યામ ભૂમિતિ અભ્યાસ માર્ગદર્શિકા
Definition (વ્યાખ્યા): The slope (gradient) of a line represents its steepness. The two-point form gives the equation of a line passing through any two distinct points on a 2D plane. / ઢાળ એ રેખાની ઢોળાવની તીવ્રતા દર્શાવે છે. દ્વિ-બિંદુ સ્વરૂપ એ 2D સમતલ પરના કોઈપણ બે ભિન્ન બિંદુઓમાંથી પસાર થતી રેખાનું સમીકરણ આપે છે.
Application (વાસ્તવિક ઉપયોગ): Used in Civil Engineering to determine the correct slope (gradient) of roads and drainage systems for water flow. / રોડ અને ગટર વ્યવસ્થામાં પાણીના પ્રવાહ માટે યોગ્ય ઢાળ નક્કી કરવા માટે સિવિલ એન્જિનિયરિંગમાં ઉપયોગ થાય છે.
Detailed Explanation / વિસ્તૃત સમજૂતી:
The slope, usually denoted by $m$, defines how much $y$ changes for a unit change in $x$. If a line passes through points $(x_1, y_1)$ and $(x_2, y_2)$, the slope is the "rise over run".
ઢાળને સામાન્ય રીતે $m$ વડે દર્શાવવામાં આવે છે, જે $x$ માં થતા એક એકમ ફેરફાર સામે $y$ માં થતો ફેરફાર દર્શાવે છે. જો રેખા $(x_1, y_1)$ અને $(x_2, y_2)$ બિંદુઓમાંથી પસાર થતી હોય, તો તેનો ઢાળ શોધી શકાય છે.
Find the equation of the line passing through points $(1,2)$ and $(2,1)$.
Step-by-step:
Given points are $A(1,2)$ and $B(2,1)$.
Slope / ઢાળ $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{2 - 1} = \frac{-1}{1} = -1$
Equation of line / રેખાનું સમીકરણ:
$$ y - y_1 = m(x - x_1) $$
$$ y - 2 = -1(x - 1) $$
$$ y - 2 = -x + 1 \implies x + y - 3 = 0 $$
Find the equation of the line passing through the points $(1, 6)$ and $(-2, 5)$. Also find the slope of the line.
Step-by-step:
Slope $m = \frac{5 - 6}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3}$
Equation / સમીકરણ:
$$ y - y_1 = m(x - x_1) $$
$$ y - 6 = \frac{1}{3}(x - 1) $$
$$ 3y - 18 = x - 1 \implies x - 3y + 17 = 0 $$
If a line makes an angle $45^\circ$ with positive x-axis then find slope of the line.
Step-by-step:
Slope formula based on angle / ખૂણા પરથી ઢાળનું સૂત્ર: $m = \tan \theta$
$$ m = \tan(45^\circ) $$
$$ m = 1 $$
Find the equation of a line passing through $(-6,2)$ and parallel to the line joining the points $(-3,1)$ and $(-5,2)$.
Step-by-step:
Slope of points $(-3,1)$ and $(-5,2)$ / આપેલ બિંદુઓનો ઢાળ:
$$ m_1 = \frac{2 - 1}{-5 - (-3)} = \frac{1}{-2} = -\frac{1}{2} $$
Since line is parallel, $m_2 = m_1 = -1/2$.
Equation through $(-6,2)$ / સમીકરણ: $y - 2 = -\frac{1}{2}(x - (-6))$
$$ 2(y - 2) = -1(x + 6) \implies 2y - 4 = -x - 6 $$
$$ x + 2y + 2 = 0 $$
Definition (વ્યાખ્યા): Different algebraic forms to represent the same straight line depending on the known properties like intercepts (where line crosses axes) or general coefficients. / રેખાને રજૂ કરવાના વિવિધ સ્વરૂપો, જે અંત:ખંડો અથવા સામાન્ય સહગુણકો જેવી જાણીતી કિંમતો પર આધારિત છે.
Application (વાસ્તવિક ઉપયોગ): Intercept form is used in economics and logistics to map resource constraints where resources cross zero-boundaries. / અર્થશાસ્ત્ર અને લોજિસ્ટિક્સમાં સંસાધનોની મર્યાદાઓ દર્શાવવા માટે અંત:ખંડ સ્વરૂપનો ઉપયોગ થાય છે.
Find out equation of the line whose X-intercept is 3 and Y-intercept is 2.
Step-by-step:
Using intercept form / અંત:ખંડ સ્વરૂપ: $a = 3, b = 2$
$$ \frac{x}{a} + \frac{y}{b} = 1 $$
$$ \frac{x}{3} + \frac{y}{2} = 1 $$
Multiply by LCM (6):
$$ 2x + 3y = 6 \implies 2x + 3y - 6 = 0 $$
Find slope of the line $2x - 5y + 3 = 0$.
Step-by-step:
From General Form $Ax + By + C = 0$, $A = 2, B = -5, C = 3$.
Slope / ઢાળ $m = -\frac{A}{B}$
$$ m = -\left(\frac{2}{-5}\right) = \frac{2}{5} $$
Find the equation of line passing through the point $(2, -3)$ and having slope 4.
Step-by-step:
Use slope-point form / ઢાળ-બિંદુ સ્વરૂપ: $m = 4$, $(x_1, y_1) = (2, -3)$
$$ y - y_1 = m(x - x_1) $$
$$ y - (-3) = 4(x - 2) $$
$$ y + 3 = 4x - 8 $$
$$ 4x - y - 11 = 0 $$
Find intercepts of a given line $4x + 7y = 0$ on axes.
Step-by-step:
The equation has no constant term ($C=0$).
Put $y=0$ for X-intercept / x-અંત:ખંડ માટે $y=0$ મુકો:
$$ 4x = 0 \implies x = 0 $$
Put $x=0$ for Y-intercept / y-અંત:ખંડ માટે $x=0$ મુકો:
$$ 7y = 0 \implies y = 0 $$
Intercepts are $(0,0)$.
Find the equation of lines passing through point $A(-8, -10)$ and product of its intercepts on both axis is $-40$.
Step-by-step:
Let intercepts be $a$ and $b$. Given $a \cdot b = -40 \implies b = \frac{-40}{a}$
Equation is $\frac{x}{a} + \frac{y}{b} = 1$. Passing through $(-8, -10)$:
$$ \frac{-8}{a} + \frac{-10}{-40/a} = 1 \implies \frac{-8}{a} + \frac{10a}{40} = 1 $$
$$ \frac{-8}{a} + \frac{a}{4} = 1 \implies -32 + a^2 = 4a \implies a^2 - 4a - 32 = 0 $$
$(a-8)(a+4) = 0 \implies a = 8 \text{ or } a = -4$. Substitute back to find equations.
Definition (વ્યાખ્યા): Two lines are parallel if their slopes are exactly equal. They are perpendicular if the product of their slopes is -1. / જો બે રેખાઓના ઢાળ સમાન હોય, તો તે સમાંતર છે. જો તેમના ઢાળનો ગુણાકાર -1 હોય, તો તે એકબીજાને લંબ છે.
Prove that lines $3x - 2y + 5 = 0$ and $2x + 3y - 7 = 0$ are perpendicular to each other.
Step-by-step:
Slope of Line 1 / રેખા 1 નો ઢાળ: $m_1 = -\frac{3}{-2} = \frac{3}{2}$
Slope of Line 2 / રેખા 2 નો ઢાળ: $m_2 = -\frac{2}{3}$
Product / ગુણાકાર: $m_1 \times m_2 = \left(\frac{3}{2}\right)\left(-\frac{2}{3}\right) = -1$
Hence, lines are perpendicular.
Show that lines $2x - 3y + 5 = 0$ and $8x - 12y + 3 = 0$ are parallel lines.
Step-by-step:
Slope of Line 1: $m_1 = -\frac{2}{-3} = \frac{2}{3}$
Slope of Line 2: $m_2 = -\frac{8}{-12} = \frac{2}{3}$
Since $m_1 = m_2$, the lines are parallel. / ઢાળ સમાન હોવાથી, રેખાઓ સમાંતર છે.
For what value of m, lines $7x + y - 1 = 0$ and $3x - my + 2 = 0$ are perpendicular to each other.
Step-by-step:
Slope $m_1 = -\frac{7}{1} = -7$
Slope $m_2 = -\frac{3}{-m} = \frac{3}{m}$
Condition / શરત: $m_1 m_2 = -1$
$$ (-7)\left(\frac{3}{m}\right) = -1 \implies \frac{-21}{m} = -1 \implies m = 21 $$
If lines $5x - py = 3$ and $2x + 3y = 4$ are parallel, find $p$.
Step-by-step:
Slope $m_1 = \frac{-5}{-p} = \frac{5}{p}$
Slope $m_2 = \frac{-2}{3}$
Condition / શરત: $m_1 = m_2$
$$ \frac{5}{p} = -\frac{2}{3} \implies p = -\frac{15}{2} $$
Definition (વ્યાખ્યા): Methods to quickly write the equation of a line that relates specifically (parallel or perpendicular) to another known line, often requiring a point to find the final constant. / બીજી કોઈ રેખાના સંદર્ભમાં સમાંતર કે લંબ રેખાનું સમીકરણ ઝડપથી શોધવાની રીત.
Find the equation of the line passing through $(2, 4)$ and perpendicular to the line $5x - 7y + 11 = 0$.
Step-by-step:
Equation of perpendicular line form / લંબ રેખાનું સ્વરૂપ:
$$ 7x + 5y + K = 0 $$
Passes through $(2,4)$ / બિંદુ $(2,4)$ મૂકતા:
$$ 7(2) + 5(4) + K = 0 \implies 14 + 20 + K = 0 \implies K = -34 $$
Equation is $7x + 5y - 34 = 0$
Find equation of line that passes through $(-3, 2)$ and parallel to line $x - y + 2 = 0$.
Step-by-step:
Equation of parallel line form / સમાંતર રેખાનું સ્વરૂપ:
$$ x - y + K = 0 $$
Passes through $(-3,2)$ / બિંદુ $(-3,2)$ મૂકતા:
$$ -3 - 2 + K = 0 \implies -5 + K = 0 \implies K = 5 $$
Equation is $x - y + 5 = 0$
Find the equation of line passing through intersection of lines $2x+y=4$ and $x-y=5$ and perpendicular to $3x+4y=0$.
Step-by-step:
Solve $2x+y=4$ and $x-y=5$: Adding yields $3x=9 \implies x=3$. Thus $y=-2$. Intersection: $(3,-2)$.
Line perpendicular to $3x+4y=0$ is $4x-3y+K=0$.
Substitute $(3,-2)$: $4(3)-3(-2)+K=0 \implies 12+6+K=0 \implies K=-18$.
Equation: $4x-3y-18=0$.
Find equation of line passing through $(-2,3)$ and perpendicular to line joining $(-5,2)$ and $(1,-2)$.
Step-by-step:
Slope of points: $m = \frac{-2-2}{1-(-5)} = \frac{-4}{6} = -\frac{2}{3}$.
Perpendicular slope $m_{\perp} = \frac{3}{2}$.
Eq: $y - 3 = \frac{3}{2}(x - (-2)) \implies 2y - 6 = 3x + 6 \implies 3x - 2y + 12 = 0$.
Definition (વ્યાખ્યા): The acute or obtuse angle formed at the intersection of two lines, calculated using their respective slopes. / બે રેખાઓના છેદબિંદુ પર રચાતો લઘુકોણ કે ગુરુકોણ, જે તેમના ઢાળની મદદથી શોધવામાં આવે છે.
Find the angle between two lines $x + y + 1 = 0$ and $2x + 3y + 7 = 0$.
Step-by-step:
$m_1 = -1$, $m_2 = -2/3$
$$ \tan \theta = \left| \frac{-1 - (-2/3)}{1 + (-1)(-2/3)} \right| = \left| \frac{-1/3}{1 + 2/3} \right| = \left| \frac{-1/3}{5/3} \right| = \frac{1}{5} $$
$$ \theta = \tan^{-1}\left(\frac{1}{5}\right) $$
Find the angle between $5x + 6y + 3 = 0$ and $x - 11y + 7 = 0$.
Step-by-step:
$m_1 = -5/6$, $m_2 = 1/11$
$$ \tan \theta = \left| \frac{-5/6 - 1/11}{1 + (-5/6)(1/11)} \right| = \left| \frac{-55/66 - 6/66}{1 - 5/66} \right| $$
$$ = \left| \frac{-61/66}{61/66} \right| = |-1| = 1 $$
$$ \theta = \tan^{-1}(1) = 45^\circ \text{ or } \frac{\pi}{4} $$
Find equations of lines passing through $A(3\sqrt{3}, 4)$ and making angle $\frac{\pi}{6}$ with line $\sqrt{3}x - 3y + 5 = 0$.
Step-by-step:
Slope of given line $m_1 = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$. Let unknown slope be $m$.
$\tan(30^\circ) = \frac{1}{\sqrt{3}} = \left|\frac{m - 1/\sqrt{3}}{1 + m/\sqrt{3}}\right|$
Solving this gives $m = \sqrt{3}$ or $m = 0$. Using point $A(3\sqrt{3}, 4)$ to find lines.
$y - 4 = \sqrt{3}(x - 3\sqrt{3})$ and $y - 4 = 0$.
Definition (વ્યાખ્યા): The locus of points at a fixed distance (radius) from a central point. / કેન્દ્રબિંદુથી સમાન અંતરે (ત્રિજ્યા) આવેલા બિંદુઓનો ગણ.
Application (વાસ્તવિક ઉપયોગ): Used in GPS and navigation to determine location zones within a specific radius of a cell tower. / ચોક્કસ સેલ ટાવરની ત્રિજ્યામાં સ્થાન નક્કી કરવા માટે GPS માં ઉપયોગ થાય છે.
Radius of the circle $x^2 + y^2 = 4$ is:
Step-by-step:
The equation is in form $(x-0)^2 + (y-0)^2 = r^2$.
$$ r^2 = 4 \implies r = 2 $$
Find out equation of the circle having center at $(2, -3)$ and radius 3.
Step-by-step:
Using $(x - h)^2 + (y - k)^2 = r^2$:
$$ (x - 2)^2 + (y - (-3))^2 = 3^2 $$
$$ (x - 2)^2 + (y + 3)^2 = 9 $$
Expand: $x^2 - 4x + 4 + y^2 + 6y + 9 = 9$
$$ x^2 + y^2 - 4x + 6y + 4 = 0 $$
Find the equation of the circle having center $(3, -1)$ and passing through $(-1, 2)$.
Step-by-step:
Find radius using distance formula / ત્રિજ્યા માટે અંતર સૂત્ર વાપરો:
$$ r = \sqrt{(3 - (-1))^2 + (-1 - 2)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = 5 $$
Equation: $(x - 3)^2 + (y + 1)^2 = 5^2$
$$ x^2 - 6x + 9 + y^2 + 2y + 1 = 25 \implies x^2 + y^2 - 6x + 2y - 15 = 0 $$
Find the equation of a circle whose centre is $(2,3)$ and circumference is $4\pi$.
Step-by-step:
Circumference / પરિઘ = $2\pi r = 4\pi \implies r = 2$.
Equation: $(x - 2)^2 + (y - 3)^2 = 2^2$
$$ x^2 - 4x + 4 + y^2 - 6y + 9 = 4 \implies x^2 + y^2 - 4x - 6y + 9 = 0 $$
Definition (વ્યાખ્યા): The expanded polynomial form of a circle equation from which center coordinates and radius can be extracted using specific coefficients. / વર્તુળના સમીકરણનું વિસ્તૃત સ્વરૂપ, જેમાંથી સહગુણકોની મદદથી કેન્દ્ર અને ત્રિજ્યા શોધી શકાય છે.
Find centre of the circle $x^2 + y^2 - 2x + 4y + 1 = 0$.
Step-by-step:
Compare with $x^2+y^2+2gx+2fy+c=0$.
$2g = -2 \implies g = -1$
$2f = 4 \implies f = 2$
Center $(-g, -f) = (1, -2)$.
Find the center and radius of $x^2 + y^2 + 6x - 4y - 3 = 0$.
Step-by-step:
$2g = 6 \implies g = 3$, $2f = -4 \implies f = -2$, $c = -3$.
Center $(-g, -f) = (-3, 2)$.
Radius $r = \sqrt{3^2 + (-2)^2 - (-3)} = \sqrt{9 + 4 + 3} = \sqrt{16} = 4$.
If the radius of the circle $4x^2 + 4y^2 - 16x - 12y + a = 0$ is 2, then find the value of a.
Step-by-step:
Divide by 4: $x^2 + y^2 - 4x - 3y + a/4 = 0$.
$g = -2$, $f = -1.5$, $c = a/4$. Given $r=2$.
$$ \sqrt{(-2)^2 + (-1.5)^2 - a/4} = 2 $$
$$ 4 + 2.25 - a/4 = 4 \implies 2.25 = a/4 \implies a = 9 $$
Find the equation of circle passing through origin and $(1,2)$ and whose center lies on X-axis.
Step-by-step:
Center on X-axis means $k=0$. Let center be $(h,0)$. Eq: $(x-h)^2 + (y-0)^2 = r^2$.
Passes through $(0,0)$: $h^2 = r^2$.
Passes through $(1,2)$: $(1-h)^2 + 4 = r^2$. Substitute $r^2 = h^2$:
$$ 1 - 2h + h^2 + 4 = h^2 \implies 5 - 2h = 0 \implies h = 5/2, r^2 = 25/4 $$
Equation: $(x - 5/2)^2 + y^2 = 25/4 \implies x^2 + y^2 - 5x = 0$.
Definition (વ્યાખ્યા): A tangent is a line that touches the circle at exactly one point. A normal is a line perpendicular to the tangent at the point of contact, always passing through the center. / સ્પર્શક એ એવી રેખા છે જે વર્તુળને માત્ર એક જ બિંદુએ સ્પર્શે છે. અભિલંબ એ સ્પર્શબિંદુ પર સ્પર્શકને લંબ રેખા છે, જે હંમેશા કેન્દ્રમાંથી પસાર થાય છે.
Find the equation of Tangent and Normal to the circle $x^2 + y^2 - 2x - 9 = 0$ at the point $(2,3)$.
Step-by-step:
$g = -1, f = 0, c = -9$. Point $(x_1, y_1) = (2, 3)$.
Tangent: $x(2) + y(3) - 1(x + 2) + 0(y + 3) - 9 = 0$
$$ 2x + 3y - x - 2 - 9 = 0 \implies x + 3y - 11 = 0 $$
Normal: Slope of tangent = $-1/3$, so normal slope = $3$.
Eq: $y - 3 = 3(x - 2) \implies y - 3 = 3x - 6 \implies 3x - y - 3 = 0$.
Find the equation of tangent and normal of the circle $x^2 + y^2 - 2x + 4y - 20 = 0$ at the point $(-2, 2)$.
Step-by-step:
$g = -1, f = 2, c = -20$. $(x_1, y_1) = (-2, 2)$.
Tangent: $x(-2) + y(2) - 1(x - 2) + 2(y + 2) - 20 = 0$
$$ -2x + 2y - x + 2 + 2y + 4 - 20 = 0 \implies -3x + 4y - 14 = 0 \implies 3x - 4y + 14 = 0 $$
Normal: Passes through center $(1, -2)$ and point $(-2, 2)$.
Slope $m = \frac{2 - (-2)}{-2 - 1} = \frac{4}{-3}$.
Eq: $y - 2 = -\frac{4}{3}(x + 2) \implies 3y - 6 = -4x - 8 \implies 4x + 3y + 2 = 0$.
Find the equation of tangent and normal to the circle $x^2 + y^2 - 4x + 2y + 3 = 0$ at the point $(-1,2)$.
Step-by-step:
$g = -2, f = 1, c = 3$. $(x_1, y_1) = (-1, 2)$.
Tangent: $x(-1) + y(2) - 2(x - 1) + 1(y + 2) + 3 = 0$
$$ -x + 2y - 2x + 2 + y + 2 + 3 = 0 \implies -3x + 3y + 7 = 0 \implies 3x - 3y - 7 = 0 $$
Normal: Center is $(2, -1)$.
Slope $m = \frac{2 - (-1)}{-1 - 2} = \frac{3}{-3} = -1$.
Eq: $y - 2 = -1(x + 1) \implies y - 2 = -x - 1 \implies x + y - 1 = 0$.
ઇન્ટરેક્ટિવ પ્રયોગશાળા: યામાક્ષ ભૂમિતિ ટૂલ
Find Eq, Slope & Distance
બે રેખાઓ વચ્ચેનો ખૂણો
Find Center & Radius