Powers with Negative Exponents: Definition, Properties and Examples

March 30, 202339 Insightful Publications

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While laying roads, one must know how steep the road will be. Similarly, when constructing a staircase, we should consider its steepness. For the same reason, anyone traveling along a hill or a bridge feels hard compared to traveling along a straight road.

All these examples illustrate one crucial aspect called steepness. The measure of steepness is called slope or gradient.

The concept of slope is important in economics because it measures the rate at which the demand for a product changes in a given period based on its price. Slope comprises of two factors, namely steepness and direction.

Note: The inclination of \(X\)-axis and every line parallel to \(X\)-axis is \({0^ \circ }.\)

The inclination of \(Y\)-axis and every line parallel to \(Y\)-axis is \({90^ \circ }.\)

Therefore, the slope of the straight line is \(m = \tan \,\tan \,\theta ,{0^ \circ } \leqslant \theta \leqslant {180^ \circ },\theta \ne {90^ \circ }\)

If one is walking up a hill, then the slope is positive.

If one is walking down a hill, then the slope is negative.

And if one is walking a flat line, the slope is zero because there is no inclination. (i.e., no vertical change).

Slope, \(m = \tan \theta \)

\(= \frac{{{\text{Opposite}}\,{\text{side}}}}{{{\text{Adjacent}}\,{\text{side}}}}\)

\( = \frac{{BC}}{{AC}}\)

We can also write it as

\(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

That is, slope, \(m = \frac{{{\text{Change}}\,{\text{in}}\,y\,{\text{coordinates}}}}{{{\text{Change}}\,{\text{in}}\,x\,{\text{coordinates}}}}\)

Therefore, by using the above formula, we can easily calculate the slope of a line between two points.

In another form, the slope of a line between two points is also said to be the rise of the line from one point to another (along \(Y\)-axis) over the run (along \(X\)–axis).

Therefore, slope, \(m = \frac{{{\text{Rise}}}}{{{\text{Run}}}}\)

The line is parallel to the positive direction of \(x\)-axis.

(ii) When \({0^ \circ } < \theta < {90^ \circ }\)

The line has a positive slope. A line with a positive slope rises from left to right.

(iii) When \({90^ \circ } < \theta < {180^ \circ }\)

The line has a negative slope. A line with a negative slope falls from left to right.

(iv) When \(\theta = {180^ \circ }\)

The line is parallel to the \(X\)-axis heading in the negative direction of \(X\)-axis.

(v) When \(\theta = {90^ \circ }\)

The slope is undefined as \(x\) coordinates are the same everywhere in the line.

Let \({l_1}\) and \({l_2}\) be two non-vertical lines with slopes \({m_1}\) and \({m_2}\) respectively.

Let the inclination of the lines with the positive direction of \(X\)-axis be \({\theta _1}\) and \({\theta _2}\) respectively.

Assume, \({l_1}\) and \({l_2}\) are parallel lines.

\({\theta _1} = {\theta _2},\) since, \({\theta _1},{\theta _2}\) are corresponding angles made by the transversal (\(X\) axis) to the parallel lines

\({\theta _1} = \tan \,\tan \,{\theta _2}\)

\( \Rightarrow {m_1} = {m_2}\)

Hence, the slopes are equal.

Therefore, non-vertical parallel lines have equal slopes.

We can find the converse also true.

Let the slopes be equal, then

\({m_1} = {m_2}\)

\( \Rightarrow \tan \,\tan \,{\theta _1} = \tan \,\tan \,{\theta _2}\)

\(\Rightarrow \,{\theta _1} = \,{\theta _2}\) (Since \(0 \leqslant {\theta _1} \leqslant {180^ \circ },0 \leqslant {\theta _2} \leqslant {180^ \circ }\)

That is, the corresponding angles are equal.

Therefore, the lines \({l_1}\) and \({l_2}\) are parallel.

Thus, non-vertical lines having equal slopes are parallel.

Hence, non-vertical lines are parallel if and only if their slopes are equal.

Let \({l_1}\) and \({l_2}\) be two non-vertical lines with slopes \({m_1}\) and \({m_2}\) respectively.

Let the inclination of the lines be \({\theta _1}\) and \({\theta _2}\) respectively.

Then \({m_1} = \tan \,\tan \,{\theta _1}\) and \({m_2} = \tan \,\tan \,{\theta _2}\)

Consider \(\Delta ABC.\)

The sum of angles of \(\Delta ABC\) is \({180^ \circ }.\)

Then, \(\angle ABC = {90^ \circ } – {\theta _1}\)

Now measuring the slope of \({l_2}\) through angles \({\theta _2}\) and \({90^ \circ } – {\theta _1},\) which are opposite to each other, we get

\(\tan \,\tan \,{\theta _2} = – \tan \,\tan \,\left({{{90}^ \circ } – {\theta _1}} \right) = \frac{{ – \sin \,\sin \,\left({{{90}^ \circ } – {\theta _1}} \right)}}{{\cos \,\cos \,\left({{{90}^ \circ } – {\theta _1}} \right)}}\)

\( = \frac{{ – \cos \cos {\theta _1}}}{{\sin \sin {\theta _1}}} = – \cot \cot {\theta _1}\)

\(\tan \tan {\theta _2} = – \frac{1}{{\tan \tan {\theta _1}}}\)

\(\operatorname{tantan} \tan {\theta _1}\tan {\theta _2} = – 1\)

\({m_1}{m_2} = – 1\)

Thus, when the line \({l_1}\) is perpendicular to the line \({l_2}\) then \({m_1}{m_2} = – 1\)

Conversely

Let \({l_1}\) and \({l_2}\) be two non-vertical lines with slopes \({m_1}\) and \({m_2}\) respectively, such that \({m_1}{m_2} = – 1.\)

\({m_1} = \tan \tan {\theta _1}\) and \({m_2} = \tan \tan {\theta _2}\)

We have, \(\tan \,tan\tan {\theta _1}\tan {\theta _2} = – 1\)

\(\tan \tan {\theta _1} = – \frac{1}{{\tan \tan {\theta _2}}}\)

\({\theta _2}\)

\(\tan \,\tan \,{\theta _1} = – \tan \,\tan \,\left({{{90}^ \circ } – {\theta _2}} \right)\)

\(\tan \tan {\theta _1} = \tan \tan \left({ – \left( {90^\circ – {\theta _2}} \right)} \right) = \tan \tan \left( {{\theta _2} – 90^\circ } \right)\,\,{\theta _1} = {\theta _2} – 90^\circ \) (Since \(0 \leqslant {\theta _1} \leqslant {180^ \circ },0 \leqslant {\theta _2} \leqslant {180^ \circ }\))

\({\theta _2} = {90^ \circ } + {\theta _1}\)

But in \(\Delta ABC,{\theta _2} = \angle C + {\theta _1}\) (In any triangle, the exterior angle is equal to the sum of the interior opposite angles)

Therefore, \(\angle C = {90^ \circ }.\)

That means there should be at least one common point through which they should pass. Thus,

Slope of line \(AB = \) Slope of line \(BC\)

Slope, \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

But for vertical lines, \({x_2} = {x_1}\)

Therefore,

\(m = \frac{{{y_2} – {y_1}}}{0} = \) undefined

In the same way, the slope of the horizontal line is equal to \(0,\) since the change in y-coordinates is zero.

\(m = \frac{0}{{{x_2} – {x_1}}} = 0\)[for horizontal line]

*Q.1. Find the slope of a line between the points* \(P = \left({0, – 5} \right)\) *and* \(Q= \left({2, 1.} \right)\) ** Ans: **Given the points \(P = \left({0, – 5} \right)\) and \(Q= \left({2, 1.} \right)\) The slope of a straight line between two points \(\left({{x_1},{y_1}} \right)\) and \(\left({{x_2},{y_2}} \right)\) is given by,

\(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

Here, \({x_1} = 0,{x_2} = 2,\,{y_1} = – 5\) and \({y_2} = 1\)

\(\therefore m = \frac{{1 – \left({ – 5} \right)}}{{2 – 0}} = \frac{6}{2} = 3\)

*Q.2. The line p passes through the points* \(\left({3, – 2} \right),\left({12,4} \right)\) *and the line q passes through the points *\(\left({6, – 2} \right)\) *and* \(\left({12, – 2.} \right)\) Is \(p\) *parallel to*\(q\)? * Ans: *Given that the line \(p\) passes through the points \(\left({3, – 2} \right),\left({12,4} \right)\) and the line \(q\) passes through the points \(\left({6, – 2} \right)\) and \(\left({12, – 2.} \right)\)

The slope of a line, \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

The slope of line \(p\) is \({m_1} = \frac{{4 – \left({ – 2} \right)}}{{12 – 3}} = \frac{6}{9} = \frac{2}{3}\)

The slope of line \(q\) is \({m_2} = \frac{{2 – \left({ – 2} \right)}}{{12 – 6}} = \frac{4}{6} = \frac{2}{3}\)

Thus, the slope of line \(p = \) slope of line \(q\)

Therefore, line \(p\) is parallel to line \(q.\)

*Q.3 Show that the points *\(\left({ – 2,5}\right),\left({6, – 1} \right),\) *and* \(\left({2,2} \right)\) *are collinear. *** Ans: **The given vertices are \(A\left({ – 2,5} \right),B\left({6, – 1} \right)\) and \(C\left({2,2} \right)\)

The slope of a line, \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

The slope of line \(AB\) is \({m_1} = \frac{{ – 1 – 5}}{{6 – \left({ – 2} \right)}} = \frac{{ – 6}}{8} = \frac{{ – 3}}{4}\)

The slope of line \(BC\) is \({m_2} = \frac{{2 – \left({ – 1} \right)}}{{2 – 6}} = \frac{3}{{ – 4}} = – \frac{3}{4}\)

Thus, the slope of line \(AB = \) slope of line \(BC\)

Therefore, the points \(A,B,C\) all lie in the same straight line.

Hence the points \(A,B\) and \(C\) are collinear.

*Q.4. Without using the Pythagoras theorem, show that the points* \(\left({1, – 4} \right),\left({2, – 3} \right)\) *and *\(\left({4, – 7} \right)\) * form a right-angled triangle. Ans: Let* the given points be \(A\left({1, – 4} \right),B\left({2, – 3} \right)\) and \(C\left({4, – 7} \right)\)

By the formula of the slope when two pints are given,

The slope of a line, \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

The slope of line \(AB\) is \({m_1} = \frac{{ – 3 + 4}}{{2 – 1}} = \frac{1}{1} = 1\)

The slope of line \(BC\) is \({m_2} = \frac{{ – 7 + 3}}{{4 – 2}} = \frac{{ – 3}}{3} = – 1\)

The slope of line \(AC\) is \({m_3} = \frac{{ – 7 + 4}}{{4 – 1}} = \frac{{ – 3}}{3} = – 1\)

Thus, the slope of line \(AB \times \) slope of line \(AC = \left( 1 \right)\left({ – 1} \right) = – 1\)

\(AB\) is perpendicular to \(AC.\angle A ={90^ \circ }\) Therefore, \(\Delta ABC\) is a right-angled triangle.

*Q.5. Find the slope of a line joining the points *\(P\left({ – 6,1} \right)\)* and* \(Q\left({ – 3,2.} \right)\) ** Ans:** Given the points \(P\left({ – 6,1} \right)\) and \(Q\left({ – 3,2.} \right)\)

The slope of a line, \(m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\)

Here, \({x_1} = – 6,{x_2} = – 3,{y_1} = 1\) and \({y_2} = 2\)

\(m = \frac{{2 – 1}}{{ – 3 + 6}} = \frac{1}{3}\).

In this article, we have learned about the meaning of the slope. We also learned the formula for calculating the slope of the line joining the two points, the slope of parallel lines, the slope of perpendicular lines, and the collinearity of three points. We also got to know about the positive and negative slope and solved some examples.

*Q.1. How to find the slope of a line?*** Ans: **The ratio of the difference between the \(y\)-coordinates and \(x\)-coordinates of the two points that form the line is called the slope of the line. It shows the rise of the line along the \(y\)-axis over the run along \(x\)-axis.

*Q.2. What do you mean by the inclination of a line?*** Ans:** The inclination of a line or the angle of inclination of a line is the angle which a straight line makes with the positive direction of \(X\)-axis measured in the counter-clockwise direction to the part of the line above the \(X\)-axis. The inclination of the line is usually denoted by \(\theta \).

*Q.2. What is the formula to find the slope of a line?*** Ans:** The formula to find the slope of a line is m=(y2-y1)/(x2-x1).

*Q.3. What is the slope between two points? *** Ans:** The slope between two points is calculated by the rate of change in \(y\)-coordinate values by the rate of change in \(x\)-coordinate values.

For example, the slope between the points \(P\left({14,10} \right)\) and \(Q\left({15, – 6} \right)\) is equal to \(m = \frac{{ – 6 – 10}}{{15 – 14}} = – \frac{{16}}{1} = – 16\).

*Q.4. When the slope of a line is undefined? *** Ans:** A vertical line has an undefined slope because all the points on the line have the same \(x\)-coordinate. As a result, the formula used to calculate slope has a denominator of \(0,\) which makes the slope undefined.

*Q.5. How do you know if two lines are perpendicular by slope?*** Ans: **Two lines are perpendicular if and only if the product of their slopes is \( – 1.\).

*We hope this detailed article on Slope of a Line helps you in your preparation. If you get stuck do let us know in the comments section below and we will get back to you at the earliest.*