Engr Jaydee's blog
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MATHalino - Engineering MathematicsenFactor trinomials mentally! Tips and tricks
https://mathalino.com/blog/factor-trinomials-mentally-tips-and-tricks
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><div class="tex2jax"><form action="/blog" method="post" id="collapse-text-dynamic-form-number-5" accept-charset="UTF-8">
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<div class="collapse-text-text">Have you ever wondered how can we factor trinomials without writing anything on paper, i.e. <strong>mentally</strong>? The best way, unfortunately, is through trial-and-error method. However, to make factoring mentally faster, the trial-and-error must be an educated and systematic one.
<p>Here are some tips and tricks in factoring the trinomial $ax^{2}+bx+c$ mentally. Once you master the techniques in this blog, you can simplify expressions and solve equations that require factoring with "lightning" speed, and impress your friends.</p>
<div class="messages status">
<ul><li>If $a+b+c=0$, then one of the factors is $(x-1)$. The other one is $(ax-c)$.</li>
<li>If $a+c=b$, then one of the factors is $(x+1)$. The other one is $(ax+c)$.</li>
<li>If $b$ and $c$ are both positive, then each factor takes a $+$ sign.</li>
<li>If $b$ is negative and $c$ is positive, then each factor takes a $-$ sign.</li>
<li>If $c$ is negative, then one of the factors take a $+$ sign, and the other takes a $-$ sign.</li>
<ul><li>When selecting combinations of signs in this case, observe it from the resulting outer and inner terms, and the middle term of the original trinomial.</li>
</ul><li>When selecting for combinations of $ax^{2}$, try to eliminate some combinations based on the parity of the middle terms (including the resulting outer and inner terms). <strong>Parity</strong> means whether a number is even or odd.</li>
<li>When selecting for combinations of $c$, try to eliminate some combinations based on the following.</li>
<ul><li>If one of the factors is still factored by the GCF method, but the original trinomial is not factorable by the GCF method.</li>
<li>If either the resulting outer or inner term is too large compared to the middle term of the original polynomial.</li>
</ul><li>Do not do trial-and-error check on all cases, only check ones that you think will yield a middle term close to the middle term of the original trinomial. </li>
<li>Repeatedly practice finding combinations mentally. This will help you find factors and combinations faster, and later on, you will be able to find factors without exerting too much thinking at all.</li>
</ul></div>
<h3>Example 1</h3>
<p>Factor the trinomial $2x^{2}-x-6$.</p>
</div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
Because the first term is $2x^{2}$, we are certain that the factorization takes the form<br />
\[<br />
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\]Taking note that the factors of $6$ are $1\times6$ and $2\times3$, we now do some deductions.
<ul><li>Since the third term $-6$ is negative, the two factors will take opposite signs of constants.</li>
<li>It is certainly not $(2x\square6)(x\square1)$. Otherwise, the first factor is still factorable by GCF as $2(x\square3)$. But we cannot take out $2$ via GCF from the original trinomial.</li>
<li>It is also certainly not $(2x\square1)(x\square6)$, because the outer terms multiply to $(2x)(6)=12x$, which is too large for the middle term $-x$ of the original trinomial. Also, notice that the inner terms multiply to $1(x)=x$, which is too small to reduce $12x$ to $-x$.</li>
</ul><p>This rules out $6=1\times6$. Hence, we are left with $2\times3$ as the factors of $6$.</p>
<ul><li>It is certainly not $(2x\square2)(x\square3)$ for the same reason as $(2x\square6)(x\square1)$.</li>
</ul><p>Hence, we are left with $(2x\square3)(x\square2)$. The outer terms multiply to $2x(2)=4x$, while the inner terms multiply to $3(x)=3x$.<br />
\[<br />
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{4x})<br />
\] Because we need to get the middle term $-x$, and because the two factors take opposite signs, the only possibility is that $4x$ must be negative and $3x$ must be positive.<br />
\[<br />
-4x+3x=-x<br />
\] leading to<br />
\[<br />
(\overbrace{2x\square\underbrace{3)(x}_{3x}\square2}^{-4x})<br />
\] Therefore, the second square must be negative, and the first square must be positive.<br />
\[<br />
(\overbrace{2x+\underbrace{3)(x}_{3x}-2}^{-4x})<br />
\] Therefore, $2x^{2}-x-6=\boxed{(2x+3)(x-2)}$. Take note that all of these deductions are happening mentally.</p>
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<p></p></fieldset><div class="collapse-text-text">
<h3>Example 2</h3>
<p>Factor the trinomial $4x^{2}-19x+21$.</p>
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<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
Because the first term is $4x^{2}$, one possible combination is<br />
\[<br />
(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\] Because the leading coefficient of each factor is $2$, an even number, we are sure that the outer and inner terms are even, and consequently the resulting middle term is even. This is a contradiction, because the middle term of the original polynomial is $-19x$, which is an odd number. Hence, we reject this one and embrace the remaining possible combination<br />
\[<br />
(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\] Because the middle term is negative and the last term is positive, each of the factors must take negative signs.<br />
\[<br />
(4x-\square)(x-\square)<br />
\] The last term $21$ has the possible factors $1\times21$ and $7\times3$. We now do some deductions.
<ul><li>Definitely, $(x-1)$ is not one of the factors because the sum of the coefficients of the terms of the trinomial is $4-19+21=6\neq0$. So, $(4x-21)(x-1)$ is out.</li>
<li>Also, it is definitely not $(4x-1)(x-21)$, because the outer terms multiply to $4x(-21)=-84x$, which is too large.</li>
</ul><p>Hence, we are left with $21=7\times3$.</p>
<ul><li>It is not $(4x-3)(x-7)$, because the outer terms multiply to $(4x)(-7)=-28x$, which is too large compared to the middle term $-19x$.</li>
</ul><p>Therefore, we are left with the factorization $(4x-7)(x-3)$. A customary check of the middle term gives<br />
\[<br />
4x(-3)+(-7)(x)=-12x-7x=-19x<br />
\] which checks. Therefore, $4x^{2}-19x+21=\boxed{(4x-7)(x-3)}$. Take note that all of these deductions are happening mentally.
</p></div>
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<p></p></fieldset><div class="collapse-text-text">
<h3>Example 3</h3>
<p>Factor the trinomial $45x^{2}+79x-124$.</p>
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<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
This trinomial looks scary because of large coefficients. However, notice that the sum of the coefficients and constant terms is zero.<br />
\[<br />
45+79-124=0<br />
\] Therefore, one of the factors is automatically $(x-1)$. The other one is $({\color{blue}{45}}x{\color{red}{\,+\,124}})$. Therefore,<br />
\[<br />
{\color{blue}{45}}x^{2}+79x{\color{red}{\,-\,124}}=\boxed{(x-1)({\color{blue}{45}}x{\color{red}{\,+\,124}})}<br />
\] Observe carefully the colored numbers.
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<p></p></fieldset><div class="collapse-text-text">
<h3>Example 4</h3>
<p>Factor the trinomial $10x^{2}-17xy-27y^{2}$.
</p></div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
Observe that the sum of the coefficients of the first and last terms equals the middle term.<br />
\[<br />
10+(-27)=-17<br />
\] Therefore, one of the factors is automatically $(x+1y)$. The other one is $({\color{blue}{10}}x{\color{red}{\,-\,27}}y)$. Therefore,<br />
\[<br />
{\color{blue}{10}}x^{2}-17xy{\color{red}{\,-\,27}}y^{2}=\boxed{(x+y)({\color{blue}{10}}x{\color{red}{\,-\,27}}y)}<br />
\]Again, observe carefully the colored numbers.
</div>
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<p></p></fieldset><div class="collapse-text-text">
<h3>Example 5</h3>
<p>Factor the trinomial $12x^{2}+7x-12$.
</p></div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Here's the thought process when factoring this mentally.</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
The leading coefficient $12$ has factors $1\times12$, $2\times6$ and $3\times4$. The constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.
<ul><li>We will eliminate $12=2\times6$ immediately, because both $2$ and $6$ are even but the middle term $7x$ is odd.</li>
<li>We will eliminate $12=1\times12$ also, because the only possible combinations which will yield an odd middle term are $(x\square4)(12x\square3)$ and $(x\square12)(12x\square1)$.</li>
<ul><li>$(x\square4)(12x\square3)$ is out, because $(12x\square3)$ is still factorable by GCF, but the original trinomial is not.</li>
<li>$(x\square12)(12x\square1)$ is out, because the inner term gives $12(12x)=144x$, which is too large compared to the middle term $7x$.</li>
</ul></ul><p>So, we are left with $12=3\times4$.<br />
\[<br />
(3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)(4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)<br />
\]Recall that the constant term $-12$ also has factors $1\times12$, $2\times6$ and $3\times4$, signs not included.</p>
<ul><li>Because the constant term is negative, the two factors take opposite signs.</li>
<li>The factor with $4x$ can only take $\pm1$ and $\pm3$ as the constant term. Otherwise, this factor is still factorable via GCF, while the original trinomial is not, a contradiction.</li>
<ul><li>If the factor with $4x$ takes $\pm1$ as the constant term, then the factor with $3x$ takes $\pm12$ as the constant term, since $12=1\times12$. This is a contradiction, because the factor $(3x\square12)$ will now be factorable by GCF. Therefore, we eliminate $\pm1$.</li>
</ul></ul><p>Therefore, we are left with $\pm3$ as the constant term of the factor with $4x$, and $\pm4$ as the constant term of the factor with $3x$, since $12=3\times4$.<br />
\[<br />
(\overbrace{4x\square\underbrace{3)(3x}_{9x}\square4}^{16x})<br />
\]To get the middle term $7x$, we must have $16x+(-9x)=7x$.<br />
\[<br />
(\overbrace{4x\square\underbrace{3)(3x}_{-9x}\square4}^{16x})<br />
\]Therefore, the signs in the squares must be<br />
\[<br />
(\overbrace{4x-\underbrace{3)(3x}_{-9x}+4}^{16x})<br />
\]Hence, $12x^{2}+7x-12=\boxed{(4x-3)(3x+4)}$, and we are done.
</p></div>
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<div class="messages warning">
<h3>Practice problems</h3>
<p>Here are some more trinomials to practice what you have learned in this blog.</p>
<ol><li>$ 3x^2 - 13x - 10 $</li>
<li>$ 40x^2 - 79x + 39 $</li>
<li>$ 8x^2 + 26x + 15 $</li>
<li>$ 16x^2 - 33x - 49 $</li>
<li>$ 18x^2 + 9xy - 20y^2 $</li>
</ol></div>
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</div></div></div></div><div class="field field-name-field-blog-tags field-type-taxonomy-term-reference field-label-inline clearfix"><div class="field-label">Tags: </div><div class="field-items"><div class="field-item even"><a href="/tag/blog/algebra" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Algebra</a></div><div class="field-item odd"><a href="/tag/blog/polynomials" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">polynomials</a></div><div class="field-item even"><a href="/tag/blog/trinomials" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">trinomials</a></div><div class="field-item odd"><a href="/tag/blog/factoring" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">factoring</a></div></div></div>Wed, 15 Sep 2021 09:42:25 +0000Engr Jaydee18754 at https://mathalino.comhttps://mathalino.com/blog/factor-trinomials-mentally-tips-and-tricks#commentsEquations of lines in general form directly from slopes
https://mathalino.com/blog/equations-lines-general-form-directly-slopes
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<div class="collapse-text-text">I always joined the MMC (Metrobank Math Challenge) when I was in high school. MMC always requires that equations of lines be expressed in general ($ax+by+c=0$) form. One way to achieve this is to use classical methods (i.e. point-slope, two-slope or slope-intercept formulas) to get the equations of the lines, and then rewrite them in general form. This was sometimes slow for me, especially in timed questions, and sometimes caused me carelessness in solving. So, when I first learned of this technique in third year high school, I always used this ever since.
<p>In this blog, I will discuss the technique will easily allow us to write the equations of lines <strong>directly</strong> to its general form, without going through the classical methods. All we need is its slope.</p>
<h4>Given a slope and a point</h4>
<p>Constructing an equation of a line from the slope is pretty straightforward. The numerator $A$ of the slope becomes the coefficient of $x$, and the denominator $B$ of the slope becomes the coefficient of $y$. If the slope $A/B$ is positive, the sign between $Ax$ and $By$ is negative (i.e. $Ax-By$). If the slope $A/B$ is negative, the sign between $Ax$ and $By$ is positive (i.e. $Ax+By$). Finally, substitute $(x_1,y_1)$ to $Ax-By$ or $Ax+By$ to get the constant term of the equation of the line.</p>
<p>Mathematically, this method is stated as follows:</p>
<div class="messages status">Given a line in the <em>xy</em>-plane with slope $\displaystyle \frac{A}{B}$ and passing through the point $(x_1, y_1)$, the equation of the line is $$Ax-By=Ax_1-By_1$$
</div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Derivation</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">Using the point-slope formula, we have $$ y-y_1=m(x-x_1) \Longrightarrow y-y_1 = \frac{A}{B}(x-x_1)\Longrightarrow By-By_1=Ax-Ax_1$$giving $\color{red}{\boxed{Ax-By=Ax_1-By_1}}$</div>
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<p></p></fieldset><div class="collapse-text-text"></div>
<p>This method has the advantage that the equation of the line you will produce is already in standard ($ax+by=c$) or general form. Many contests such as MMC require that all equations of lines be expressed in general form. Also, some questions in engineering board exams have all equations of lines in the choices expressed in standard or general form. In both of these cases, this method is faster than using classical methods and then rewriting to standard or general form. You can actually do this method mentally with practice!</p>
<div class="messages warning">If the question requires that the line be expressed in slope-intercept form ($y=mx+b$), or if the choices are expressed in slope-intercept form, then the slope-intercept form and point-slope forms are definitely faster than this method. So, I suggest learning <strong>both</strong> this method and the classical methods to be able to apply them in the appropriate cases.</div>
<p>I hope the short examples below will give you an idea as to how to use this method. I color-highlighted the numbers for us to see where the coefficients go and what substitutions are made.
</p></div>
<p></p><fieldset class="collapse-text-fieldset collapsible form-wrapper"><legend><span class="fieldset-legend">Example 1</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
<table><tr><th>
<div align="center">Given</div>
</th>
<th>
<div align="center">Calculation</div>
</th>
<th>
<div align="center">Equation of the line</div>
</th>
</tr><td>
<div align="center">slope = $\color{red}3/\color{blue}5$, passes through $(\color{magenta}2, \color{magenta}{-6})$</div>
</td>
<td>
<div align="center">$\color{red}3x-\color{blue}5y=\color{red}3(\color{magenta}2)-\color{blue}5(\color{magenta}{-6})=36$</div>
</td>
<td>
<div align="center">$3x-5y-36=0$</div>
</td>
<tr><td>
<div align="center">slope = $-\color{red}{4}/\color{blue}7$, passes through $(\color{magenta}9, \color{magenta}1)$</div>
</td>
<td>
<div align="center">$\color{red}4x+\color{blue}7y=\color{red}4(\color{magenta}9)+\color{blue}7(\color{magenta}1)=43$</div>
</td>
<td>
<div align="center">$4x+7y-43=0$</div>
</td>
</tr><tr><td>
<div align="center">slope = $\color{red}3/\color{blue}2$, passes through $(\color{magenta}{-8}, \color{magenta}0)$</div>
</td>
<td>
<div align="center">$\color{red}3x-\color{blue}2y=\color{red}3(\color{magenta}{-8})-\color{blue}2(\color{magenta}0)=-24$</div>
</td>
<td>
<div align="center">$3x-2y+24=0$</div>
</td>
</tr><tr><td>
<div align="center">slope = -6, passes through $(\color{magenta}{-4}, \color{magenta}{-10})$</div>
</td>
<td>
<div align="center">$\color{red}6x+\color{blue}1y=\color{red}6(\color{magenta}{-4})+\color{blue}1(\color{magenta}{-10})=-34$</div>
</td>
<td>
<div align="center">$6x+y+34=0$</div>
</td>
</tr></table><div class="messages warning">If the slope is an integer, rewrite it as a fraction with denominator 1. For the fourth example, $\displaystyle-6=-\frac{\color{red}6}{\color{blue}1}$.</div>
</div>
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<p></p></fieldset><div class="collapse-text-text">
<h4>Finding the slope of the line in general form</h4>
<p>We can reverse engineer the above method to do the reverse case, i.e. given the equation of the line in general or standard form, find its slope. No need to express them in slope-intercept form!</p>
<div class="messages status">The slope of the line $ax+by+c=0$ is given by $\displaystyle m=-\frac{a}{b}$</div>
</div>
<p></p><fieldset class="collapse-text-fieldset collapsible collapsed form-wrapper"><legend><span class="fieldset-legend">Derivation</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">Let us rewrite the equation to its slope-intercept form. We have<br />
$$ax+by+c=0\Longrightarrow by=-ax-c \Longrightarrow y=-\frac{a}{b}x-\frac{c}{b}$$Therefore, the slope of the line is the coefficient of $x$. That is, $\displaystyle \color{red}{\boxed{m=-\frac{a}{b}}}$.
</div>
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<p></p></fieldset><div class="collapse-text-text">
<p>As with our above method, the numerator of the slope is the coefficient of $x$, while the denominator is the coefficient of $y$. The examples below will demonstrate how to apply this formula.</p>
</div>
<p></p><fieldset class="collapse-text-fieldset collapsible form-wrapper"><legend><span class="fieldset-legend">Example 2</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
<table><tr><th>
<div align="center">Equation</div>
</th>
<th>
<div align="center">Slope</div>
</th>
</tr><tr><td>
<div align="center">$\color{red}9x\color{blue}{+15}y-20=0$</div>
</td>
<td>
<div align="center">$\displaystyle m=-\frac{\color{red}9}{\color{blue}{+15}}=-\frac{3}{5}$</div>
</td>
</tr><tr><td>
<div align="center">$\color{red}8x\color{blue}{-4}y-2021=0$</div>
</td>
<td>
<div align="center">$\displaystyle m=-\frac{\color{red}8}{\color{blue}{-4}}=2$</div>
</td>
</tr></table></div>
</div>
<p></p></fieldset><div class="collapse-text-text">
<h4>Parallel and perpendicular lines</h4>
<p>Recall that if two lines in the $xy$-plane are parallel, then they have the same slope. It follows from the technique that we introduced that these two lines will have the same "form" in the general form, i.e. if one line has the form $ax+by$, then all lines parallel to it will also have the form $ax+by$. They will only differ in the constant term, which can be obtained by substituting the given point.</p>
<div class="messages status">The equation of the line that is parallel to the line $Ax+By=C$ and passing through the point $(x_1,y_1)$ is given by $$Ax+By=Ax_1+By_1$$</div>
<p>Also, recall that if two lines in the $xy$-plane are perpendicular, then the slope of the perpendicular line is the negative reciprocal of the original line. For example, if the original line has slope $a/b$, then the perpendicular line has slope $-b/a$. Using the technique, the original line will have the form $\color{red}ax-\color{blue}by$, whereas the perpendicular line will have the form $\color{red}bx+\color{blue}ay$.</p>
<p>Take note of the difference between $\color{red}ax-\color{blue}by$ and $\color{red}bx+\color{blue}ay$. Notice that the coefficients of $x$ and $y$ switched. Also, notice that the sign between them changed. This gives us the technique for finding equations of perpendicular lines.</p>
<div class="messages status">The equation of the line that is perpendicular to the line $Ax+By=C$ and passing through the point $(x_1,y_1)$ is given by $$Bx-Ay=Bx_1-Ay_1$$</div>
<p>Now, for some examples.</p>
</div>
<p></p><fieldset class="collapse-text-fieldset collapsible form-wrapper"><legend><span class="fieldset-legend">Example 3</span></legend><br /><div class="fieldset-wrapper">
<div class="collapse-text-text">
<table><tr><th>
<div align="center">Given</div>
</th>
<th>
<div align="center">Calculation</div>
</th>
<th>
<div align="center">Equation of the line</div>
</th>
</tr><tr><td>
<div align="center"><ins>parallel</ins> to $\color{red}3x+\color{blue}5y=12$, passes through $(-2, 7)$</div>
</td>
<td>
<div align="center">$\color{red}3x+\color{blue}5y=\color{red}3(-2)+\color{blue}5(7)=29$</div>
</td>
<td>
<div align="center">$3x+5y-29=0$</div>
</td>
</tr><tr><td>
<div align="center"><ins>perpendicular</ins> to $\color{red}9x+\color{blue}2y=0$, passes through $(4, -6)$</div>
</td>
<td>
<div align="center">$\color{blue}2x-\color{red}9y=\color{blue}2(4)-\color{red}9(-6)=62$</div>
</td>
<td>
<div align="center">$2x-9y-62=0$</div>
</td>
</tr></table></div>
</div>
<p></p></fieldset><div class="collapse-text-text">
<h4>Other cases</h4>
<p>The above method can also be used to solve other cases in the equations of lines. Always remember that for this method to work, you always need a slope and a point.</p>
<ul><li><strong>given a slope and a $y$-intercept:</strong> The method directly applies. After all, $y$-intercept is also a point.</li>
<li><strong>given two points:</strong> Find the slope from two points. After that, the method directly applies, using either of the two points.</li>
<li><strong>given two intercepts:</strong> The case of given two points applies. After all, intercepts are just points. Although the two-intercept formula $x/a+y/b=1$ works faster than this technique.</li>
</ul></div>
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</div></div></div></div><div class="field field-name-field-blog-tags field-type-taxonomy-term-reference field-label-inline clearfix"><div class="field-label">Tags: </div><div class="field-items"><div class="field-item even"><a href="/tag/blog/analytic-geometry" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Analytic geometry</a></div><div class="field-item odd"><a href="/tag/blog/slope" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">slope</a></div><div class="field-item even"><a href="/tag/blog/lines" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">lines</a></div></div></div>Fri, 18 Jun 2021 04:01:06 +0000Engr Jaydee18533 at https://mathalino.comhttps://mathalino.com/blog/equations-lines-general-form-directly-slopes#comments