \(CD\) is in the standard form \(y = mx + c\)

\brainstorm{align*} 2x - y & = 2 \\ y & = 2x - 2 \\ \therefore {m}_{CD} & = 2 \cease{align*}

Write the final answer

\[{m}_{AB} = {g}_{CD}\]

therefore line \(AB\) is parallel to line \(CD\).

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Worked example 7: Perpendicular lines

Line \(AB\) is perpendicular to line \(CD\). Find \(y\) given \(A\left(2;-3\right)\), \(B\left(-2;half dozen\right)\), \(C\left(4;iii\right)\) and \(D\left(7;y\right)\).

Draw a sketch

Write downwards the human relationship betwixt the gradients of the perpendicular lines \(AB\perp CD\)

\brainstorm{align*} {m}_{AB} \times {m}_{CD} & = -1 \\ \frac{{y}_{B} - {y}_{A}}{{x}_{B} - {x}_{A}} \times \frac{{y}_{D} - {y}_{C}}{{x}_{D} - {x}_{C}} & = -1 \stop{align*}

Substitute values and solve for \(y\)

\begin{align*} \frac{half dozen - \left(-3\right)}{-ii - 2}\times \frac{y - iii}{7 - iv}& = -1 \\ \frac{9}{-4} \times \frac{y - 3}{3} & = -1 \\ \frac{y - 3}{3} & = -ane \times \frac{-4}{9} \\ \frac{y - iii}{three} & = \frac{4}{9} \\ y - three & = \frac{four}{9} \times 3 \\ y - 3 & = \frac{iv}{three} \\ y & = \frac{4}{3} + iii \\ & = \frac{4 + 9}{3} \\ & = \frac{13}{3} \\ & = 4\frac{i}{three} \end{align*}

Write the last answer

Therefore the coordinates of \(D\) are \(\left(7;4\frac{1}{3}\right)\).

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Horizontal and vertical lines (EMA6F)

A line that runs parallel to the \(x\)-axis is called a horizontal line and has a gradient of nix. This is considering in that location is no vertical change:

\[m = \frac{\text{alter in } y}{\text{change in } 10} = \frac{0}{\text{modify in } x} = 0\]

A line that runs parallel to the \(y\)-axis is chosen a vertical line and its gradient is undefined. This is because there is no horizontal modify:

\[m = \frac{\text{modify in } y}{\text{change in } x } = \frac{\text{change in } y}{0} = \text{ undefined}\]

Points on a line (EMA6G)

A directly line is a set of points with a constant gradient between whatever of the two points. There are two methods to show that points lie on the same line: the gradient method and a method using the distance formula.

If two points lie on the same line then the two points are said to be collinear.

Worked example 8: Points on a line

Prove that \(A\left(-3;3\correct)\), \(B\left(0;5\right)\) and \(C\left(three;seven\right)\) are on a direct line.

Describe a sketch

Summate two gradients between any of the three points

\brainstorm{marshal*} m & = \frac{{y}_{2} - {y}_{one}}{{ten}_{ii} - {x}_{i}} \\ {g}_{AB} & = \frac{five - 3}{0 - \left(-three\correct)} = \frac{two}{iii} \stop{align*}

and

\[{thou}_{BC} = \frac{7 - 5}{3 - 0} = \frac{2}{3}\]

OR

\[{m}_{AC} = \frac{3-7}{3-three} = \frac{-4}{-6} = \frac{2}{3}\]

and

\[{g}_{BC} = \frac{7-v}{3-0} = \frac{2}{3}\]

Explicate your answer

\[{m}_{AB} ={1000}_{BC} ={grand}_{Air conditioning}\]

Therefore the points \(A\), \(B\) and \(C\) are on a straight line.

To evidence that three points are on a direct line using the altitude formula, we must calculate the distances between each pair of points and then prove that the sum of the two smaller distances equals the longest distance.

Worked example 9: Points on a straight line

Prove that \(A\left(-3;3\correct)\), \(B\left(0;5\right)\) and \(C\left(3;7\right)\) are on a straight line.

Draw a sketch

Calculate the three distances \(AB\), \(BC\) and \(Air conditioning\)

\begin{align*} {d}_{AB} & = \sqrt{{\left(-3 - 0\right)}^{two} + {\left(3-5\right)}^{2}} \\ & = \sqrt{{\left(-3\right)}^{2} + {\left(-2\right)}^{ii}} \\ & = \sqrt{9 + 4} \\ & = \sqrt{13} \\ \\ {d}_{BC} & = \sqrt{{\left(0 - three\right)}^{two} + {\left(5 - seven\right)}^{2}} \\ & = \sqrt{{\left(-iii\right)}^{ii} + {\left(-2\correct)}^{2}} \\ & = \sqrt{9 + 4} \\ & = \sqrt{thirteen} \\ \\ {d}_{Air-conditioning} & = \sqrt{{\left(\left(-3\right) - 3\right)}^{ii} + {\left(3 - 7\correct)}^{2}} \\ & = \sqrt{{\left(-half dozen\right)}^{2} + {\left(-4\right)}^{2}} \\ & = \sqrt{36+xvi} \\ & = \sqrt{52} \end{align*}

Find the sum of the two shorter distances

\[{d}_{AB} + {d}_{BC} = \sqrt{thirteen} + \sqrt{xiii} = ii\sqrt{13} = \sqrt{4\times 13} = \sqrt{52}\]

Explain your reply

\[{d}_{AB} + {d}_{BC} = {d}_{AC}\]

therefore points \(A\), \(B\) and \(C\) lie on the same straight line.

Textbook Practice 8.4

\(A(3;-4)\), \(B(v;2)\), \(C(-1;-one)\), \(D(7;23)\)

Nosotros demand to calculate the gradients of lines \(AB\) and \(CD\). Then we can compare the gradients to determine if the lines are parallel (the gradients are equal), perpendicular (the gradients are the negative inverses of each other) or neither.

\begin{marshal*} m_{AB} &= \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{two-(-four)}{5-iii}\\ & = \frac{six}{2}\\ & = 3 \end{align*}

And:

\brainstorm{align*} m_{CD} &= \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{23-(-one)}{seven-(-1)}\\ & = \frac{24}{8}\\ & = three \end{align*}

Therefore:

\brainstorm{marshal*} m_{AB} & = m_{CD}\\ \therefore AB & \parallel CD \end{align*}

Lines \(AB\) and \(CD\) are parallel.

\(A(3;-four)\), \(B(five;2)\), \(C(-1;-1)\), \(D(0;-4)\)

We need to calculate the gradients of lines \(AB\) and \(CD\). Then we tin can compare the gradients to decide if the lines are parallel (the gradients are equal), perpendicular (the gradients are the negative inverses of each other) or neither.

\begin{align*} m_{AB} & = \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{2 - (-4)}{5 - three}\\ & = \frac{6}{2}\\ & = 3 \end{align*}

And:

\begin{align*} m_{CD} & = \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{-iv-(-1)}{0-(-ane)}\\ & = \frac{-3}{ane}\\ & = -three \end{align*}

So \(m_{AB}\ne m_{CD}\). Therefore \(AB\) is non parallel to \(CD\).

And \(m_{AB}\times\frac{1}{m_{CD}} \ne -one\). Therefore \(AB\) and \(CD\) are not perpendicular.

Lines \(AB\) and \(CD\) are neither parallel nor perpendicular.

\(A(3;-iv)\), \(B(five;2)\), \(C(-i;3)\), \(D(-2;2)\)

Nosotros need to calculate the gradients of lines \(AB\) and \(CD\). Then we can compare the gradients to determine if the lines are parallel (the gradients are equal), perpendicular (the gradients are the negative inverses of each other) or neither.

\brainstorm{align*} m_{AB} & = \frac{y_2-y_1}{x_2-x_1}\\ & = \frac{2 - (-4)}{v - 3}\\ & = \frac{6}{2}\\ & = 3 \stop{align*}

And:

\brainstorm{align*} m_{CD} &= \frac{y_2-y_1}{x_2-x_1}\\ &= \frac{2 - 3}{one - (-2)}\\ &= \frac{-1}{3} \end{align*}

So \(m_{AB}\ne m_{CD}\). Therefore \(AB\) is not parallel to \(CD\).

And \(m_{AB} \times \frac{1}{m_{CD}} = -1\). Therefore \(AB\) and \(CD\) are perpendicular.

\(E(0;three)\), \(F(-ii;5)\), \(Thou(ii;1)\)

\brainstorm{marshal*} m_{EF} &= \frac{y_2-y_1}{x_2-x_1}\\ &= \frac{5-three}{-2-0}\\ &= \frac{two}{-2}\\ &= -1 \end{align*}

And,

\brainstorm{align*} m_{FG} &=\frac{y_2-y_1}{x_2-x_1}\\ &= \frac{ane-5}{2-(-2)}\\ &= \frac{-4}{iv}\\ &= -ane \finish{align*}

And then \(m_{EF} = m_{FG}\) and \(F\) is a mutual signal. Therefore \(E\), \(F\) and \(G\) are collinear (they lie on the same line).

\(H(-3;-v)\), \(I(0;0)\), \(J(6;x)\)

\begin{align*} m_{HI} &= \frac{y_2-y_1}{x_2-x_1}\\ &= \frac{0-(-v)}{0-(-iii)}\\ &= \frac{five}{three} \terminate{align*}

And,

\begin{marshal*} m_{IJ} &= \frac{y_2-y_1}{x_2-x_1}\\ &= \frac{10-0}{6-0}\\ &= \frac{ten}{half dozen}\\ &= \frac{5}{three} \end{marshal*}

And so \(m_{Hello} = m_{IJ}\) and \(I\) is a common point. Therefore \(H\), \(I\) and \(J\) are collinear (they lie on the same line).

\(M(-6;2)\), \(L(-3;1)\), \(1000(1;-ane)\)

\begin{align*} m_{KL} &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{1 - ii)}{-3 - (-6)}\\ &= -\frac{1}{iii} \cease{align*}

And,

\begin{align*} m_{LM} &= \frac{y_2-y_1}{x_2-x_1}\\ &= \frac{-1-ane}{one-(-iii)}\\ &= \frac{-two}{four}\\ &= -\frac{1}{2} \cease{align*}

Then \(m_{KL} \ne m_{LM}\). Therefore \(Yard\), \(L\) and \(M\) are not collinear (they do not lie on the same line).

You lot are given the following diagram:

Calculate the equation of the line \(AB\).

To summate the equation of the straight line, we first summate the gradient (\(m\)) of the line \(AB\):

\(\begin{aligned} m & = \frac{y_B - y_A}{x_B - x_A} \\ chiliad & = \frac{(-\text{iii,v}) - (\text{2,v})}{(\text{two}) - (-\text{1})} \\ m & = -\text{two} \\ \cease{aligned}\)

Secondly, we calculate the value of the \(y\)-intercept (\(c\)) of the line \(AB\). We do this past substituting either point \(A\) or \(B\) into the general form for a directly line. Nosotros will use indicate \(A\).

\(\begin{aligned} y & = mx + c \\ (\text{2,5}) & = (-\text{ii}) \times (-\text{one}) + c \\ c & = \text{0,five} \\ \end{aligned}\)

Therefore, the equation of the line \(AB\) is as follows:

\(\begin{aligned} y & = -\text{ii} x + \text{0,5} \end{aligned}\)

You are given the following diagram:

Calculate the equation of the line \(AB\).

To calculate the equation of the straight line, nosotros kickoff calculate the gradient (\(yard\)) of the line \(AB\):

\(\begin{aligned} g & = \frac{y_B - y_A}{x_B - x_A} \\ m & = \frac{(\text{1,5}) - (-\text{ane})}{(\text{4}) - (-\text{1})} \\ 1000 & = \text{0,5} \\ \end{aligned}\)

Secondly, we calculate the value of the \(y\)-intercept (\(c\)) of the line \(AB\). We do this by substituting either point \(A\) or \(B\) into the general form for a direct line. Nosotros will use point \(A\).

\(\begin{aligned} y & = mx + c \\ (-\text{ane}) & = (\text{0,5}) \times (-\text{ane}) + c \\ c & = -\text{0,5} \\ \end{aligned}\)

Therefore, the equation of the line \(AB\) is every bit follows:

\(\begin{aligned} y & = \text{0,5} x -\text{0,five} \end{aligned}\)

Points \(P(-vi;2)\), \(Q(2;-ii)\) and \(R(-3;r)\) lie on a directly line. Find the value of \(r\).

Since the three points lie on a straight line we can use the fact that the slope of \(PQ\) is equal to the gradient of \(QR\) to find \(r\).

The gradient of \(PQ\) is:

\begin{align*} m_{PQ} & = \frac{y_2 - y_1}{x_2 - x_1} \\ & = \frac{-2 - 2}{2 - (-6)} \\ & = \frac{-4}{eight} \\ & = -\frac{1}{2} \end{align*}

And the gradient of \(QR\) in terms of \(r\) is:

\begin{align*} m_{QR} & = \frac{y_2 - y_1}{x_2 - x_1} \\ & = \frac{r - (-2)}{-iii - ii} \\ & = \frac{r + 2}{-v} \end{align*}

Now we permit \(m_{PR} = m_{QR} = -\frac{1}{2}\) and solve for \(r\):

\begin{align*} -\frac{i}{2} & = \frac{r + 2}{-v} \\ (-1)\times(-five) & = ii(r + 2) \\ 5 & = 2r + iv \\ 5 - 4 & = 2r \\ 1 & = 2r \\ r & = \frac{i}{ii} \end{align*}

Line \(PQ\) with \(P(-1;-seven)\) and \(Q(q;0)\) has a gradient of \(\text{1}\). Detect \(q\).

\begin{marshal*} m_{PQ} & = \frac{y_2 - y_1}{x_2 - x_1} \\ 1 & = \frac{0 - (-7)}{q - (-i)} \\ q + 1 & = 7 \\ q & = half dozen \stop{marshal*}

Yous are given the following diagram:

You are as well told that line \(AB\) runs parallel to the following line: \(y = x - \text{five}\). Point \(A\) is at \(\left(-2 ; -four \right)\).

Observe the equation of the line \(AB\).

We are told that line \(AB\) is parallel to \(y = x - 5\), so the gradient of line \(AB\) is equal to the gradient of \(y = x - 5\). The gradient of \(y = ten - 5\) is 1.

Now we tin employ indicate \(A\) and the gradient of the line to detect the \(y\)-intercept of the line:

\(\begin{aligned} y & = mx + c \\ (-\text{four}) & = (\text{one})(-\text{2}) + c \\ c & = -\text{2} \\ \end{aligned}\)

Therefore, the equation of the line \(AB\) is: \(y = x - 2\).

You are given the following diagram:

You are also told that line \(AB\) runs parallel to the following line: \(y = -x + \text{4,5}\). Indicate \(A\) is at \(\left(-1 ; \text{2,5}\right)\).

Find the equation of the line \(AB\).

Nosotros are told that line \(AB\) is parallel to \(y = -x + \text{four,5}\), so the slope of line \(AB\) is equal to the gradient of \(y = -x + \text{4,5}\). The slope of \(y = -ten + \text{4,5}\) is \(-\text{1}\).

Now we tin employ bespeak \(A\) and the gradient of the line to find the \(y\)-intercept of the line:

\(\begin{aligned} y & = mx + c \\ (\text{2,5}) & = (-\text{1}) \times (-\text{1}) + c \\ c & = \text{ane,five} \\ \end{aligned}\)

Therefore, the equation of the line \(AB\) is: \(y = -x + \text{ane,five}\).

Given line \(AB\) which runs parallel to \(y = \text{0,5}x - \text{6}\). Points \(A(-ane;-\text{2,five})\) and \(B(x;0)\) are also given.

Summate the missing co-ordinate of point \(B\).

We are told that line \(AB\) is parallel to \(y = \text{0,5}10 - six\), so the gradient of line \(AB\) is equal to the slope of \(y = \text{0,5}ten - 6\). The gradient of \(y = \text{0,5}ten - 6\) is \(\text{0,5}\).

Now we can utilise indicate \(A\) and the slope of the line to observe the \(y\)-intercept of the line:

\brainstorm{align*} y & = mx + c \\ (-\text{2,5}) & = (\text{0,5})(-\text{1}) + c \\ c & = -\text{2} \\ \end{align*}

Therefore, the equation of the line \(AB\) is: \(y = \text{0,five}10 - 2\).

Now nosotros can substitute point \(B\) into the equation of line \(AB\) to solve for \(x\):

\begin{align*} y & = \text{0,5}ten - two \\ 0 & = \text{0,5}ten - 2 \\ 4 & = 10 \finish{align*}

Therefore the coordinates of point \(B\) are \((0;4)\).

Given line \(AB\) which runs parallel to \(y = -\text{1,v} ten + \text{4}\). Points \(A(-\text{2};\text{iv})\) and \(B(2;y)\) are also given.

Calculate the missing co-ordinate of signal \(B(2;y)\).

We are told that line \(AB\) is parallel to \(y = -\text{1,v}x + 4\), so the gradient of line \(AB\) is equal to the gradient of \(y = -\text{1,5}ten + 4\). The slope of \(y = -\text{one,5}x + 4\) is \(-\text{1,5}\).

Now we can employ point \(A\) and the gradient of the line to find the \(y\)-intercept of the line:

\begin{align*} y & = mx + c \\ (\text{4}) & = (-\text{1,5})(-\text{2}) + c \\ c & = \text{1} \\ \finish{align*}

Therefore, the equation of the line \(AB\) is: \(y = -\text{i,5}x + ane\).

At present we tin substitute point \(B\) into the equation of line \(AB\) to solve for \(x\):

\begin{marshal*} y & = -\text{i,5}x + 1 \\ y & = -\text{1,5}(2) + one \\ y & = -ii \end{align*}

Therefore the coordinates of signal \(B\) are \((2;-2)\).

The graph here shows line \(AB\). The blue dashed line is perpendicular to \(AB\).

The equation of the blue dashed line is \(y = x + i\). Indicate \(A\) is at \((-two; four)\).

Make up one's mind the equation of line \(AB\).

The general form of a directly line is: \(y = mx + c\).

Line \(AB\) is perpendicular to the blue dashed line so \(m_{AB} = \dfrac{-ane}{m_{\text{blue line}}}\).

\begin{marshal*} y & = mx + c \\ y & = \left( \frac{-1}{m_{\text{blue line}}} \right) x + c \\ y & = \left( \frac{-1}{\text{1}} \right) x + c \\ y & = -x + c \end{align*}

Now we can substitute in the coordinates of point \(A\) to observe the \(y\)-intercept:

\begin{align*} y & = mx + c \\ (\text{iv}) & = (-\text{i}) (-\text{2}) + c \\ c & = \text{2} \\ \cease{align*}

Therefore, the equation of the line \(AB\) is: \(y = -x + two\).

The graph hither shows line \(AB\). The blue dashed line is perpendicular to \(AB\).

The equation of the bluish dashed line is \(y = -\text{0,v} 10 - \text{0,5}\). Indicate \(A\) is at \((-\text{one}; -\text{3,5})\).

Determine the equation of line \(AB\).

The general form of a directly line is: \(y = mx + c\).

Line \(AB\) is perpendicular to the blue dashed line and then \(m_{AB} = \dfrac{-i}{m_{\text{blue line}}}\).

\begin{marshal*} y & = mx + c \\ y & = \left( \frac{-one}{m_{\text{bluish line}}} \right) x + c \\ y & = \left( \frac{-1}{-\text{0,5}} \right) x + c \\ y & = 2x + c \end{align*}

At present we tin can substitute in the coordinates of bespeak \(A\) to find the \(y\)-intercept:

\brainstorm{align*} y & = mx + c \\ (-\text{three,5}) & = (\text{2}) (-\text{1}) + c \\ c & = -\text{1,5} \\ \end{marshal*}

Therefore, the equation of the line \(AB\) is: \(y = 2x - \text{i,5}\).

Given line \(AB\) which runs perpendicular to line \(CD\) with equation \(y = -2x + 1\). Points \(A(-5;-1)\) and \(B(3;a)\) are as well given.

Calculate the missing co-ordinate of point \(B\).

The general form of a straight line is: \(y = mx + c\).

Line \(AB\) is perpendicular to line \(CD\) and so \(m_{AB} = \dfrac{-i}{m_{CD}}\).

\begin{align*} y & = mx + c \\ y & = \left(\frac{-i}{m_{CD}}\right)x + c \\ y & = \left(\frac{-1}{-\text{two}}\right)x + c \\ y & = \text{0,five} x + c \finish{align*}

Now nosotros can substitute point \(A\) into the equation to find the \(y\)-intercept:

\begin{align*} y & = \text{0,5}x + c \\ -\text{1} & = (\text{0,5})(-\text{v}) + c \\ c & = \text{1,v} \cease{align*}

Next we can substitute in point \(B\) to find the missing coordinate:

\brainstorm{align*} y & = \text{0,5}10 + \text{1,5} \\ a & = (\text{0,5})(\text{iii}) + \text{one,5} \\ & = \text{iii} \end{marshal*}

Therefore the missing coordinate is \(B(3;3)\).

Given line \(AB\) which runs perpendicular to line \(CD\) with equation \(y = 2x - \text{0,75}\). Points \(A(-5;one)\) and \(B(a;-\text{2,5})\) are also given.

Calculate the missing according of point \(B\).

The general grade of a straight line is: \(y = mx + c\).

Line \(AB\) is perpendicular to line \(CD\) and so \(m_{AB} = \dfrac{-1}{m_{CD}}\).

\begin{align*} y & = mx + c \\ y & = \left(\frac{-1}{m_{CD}}\right)x + c \\ y & = \left(\frac{-1}{2}\right)ten + c \\ y & = -\text{0,5}x + c \finish{align*}

Now we can substitute point \(A\) into the equation to observe the \(y\)-intercept:

\brainstorm{align*} y & = -\text{0,5}x + c \\ \text{ane} & = (\text{0,5})(-\text{5}) + c \\ c & = -\text{1,five} \end{align*}

Adjacent nosotros tin substitute in point \(B\) to find the missing coordinate:

\brainstorm{align*} y & = \text{0,5}x - \text{1,5} \\ -\text{2,5} & = \text{0,v}a - \text{i,5} \\ a & = -\text{two} \end{align*}

Therefore the missing coordinate is \(B(-2;-\text{2,5})\).

You are given the following diagram:

Yous are also told that line \(AB\) has the following equation: \(y = -\text{0,v} x + \text{1,5}\).

Calculate the missing according of point \(B\).

Nosotros substitute the known value for point \(B\) into the equation and solve for the unknown value:

\(\begin{aligned} y & = (-\text{0,5})x + \text{1,5} \\ a & = (-\text{0,v})(\text{2}) + \text{1,5} \\ & = \text{0,5} \end{aligned}\)

You are given the following diagram:

You are also told that line \(AB\) has the following equation: \(y = \text{0,v}ten - \text{1}\).

Calculate the missing coordinate of point \(B\).

Nosotros substitute the known value for point \(B\) into the equation and solve for the unknown value:

\(\begin{aligned} y & = (\text{0,5})x - \text{1} \\ a & = (\text{0,5})(\text{1}) - \text{1} \\ & = -\text{0,5} \cease{aligned}\)

\(A\) is the point \((-3;-5)\) and \(B\) is the signal \((n;-11)\). \(AB\) is perpendicular to line \(CD\) with equation \(y = \frac{3}{ii}x - five\). Notice the value of \(north\).

Line \(AB\) is perpendicular to line \(CD\) and and then \(m_{AB} = \dfrac{-ane}{m_{CD}}\).

\begin{marshal*} m_{AB} & = -1 \div \frac{iii}{2} \\ & = \frac{-2}{3} \cease{marshal*}

Therefore:

\begin{align*} \frac{-2}{3} & = \frac{-11 - (-5)}{northward - (-3)} \\ \frac{-2}{iii} & = \frac{-6}{n + 3} \\ \frac{-2}{3}(n + 3) & = -half dozen \\ -ii(n + three) & = -eighteen \\ northward + three & = ix \\ northward & = 6 \end{align*}

The points \(A(four;-iii)\), \(B(-5;0)\) and \(C(-iii;p)\) are given. Make up one's mind the value of \(p\) if \(A\), \(B\) and \(C\) are collinear.

We are told that \(A\), \(B\) and \(C\) are collinear, therefore \(m_{AB} = m_{BC}\).

\begin{align*} \frac{0 + iii}{-5 - iv} & = \frac{p}{-3 + 5} \\ \frac{3}{-9} & = \frac{p}{ii} \\ \therefore p & = \frac{6}{-9} \\ & = \frac{-2}{iii} \end{align*}

Testify that \(\triangle ABC\) is right-angled. Show your working.

To show that \(\triangle ABC\) is right angled nosotros need to show that \(AB \perp AC\) or that \(AC \perp BC\) or \(AB \perp BC\). We tin exercise this by calculating the gradients of \(AB\), \(Air-conditioning\) and \(BC\) and and then seeing if any of these gradients is the negative inverse of any of the other two gradients.

\begin{align*} m_{AB} & = \frac{-4 - two}{ii - (-v)} \\ & = \frac{-vi}{7} \\\\ m_{BC} & = \frac{-4 - 3}{2 - 8} \\ & = \frac{-vii}{-vi} \\ & = \frac{7}{half dozen} \\\\ m_{Air conditioning} & = \frac{three - 2}{8 - (-5)} \\ & = \frac{one}{13} \cease{marshal*}

Now we annotation that:

\begin{marshal*} m_{AB} \times m_{BC} & = \frac{-6}{vii} \times \frac{vii}{6} = -1 \\ \therefore AB \perp BC \\ \therefore \triangle ABC \text{ is right-angled} \finish{align*}

Find the area of \(\triangle ABC\).

This is a right-angled triangle (with right-angle \(A\hat{B}C\)) and and so the perpendicular height is the length of one of the sides. We will use side \(BC\) as the perpendicular height and side \(AB\) as the base.

Note that nosotros cannot use side \(AC\) for either the base or the height equally we would so need to construct a new perpendicular peak from \(A\) to \(B\) and summate the length of that line.

The length of \(BC\) is:

\brainstorm{align*} d & = \sqrt{(-5-2)^{ii} + (2+4)^{two}} \\ & = \sqrt{49 + 36} \\ & = \sqrt{85} \finish{align*}

The length of \(AB\) is:

\begin{align*} d & = \sqrt{(8-2)^{2} + (3+4)^{ii}} \\ & = \sqrt{36 + 49} \\ & = \sqrt{85} \end{align*}

Therefore the surface area of \(\triangle ABC\) is:

\begin{align*} A & = \frac{1}{2}bh \\ & = \frac{1}{2}(AB)(BC) \\ & = \frac{1}{2}\left(\sqrt{85}\right)\left(\sqrt{85}\correct) \\ & = \frac{one}{2}\left(85\right) \\ & = \text{42,5} \end{align*}

Evidence that triangle \(ABC\) is a right-angled triangle.

Nosotros first draw a sketch:

To testify that \(\triangle ABC\) is right angled nosotros need to testify that \(AB \perp AC\) or that \(Ac \perp BC\) or \(AB \perp BC\). We can do this by calculating the gradients of \(AB\), \(AC\) and \(BC\) and then seeing if any of these gradients is the negative inverse of any of the other two gradients.

\begin{align*} m_{AB} & = \frac{-2 - ane}{3 - (-3)} \\ & = \frac{-3}{6} \\ & = \frac{-ane}{ii} \\\\ m_{BC} & = \frac{10 - (-2)}{9 - 3} \\ & = \frac{12}{six} \\ & = 2 \\\\ m_{Ac} & = \frac{10 - 1}{nine - (-3)} \\ & = \frac{nine}{12} \\ & = \frac{three}{4} \cease{align*}

At present we note that:

\begin{align*} m_{AB} \times m_{BC} & = \frac{-1}{2} \times two = -1 \\ \therefore AB \perp BC \\ \therefore \triangle ABC \text{ is right-angled} \end{marshal*}

Find the coordinates of \(D\), if \(ABCD\) is a parallelogram.

A parallelogram has both sides equal in length and parallel. Therefore \(CD \parallel AB\) and \(Advertizement \parallel BC\). As well \(CD = AB\) and \(Advertising = BC\).

Let the coordinates of \(D\) be \((x;y)\).

Since \(Advertisement \parallel BC\), \(m_{AD} = m_{BC}\).

From the previous question we know that \(m_{BC} = 2\). Therefore the gradient of \(Advertizing\) is:

\begin{align*} m_{AD} & = \frac{y_{2} - y_{one}}{x_{2} - x_{ane}} \\ two & = \frac{y - 1}{x - (-3)} \\ 2(x + 3) & = y - 1 \\ 2x + 6 & = y - 1 \\ 2x + 7 & = y \end{align*}

Since \(CD \parallel AB\), \(m_{CD} = m_{AB}\).

From the previous question we know that \(m_{AB} = \frac{-1}{2}\). Therefore the gradient of \(CD\) is:

\brainstorm{align*} m_{CD} & = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ \frac{-1}{ii} & = \frac{y - x}{x - nine} \\ \frac{-1}{2}(ten - ix) & = y - x \\ -10 + 9 & = 2y - 20 \\ -x + 29 & = 2y \end{align*}

At present we have ii equations with two unknowns. Nosotros can equate the ii equations and solve for \(x\):

\brainstorm{marshal*} 4x + xiv & = -x + 29 \\ 5x & = xv \\ x & = three \stop{align*}

At present we can solve for \(y\):

\begin{align*} 2x + 7 & = y \\ 2(3) + seven & = y \\ 13 & = y \cease{align*}

Therefore the coordinates of \(D\) are \((3;13)\).

Find the equation of a line parallel to the line \(BC\), which passes through the point \(A\).

We know from the offset question that the gradient of line \(BC\) is ii. Nosotros also know that bespeak \(A\) is at \((-iii;1)\).

The line parallel to \(BC\) will have the same gradient as \(BC\) so nosotros tin write the equation of this line as: \(y = 2x + c\).

Now nosotros tin can use point \(A\) to find the \(y\)-intercept of the line:

\brainstorm{marshal*} 2 & = 2(-5) + c \\ 2 & = -10 + c \\ 12 & = c \stop{align*}

Therefore the equation of the line parallel to \(BC\) and passing through \(A\) is \(y = 2x + 12\).

Source: https://www.siyavula.com/read/maths/grade-10/analytical-geometry/08-analytical-geometry-02

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