mathematics-class-11-unit-11-chapter-1-straight-lines-l-1-6-68ln5ln-wh0

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>$x x^{\prime} \rightarrow x \text {-axis}$</p>
<p>$y y^{\prime} \rightarrow y \text {-axis } .$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Distance formula</B></Primary></p>
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<p>In right angle $\triangle P Q R$.</p>
<p>$P Q^2=P R^2+Q R^2$</p>
<p>$\Rightarrow d^2=(x_2-x_1)^2+(y_2-y_1)^2$</p>
<p>$\Rightarrow d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 1</B></Primary></p>
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<p>Find the distance between the point P(1,2) and Q(-2,1).</p>
<p><strong>Solution:</strong></p>
<p>By the distance formula</p>
<p>$P Q=\sqrt{\left(x_2-x_1\right)^2+\left(x_2-y_1\right)^2}$</p>
<p>$ =\sqrt{(1+2)^2+(3-1)^2}$</p>
<p>$ =\sqrt{9+4}$</p>
<p>$ =\sqrt{13} $</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 2</B></Primary></p>
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<p>Show that the points $(-3,1),(2,4)$ and $(0,-4)$ are vertices of a right triangle.</p>
<p><strong>Solution:</strong></p>
<p>$ A B^2 =(2+3)^2+(4-1)^2$
$ =25+9=34$</p>
<p>$B C^2 =(2-0)^2+(4+4)^2$
$ =4+64=68$</p>
<p>$A C^2 =(0+3)^2+(-4+1)^2$
$ =9+25=34$</p>
<p>$A B^2+A C^2=34+34=68=B C^2$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>Try this:</p>
<p><strong>1.</strong> Find distance between $(a, b)$ and $(a+c, b+d)$.</p>
<p><strong>2.</strong> $A(1,0), B(-2,3), C(2,-1)$ and $D(5,2)$ are vertices of a parallelogram.</p>
<p><strong>3.</strong> The point $(x, y)$ is on the $x$-axis and 6 units away from the point $(1,4)$, find $x$ & $y$.</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Segment of line</B></Primary></p>
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<p>$\triangle P R S \sim \triangle P Q T$.</p>
<p>$\frac{PS}{PT} = \frac{PR}{PQ}$</p>
<p>$\Rightarrow \frac{x-x_1}{x_2-x_1}=\frac{m}{m+n}$</p>
<p>$\Rightarrow x-x_1=\frac{m\left(x_2-x_1\right)}{m+n}$</p>
<p>$\Rightarrow x=x_1+\frac{m\left(x_2-x_1\right)}{m+n}$</p>
<p>$=\frac{m x_2+n x_1}{m+n}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Segment of line</B></Primary></p>
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<p>similarly $y=\frac{m y_2+n y_1}{m+n}$</p>
<p>when are intersect PQ externally</p>
<p>$x=\frac{m x_2-n  x_1}{m-n} \text { or } \frac{n x_1-m x_2}{n-m}$</p>
<p>$y=\frac{m y_2-n y_1}{m-n} \text { or } \frac{n y_1-m y_2}{n-m}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>$x=\frac{1 \cdot x_2+1 \cdot x_1}{1+1}$</p>
<p>$=\frac{x_1+x_2}{2}$</p>
<p>$y=\frac{y_1+y_2}{2}$</p>
<p>$R(x, y)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 3</B></Primary></p>
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<p>Find the co-ordinates of the point $R$ which divides the segment $P(1,3), Q(-2,1)$ in the ratio</p>
<p>$1:3 .$</p>
<p><strong>Solution:</strong></p>
<p>By section formula</p>
<p>$x=\frac{m x_2+n x_1}{m+n}$
$=\frac{1.1+3 \times(-2)}{1+3}$
$=-\frac{5}{4}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>$y=\frac{m y_2+n y_1}{m+n}$</p>
<p>$=\frac{1 \cdot 3+3 \cdot 1}{1+3}$</p>
<p>$=\frac{6}{4}$</p>
<p>$=\frac{3}{2}$</p>
<p>$\therefore \ R(x, y)=\left(-\frac{5}{4}, \frac{3}{2}\right)$</p>
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<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 4</B></Primary></p>
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<p>Find the coordinates of the Mid point of the segment- $A(4,1), B(3,2)$.</p>
<p><strong>Solution:</strong></p>
<p>Mid Point formula</p>
<p>$M(x, y)=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$</p>
<p>$M(x, y)=\left(\frac{4+3}{2}, \frac{1+2}{2}\right)$</p>
<p>$=\left(\frac{7}{2}, \frac{3}{2}\right)$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 5</B></Primary></p>
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<p>Find the ratio in which the line joining $(-2,2)$ and $(4,5)$ is cut by the $y$-axis?</p>
<p><strong>Solution:</strong></p>
<p>Let $R(0, a)$ divides</p>
<p>$P Q$ in $k: 1$ ratio</p>
<p>$0=\frac{k \times 4+1 \times(-2)}{k+1}$
$ \Rightarrow \quad 4 k-2=0$
$ \Rightarrow \quad k=\frac{1}{2}$</p>
<p>$\therefore$ ratio $=1: 2$</p>
<p>$a=\frac{1 \times 5+2 \times 2}{1+2}=\frac{9}{3}=3$</p>
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<p>$\Delta=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$</p>
<p>$=a_1 (b_2 c_3-b_3 c_2)-a_2 (b_1 c_3-b_3 c_1)$
$ +a_3 (b_1 c_2-b_2 c_1)$</p>
<p>Area of $\Delta$ ABC</p>
<p>$\Delta=\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>Condition for collinearity of three points $(x_1, y_1),(x_2, y_2)$ & $(x_3, y_3)$</p>
<p>$\triangle=\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|=0$</p>
<p>$\Rightarrow$ Points $(x_1, y_1)$ & $(x_2, y_2)$ & $(x_3, y_3)$ are collinear points.</p>
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<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 6</B></Primary></p>
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<p>Show that the points $(2,6),(-8,1)$ and $(-2,4)$ are collinear.</p>
<p><strong>Solution:</strong></p>
<p>Given points $A(2,6), B(-8,1)$ & $C(-2,4)$</p>
<p>$\Delta=\left|\begin{array}{ccc}2 & 6 & 1 \\ -8 & 1 & 1 \\ -2 & 4 & 1\end{array}\right|$</p>
<p>$=2(|x|-\mid x 4)-6(-8 \times 1-(-2) \times 1) +1(-8 \times 4-(-2) \times 1)$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
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<p>$=2(1-4)-6(-8+2)+1(-32+2)$</p>
<p>$=-6+36-30 =-36+36 =0 $</p>
<p>$\therefore \ A, B \ \& \ C$  are collinear point.</p>
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<p>Slope (m) = tan$\theta$</p>
<p>in $\triangle P Q R, \angle R=90^{\circ}$</p>
<p>$\therefore \tan \theta=\frac{Q R}{P R}=\frac{y_2-y_1}{x_2-x_1}$</p>
<p>(i) $ \theta=0 $ $ \ \Rightarrow P Q ||$ x -axis</p>
<p>$\Rightarrow$  slope of $P Q=0$</p>
<p>(ii)$\theta=90^{\circ} $ $ \ \Rightarrow R Q ||$ y-axis</p>
<p>$\Rightarrow$ slope of $PQ=n \cdot d$.</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line lecture 1</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 7</B></Primary></p>
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<p>Find the slope of the line passing through the points $(2,3)$ and $(4,9)$.</p>
<p><strong>Solution:</strong></p>
<p>Slope of PQ</p>
<p>in $m=\frac{y_2-y_1}{x_2-x_1}$</p>
<p>$=\frac{9-3}{4-2}$</p>
<p>$=\frac{6}{2}=3$</p>
<p>$\therefore$ Slope of line $P Q=3$</p>
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