{"id":626095,"date":"2023-09-09T08:27:23","date_gmt":"2023-09-09T08:27:23","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626095"},"modified":"2023-09-09T08:27:34","modified_gmt":"2023-09-09T08:27:34","slug":"kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/","title":{"rendered":"KC Sinha: Exercise 18.1 – Mathematics Solution Class 11 Chapter 18 \u0938\u0930\u0932 \u0930\u0947\u0916\u093e"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Question 1 <\/h4>\n\n\n\n

\u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u091d\u0941\u0915\u093e\u0935 \u0939\u0948 :
[Find the slope of the line whose inclination is]<\/strong><\/p>\n\n\n\n

(i)<\/strong> 0\u00b0
Sol :
\u03b8=0\u00b0<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932 ,m=tan0\u00b0=0<\/p>\n\n\n\n

(ii)<\/strong> 60\u00b0
Sol :<\/p>\n\n\n\n

(iii)<\/strong> 150\u00b0
Sol :
\u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932=tan150\u00b0
=tan(180\u00b0-30\u00b0)
=-tan30\u00b0
$=-\\frac{1}{\\sqrt{3}}$<\/p>\n\n\n\n

(iv)<\/strong> 45\u00b0
Sol :<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092c\u093f\u0928\u094d\u0926\u0941\u0913 \u0938\u0947 \u091c\u093e\u0924\u0940  \u0939\u0941\u0908 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093c\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the slope of the line through the points]<\/strong><\/p>\n\n\n\n

(i)<\/strong> (6,3) \u0924\u0925\u093e (9,3)
Sol :
\u092e\u093e\u0928\u093e A(6,3) ,B(9,3) \u0938\u0947 \u0917\u0941\u091c\u0930\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e AB \u0939\u0948\u0964<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932$=\\frac{3-3}{9-6}=\\frac{0}{3}$
=0<\/p>\n\n\n\n

(ii)<\/strong> (1,2) \u0924\u0925\u093e (4,2)
Sol :<\/p>\n\n\n\n

(iii)<\/strong> (0,9) \u0924\u0925\u093e (-3,0)
Sol :<\/p>\n\n\n\n

(iv)<\/strong> (0,-4) \u0924\u0925\u093e (-6,2)
Sol :<\/p>\n\n\n\n

(v)<\/strong> (3,-2) \u0924\u0925\u093e (3,4)
Sol :<\/p>\n\n\n\n

(vi)<\/strong> (3,-2) \u0924\u0925\u093e (-1,4)
Sol :
$=\\frac{2-(-4)}{-6-0}=\\frac{6}{-6}$
=-1<\/p>\n\n\n\n

(vii)<\/strong> (3,-2) \u0924\u0925\u093e (7,-2)
Sol :<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u092c\u093f\u0928\u094d\u0926\u0941\u0913 (5,6) \u0924\u0925\u093e (2,3) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e (9,-2) \u0924\u0925\u093e (6,-5) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e \u0915\u0947 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u0939\u0948 \u0964
[Show that the line joining (5,6) and (2,3) is parallel to the line through (9,-2) and (6,-5)]
<\/strong> Sol :
\u092e\u093e\u0928\u093e A(5,6),B(2,3) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e AB \u0939\u0948 \u0964
\u0914\u0930 C(9,-2) , D(6,-5) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e CD \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093e\u0932 $=\\frac{3-6}{2-5}=\\frac{-3}{-3}$
=1<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e CD \u0915\u093e \u0922\u093e\u0932 $=\\frac{-5-(-2)}{6-9}=\\frac{-3}{-3}$<\/p>\n\n\n\n

=1<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e CD \u0915\u093e \u0922\u093c\u093e\u0932 $=\\frac{-5-(-2)}{6-9}=\\frac{-3}{-3}$
=1<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932 =\u0930\u0947\u0916\u093e CD \u0915\u093e \u0922\u093c\u093e\u0932
AB||CD<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (2,-5) \u0924\u0925\u093e (-2,5) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e (6,3) \u0924\u0925\u093e (1,1) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e \u092a\u0930 \u0932\u092e\u094d\u092c \u0939\u0948 \u0964
[Show that the line through (2,-5) and (-2,5) is perpendicular to the line through (6,3) and (1,1) ]
<\/strong> Sol :
\u092e\u093e\u0928\u093e A(2,-5) , (-2,5) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e AB \u0939\u0948 \u0964
\u0914\u0930 C(6,3) ,D(1,1) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e CD \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932$=\\frac{5-(-5)}{-2-(2)}=\\frac{10}{-4}$
$=\\frac{-5}{2}$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e CD \u0915\u093e \u0922\u093c\u093e\u0932 $=\\frac{1-3}{1-6}=\\frac{-2}{-5}=\\frac{2}{5}$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932\u00d7\u0930\u0947\u0916\u093e CD \u0915\u093e \u0922\u093c\u093e\u0932
$=\\frac{-5}{2} \\times \\frac{2}{5}$
=-1<\/p>\n\n\n\n

m(AB)\u00d7m(CD)
$=\\frac{-5}{2} \\times \\frac{2}{5}$
=-1<\/p>\n\n\n\n

\u2234AB\u27c2CD<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

\u091c\u093e\u0901\u091a \u0915\u0930\u0947 \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092e\u0947 \u0926\u094b \u0930\u0947\u0916\u093e\u090f\u0901 , \u092a\u0930\u0938\u094d\u092a\u0930 \u0932\u092e\u094d\u092c \u092f\u093e \u0928 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u0914\u0930 \u0928 \u0932\u092e\u094d\u092c \u0939\u0948 \u0964
[Examine whether the two lines in each of the following are parallel perpendicular or neither parallel nor perpendicular]<\/strong><\/p>\n\n\n\n

(i)<\/strong> (-2,6) \u0924\u0925\u093e (4,8) \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0941\u0908
[through (-2,6) and(4,8)]
(8,12) \u0924\u0925\u093e (4,24) \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0941\u0908
<\/p>\n\n\n

[through (8,12) and(4,24)]<\/p>\n\n\n\n

(ii)<\/strong> (9,5) \u0924\u0925\u093e (-1,1) \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0941\u0908
[through (9,5) and (-1,1)]
(8,-3) \u0924\u0925\u093e (3,-5) \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0941\u0908
[through (8,-3) and(3,-5)]
Sol :<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

A(5,-3) , B(8,2) , C(0,0) \u0915\u093f\u0938\u0940 \u0924\u094d\u0930\u093f\u092d\u0941\u091c \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f A \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0941\u0908 \u092e\u0927\u094d\u092f\u093f\u0915\u093e \u092d\u0941\u091c\u093e BC \u092a\u0930 \u0932\u092e\u094d\u092c \u0939\u0948 \u0964
[A(5,-3) , B(8,2) , C(0,0) are the vertices of a triangles. Show that the median from A is perpendicular to the side BC.]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e A(5,-3) , B(8,2) , C(0,0)
\u0915\u093f\u0938\u0940 \u0394ABC \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964<\/p>\n\n\n\n

AD, \u0394ABC \u0915\u0940 \u092e\u0927\u094d\u092f\u093f\u0915\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2234D,BC \u0915\u093e \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

D \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936$=\\left(\\frac{8+0}{2}+\\frac{2+0}{2}\\right)$<\/p>\n\n\n\n

$=\\left(\\frac{8}{2},\\frac{2}{2}\\right)$
=(4,1)<\/p>\n\n\n\n

$m(BC)=\\frac{0-2}{0-8}=\\frac{-2}{-8}=\\frac{1}{4}$<\/p>\n\n\n\n

$m(AD)=\\frac{1-(-3)}{4-5}=\\frac{4}{-1}$
=-4<\/p>\n\n\n\n

m(AD)\u00d7m(BC)$=-4 \\times \\frac{1}{4}$
=-1<\/p>\n\n\n\n

\u2234AD\u27c2BC<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

(i) y \u0915\u093e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947 \u0924\u093e\u0915\u093f (3,y) \u0924\u0925\u093e (2,7) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e (-1,4) \u0924\u0925\u093e (0,6) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e \u0915\u0947 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u0939\u0948 ?
[Determine y so that the line through (3,y) and (2,7) is parallel to the line through (-1,4) and (0,6) ?]
<\/strong> Sol :
\u092e\u093e\u0928\u093e A(3,y) \u0924\u0925\u093e B(2,7) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e AB, \u092c\u093f\u0928\u094d\u0926\u0941\u0913 C(-1,4) \u0924\u0925\u093e D(0,6) \u0938\u0947 \u091c\u093e\u0924\u0940 \u0939\u0941\u0908 \u0930\u0947\u0916\u093e CD \u0915\u0947 \u0938\u092e\u093e\u0902\u0924\u0930 \u0939\u0948 \u0964<\/p>\n\n\n\n

m(AB)=m(CD)<\/p>\n\n\n\n

$\\frac{7-y}{2-3}=\\frac{6-4}{0-(-1)}$<\/p>\n\n\n\n

$\\frac{7-y}{-1}=\\frac{2}{1}$<\/p>\n\n\n\n

7-y=-2
7+2=y
y=9<\/p>\n\n\n\n

(iii)<\/strong> x \u0915\u093e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0947 \u0932\u093f\u090f \u092c\u093f\u0928\u094d\u0926\u0941 (x,-1) ,(2,1) \u0924\u0925\u093e (4,5) \u0938\u0902\u0930\u0947\u0916 \u0939\u0948 \u0964
[Find the value of x fro which the points (x,1),(2,1) and (4,5) arc collinear]
Sol :
\u092e\u093e\u0928\u093e A,B \u0924\u0925\u093e C \u0924\u0940\u0928 \u0938\u0902\u0930\u0947\u0916\u0940 \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932=\u0930\u0947\u0916\u093e BC \u0915\u093e \u0922\u093c\u093e\u0932
=\u0930\u0947\u0916\u093e AC \u0915\u093e \u0922\u093e\u0932<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e A(x,-1) , B(2,1) \u0924\u0925\u093e C(4,5) \u0924\u0940\u0928 \u0938\u0902\u0930\u0947\u0916\u0940 \u092c\u093f\u0902\u0926\u0941  \u0939\u0948\u0964<\/p>\n\n\n\n

m(AB)=n(BC)<\/p>\n\n\n\n

$\\frac{1-(-1)}{2-x}=\\frac{5-1}{4-2}$<\/p>\n\n\n\n

$\\frac{2}{2-x}=\\frac{4}{2}$<\/p>\n\n\n\n

4-2x=2
-2x=2-4
-2x=-2
$x=\\frac{-2}{-2}=1$<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

\u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093c\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u094b y-\u0905\u0915\u094d\u0937\u0902 \u0915\u0940 \u0927\u0928\u093e\u0924\u094d\u092e\u0915 \u0926\u093f\u0936\u093e \u0938\u0947 \u0918\u0921\u093c\u0940 \u0915\u0940 \u0938\u0941\u0908 \u0915\u0940 \u0935\u093f\u092a\u0930\u093f\u0924 \u0926\u093f\u0936\u093e \u092e\u0947 30\u00b0 \u0915\u093e \u0915\u094b\u0923 \u092c\u0928\u093e\u0924\u0940 \u0939\u0948 \u0964
[Find the slope of the line , which makes an angle 30\u00b0 will the positive direction of y-axis measured anti-clockwise]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u0930\u0947\u0916\u093e l \u0926\u094d\u0935\u093e\u0930\u093e x-\u0905\u0915\u094d\u0937 \u092a\u0930,<\/p>\n\n\n\n

\u091d\u0942\u0915\u093e\u0935=90\u00b0+36
=120\u00b0<\/p>\n\n\n\n

m(l)=tan120\u00b0
=tan(90\u00b0+30\u00b0)
=-cot30\u00b0
=-\u221a3<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093c\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u094b \u092e\u0942\u0932 \u092c\u093f\u0928\u094d\u0926\u0941 A(0,-4) \u0914\u0930 B(8,0) \u0915\u0947 \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0948 ?
[Find the slope of a line , which passes through the origin and the mid-point of the line segment joinning the points A(0,-4) and B(8,0)]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e \u092e\u0942\u0932 \u092c\u093f\u0902\u0926\u0941 O(0,0) \u0939\u0948 \u0964<\/p>\n\n\n\n

C,AB \u0915\u093e \u092e\u0926\u094d\u0927\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

C \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 $=\\left(\\frac{0+8}{2},\\frac{-4+0}{2}\\right)$<\/p>\n\n\n\n

$=\\left(\\frac{8}{2},\\frac{-4}{2}\\right)$<\/p>\n\n\n\n

=(4,-2)<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e OC \u0915\u093e \u0922\u093c\u093e\u0932 $=\\frac{-2-0}{4-0}=\\frac{-2}{4}$<\/p>\n\n\n\n

$=-\\frac{1}{2}$<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u092c\u093f\u0928\u094d\u0926\u0941\u0913 (3,-1) \u0924\u0925\u093e (4,-2) \u0915\u094b \u092e\u093f\u0932\u093e\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e \u0915\u0947 \u092c\u0940\u091a \u0915\u093e \u0915\u094b\u0923 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the angle between the x-axis and the line joining the joining the points (3,-1) and (4,-2)]
<\/strong> Sol :
\u092e\u093e\u0928\u093e x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u092c\u093f\u0928\u094d\u0926\u0941\u0913 (3,-1) \u0924\u0925\u093e (4,-2) \u0915\u094b \u092e\u093f\u0932\u093e\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e \u03b8 \u0915\u094b\u0923 \u092c\u0928\u0924\u0940 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093c\u093e\u0932 $=\\frac{-2-(-1)}{4-3}$<\/p>\n\n\n\n

tan\u03b8$=\\frac{-2+1}{1}$<\/p>\n\n\n\n

tan\u03b8=-1<\/p>\n\n\n\n

tan\u03b8=-tan45\u00b0<\/p>\n\n\n\n

tan\u03b8=tan(90\u00b0+45\u00b0)<\/p>\n\n\n\n

tan\u03b8=tan135\u00b0<\/p>\n\n\n\n

\u03b8=135\u00b0<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u090f\u0915 \u0930\u0947\u0916\u093e (x1<\/sub>,y1<\/sub>) \u0924\u0925\u093e (h,k) \u0938\u0947 \u0917\u0941\u091c\u0930\u093e\u0924\u0940 \u0939\u0948 \u0964 \u092f\u0926\u093f \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093e\u0932 m \u0939\u0948 \u0924\u094b \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f
[A Line passes through (x1<\/sub>,y1<\/sub>) and (h,k). If slope of the line is m , show that]
k-y1<\/sub>=m(h-x1<\/sub>)
<\/strong> Sol :
\u092e\u093e\u0928\u093e \u092c\u093f\u0902\u0926\u0941 A(x1<\/sub>,y1<\/sub>) \u0924\u0925\u093e B(h,k) \u0938\u0947 \u0917\u0941\u091c\u0930\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e AB \u0915\u093e \u0922\u093e\u0932 m \u0939\u0948 \u0964<\/p>\n\n\n\n

$m=\\frac{k-y_1}{h-x_1}$<\/p>\n\n\n\n

m(h-x1<\/sub>)=k-y1<\/sub><\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u0922\u093e\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u0908 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (1,1),(2,3) \u0924\u0925\u093e (3,5) \u0938\u0902\u0930\u0947\u0916 \u0939\u0948 \u0964
[Using slopes , show that the points (1,1),(2,3) and (3,5) are collinear]
<\/strong> Sol :
\u092e\u093e\u0928\u093e A(1,1),B(2,3) \u0924\u0925\u093e C(3,5) \u0924\u0940\u0928 \u092c\u093f\u0902\u0926\u0941\u090f\u0901 \u0939\u0948 \u0964<\/p>\n\n\n\n

m(AB)$=\\frac{3-1}{2-1}=\\frac{2}{1}$
=2<\/p>\n\n\n\n

m(BC)$=\\frac{5-3}{3-2}=\\frac{2}{1}$
=2<\/p>\n\n\n\n

m(AC)$=\\frac{5-1}{3-1}=\\frac{4}{2}$
=2<\/p>\n\n\n\n

m(AB)=m(BC)=m(AC)<\/p>\n\n\n\n

\u2234A(1,1),B(2,3),C(3,5) \u0938\u0902\u0930\u0947\u0916 \u0939\u0948 \u0964<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

A(3,4), B(-3,0) \u0924\u0925\u093e C(7,-4) \u0915\u093f\u0938\u0940 \u0924\u094d\u0930\u093f\u092d\u0941\u091c \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f AB \u0924\u0925\u093e AC \u0915\u0947 \u092e\u0927\u094d\u092f \u092c\u093f\u0928\u094d\u0926\u0941\u0913 \u0915\u094b \u092e\u093f\u0932\u093e\u0928\u0947 \u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e BC \u0915\u0947 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u0924\u0925\u093e \u0906\u0927\u0940 \u0939\u0948 \u0964
[A(3,4) ,B(-3,0) and C(7,-4) are the vertices of a triangle . Show that the line joining the mid-points of AB and AC is parallel to BC and is half of it]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e \u0394ABC \u092e\u0947 , AB \u0924\u0925\u093e AC \u0915\u093e \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0915\u094d\u0930\u092e\u0936 D \u0924\u0925\u093e E \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0924\u094b \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0928\u093e \u0939\u0948 \u0903<\/p>\n\n\n\n

\u2235D \u0924\u0925\u093e E \u0915\u094d\u0930\u092e\u0936\u0903 AB \u0924\u0925\u093e AC \u0915\u093e \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

D \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915$=\\left(\\frac{-3+3}{2},\\frac{0+4}{2}\\right)$
$=\\left(\\frac{0}{2},\\frac{4}{2}\\right)$
=(0,2)<\/p>\n\n\n\n

D \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915$=\\left(\\frac{7+3}{2},\\frac{-4+4}{2}\\right)$<\/p>\n\n\n\n

$=\\left(\\frac{10}{2},\\frac{0}{2}\\right)$<\/p>\n\n\n\n

=(5,0)<\/p>\n\n\n\n

m(DE)$=\\frac{0-2}{5-0}=-\\frac{2}{5}$<\/p>\n\n\n\n

m(BC)$=\\frac{-4-0}{7-(-3)}=\\frac{-4}{10}$
$=-\\frac{2}{5}$<\/p>\n\n\n\n

m(DE)=m(BC)
DE||BC<\/p>\n\n\n\n

DE$=\\sqrt{(5-0)^2+(0-2)^2}$
$=\\sqrt{25+4}=\\sqrt{25}$<\/p>\n\n\n\n

BC$=\\sqrt{(-4-0)^2+(7-(-3))^2}$
$=\\sqrt{16+100}=\\sqrt{116}$
$=\\sqrt{2\\times 2\\times 29}=2\\sqrt{29}$<\/p>\n\n\n\n

$DE=\\frac{1}{2}BC$<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

\u0924\u0940\u0928 \u092c\u093f\u0928\u094d\u0926\u0941 A(x1<\/sub>,y1<\/sub>) ,B(x2<\/sub>,y2<\/sub>) \u0924\u0925\u093e C(x,y) \u0938\u0902\u0930\u0947\u0916 \u0939\u0948 \u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f
[Three points A(x1<\/sub>,y1<\/sub>) ,B(x2<\/sub>,y2<\/sub>) and C(x,y) are collinear. Prove that (x-x1<\/sub>)(y2<\/sub>-y1<\/sub>)=(x2<\/sub>-x1<\/sub>)(y-y1<\/sub>)
<\/strong> Sol :
\u0926\u093f\u092f\u093e \u0939\u0948(Given)<\/p>\n\n\n\n

A(x1<\/sub>,y1<\/sub>) ,B(x2<\/sub>,y2<\/sub>) \u0924\u0925\u093e C(x,y) \u0938\u0902\u0930\u0947\u0916 \u0939\u0948<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

m(AC)=m(AB)<\/p>\n\n\n\n

$\\frac{y-y_1}{x-x_1}=\\frac{y_2-y_!}{x_2-x_1}$<\/p>\n\n\n\n

(x-x1<\/sub>)(y2<\/sub>-y1<\/sub>)=(x2<\/sub>-x1<\/sub>)(y-y1<\/sub>)<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

\u0922\u093e\u0932 \u0915\u0940 \u0905\u0935\u0927\u093e\u0930\u0923\u093e \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (-2,-1),(4,0),(3,3) \u0924\u0925\u093e (-3,2) \u090f\u0915 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964
[By using the concept of slope, show that (-2,-1),(4,0),(3,3) and (-3,2) are the vertices of a parallelogram]
<\/strong> Sol :
\u092e\u093e\u0928\u093e A(-2,-1) ,B(4,0),C(3,3) \u0924\u0925\u093e D(-3,2) \u091a\u093e\u0930 \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

m(AB)$=\\frac{0-(-1)}{4-(-2)}=\\frac{1}{6}$<\/p>\n\n\n\n

m(BC)$=\\frac{3-0}{3-4}=\\frac{3}{-1}$
=-3<\/p>\n\n\n\n

m(CD)$=\\frac{2-3}{-3-3}=\\frac{-1}{-6}=\\frac{1}{6}$<\/p>\n\n\n\n

m(AD)$=\\frac{2-(1)}{-3-(-2)}=\\frac{3}{-1}$<\/p>\n\n\n\n

=-3<\/p>\n\n\n\n

$\\begin{array}{l|l}m(AB)=m(CD)&m(BC)=m(AD)\\\\AB||CD&BC||AD\\end{array}$<\/p>\n\n\n\n

\u2234ABCD \u090f\u0915 \u0938\u092e\u093e\u0902\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0939\u0948 \u0964<\/p>\n\n\n\n

A(-2,-1),B(4,0),C(3,3) \u0924\u0925\u093e D(-3,2) \u0938\u092e\u093e\u0902\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

\u090f\u0915 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 (4,1),(1,7),(-6,0) \u0924\u0925\u093e (-1,-9) \u0939\u0948 \u0964 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u0907\u0938 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u092e\u0927\u094d\u092f \u092c\u093f\u0928\u094d\u0926\u0941 \u090f\u0915 \u0938\u092e\u093e\u0928\u094d\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u092c\u0928\u093e\u0924\u0947 \u0939\u0948 \u0964
[A quadrilateral has vertices (4,1),(1,7),(-6,0) and (-1,-9). Show that the mid-points of the sides of sides of this quadrilateral form a parallelogram]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e A(4,1),B(1,7),C(-6,0) and D(-1,-9) \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c ABCD \u0915\u0947 \u0936\u0940\u0930\u094d\u0937 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e P,Q,R \u0924\u0925\u093e S \u0915\u094d\u092e\u0936\u0903 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c ABCD \u0915\u0947 \u092d\u0941\u091c\u093e\u0913 AB,BC,CD \u0924\u0925\u093e AD \u0915\u0947 \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0924\u094b \u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0928\u093e \u0939\u0948 \u0903 PQRS \u090f\u0915 \u0938\u092e\u093e\u0902\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092a\u094d\u0930\u092e\u093e\u0923
P,Q,R and S \u0915\u094d\u0930\u092e\u0936\u0903 AB,BC,CD \u0924\u0925\u093e AD \u0915\u0947 \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

P \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 $=\\left(\\frac{4+1}{2},\\frac{1+7}{2}\\right)$
$=\\left(\\frac{5}{2},\\frac{8}{2}\\right)$
$=\\left(\\frac{5}{2},2\\right)$<\/p>\n\n\n\n

Q \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 $=\\left(\\frac{1-6}{2},\\frac{7+0}{2}\\right)$
$=\\left(-\\frac{5}{2},\\frac{7}{2}\\right)$<\/p>\n\n\n\n

R \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 $=\\left(-\\frac{-1+(-6)}{2},\\frac{-9+0}{2}\\right)$
$=\\left(\\frac{-3}{2},\\frac{-9}{2}\\right)$<\/p>\n\n\n\n

S \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 $=\\left(-\\frac{-1+4}{2},\\frac{-9+1}{2}\\right)$
$=\\left(\\frac{3}{2},\\frac{-8}{2}\\right)$
$=\\left(\\frac{3}{2},-4\\right)$<\/p>\n\n\n\n

m(PQ)$=\\dfrac{\\frac{7}{2}-4}{-\\frac{5}{2}-\\frac{5}{2}}$
$=\\dfrac{\\frac{7-8}{2}}{\\frac{-5-5}{2}}$
$=\\frac{-1}{-10}=\\frac{1}{10}$<\/p>\n\n\n\n

m(QR)$=\\dfrac{\\frac{-9}{2}-\\frac{7}{2}}{-\\frac{7}{2}-\\left(-\\frac{5}{2}\\right)}$
$=\\dfrac{\\frac{-16}{2}}{\\frac{-2}{2}}$
=8<\/p>\n\n\n\n

m(RS)$=\\dfrac{\\frac{-9}{2}-4}{-\\frac{7}{2}-\\frac{3}{2}}$
$=\\dfrac{\\frac{-9+8}{2}}{\\frac{-7-3}{2}}$
$=\\frac{-1}{-10}=\\frac{1}{10}$<\/p>\n\n\n\n

m(PS)$=\\dfrac{4-(-4)}{-\\frac{5}{2}-\\frac{3}{2}}$
$=\\dfrac{8}{\\frac{2}{2}}$
=8<\/p>\n\n\n\n

$\\begin{array}{l|l}m(PQ)=m(RS)&m(QR)=m(PS)\\\\PQ||RS&QR||PS\\end{array}$<\/p>\n\n\n\n

\u2234PQRS \u090f\u0915 \u0938\u092e\u093e\u0902\u0924\u0930 \u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0939\u0948 \u0964<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f \u0938\u092e\u092c\u093e\u0939\u0941 \u0924\u094d\u0930\u093f\u092d\u0941\u091c \u0915\u0940 \u092e\u093e\u0927\u094d\u092f\u093f\u0915\u093e \u0938\u0902\u0917\u0924 \u092d\u0941\u091c\u093e \u092a\u0930 \u0932\u092e\u094d\u092c \u0939\u094b\u0924\u0940 \u0939\u0948 \u0964
[Prove that a median of an equilateral triangle is perpendicular to the corresponding side of the triangle]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e ABC \u090f\u0915 \u0938\u092e\u092c\u093e\u0939\u0941 \u0924\u094d\u0930\u093f\u092d\u0941\u091c \u0939\u0948 \u0964<\/p>\n\n\n\n

\u091c\u0939\u093e\u0901 AB=BC=AC=2a<\/p>\n\n\n\n

\u091c\u093f\u0938\u0915\u0940 \u092d\u0941\u091c\u093e BC x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e O,BC \u0915\u093e \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964 \u091c\u094b \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 \u0905\u0915\u094d\u0937\u094b \u0915\u0947 \u092e\u0942\u0932 \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

AB=AC=2a<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e A \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915=(p.q)<\/p>\n\n\n\n

AB=AC
AB2<\/sup>=AC2<\/sup>
(p+a)2<\/sup>+(q.0)2<\/sup>=(p-a)2<\/sup>+(q-0)2<\/sup>
p2<\/sup>+a2<\/sup>+2pa=p2<\/sup>+a2<\/sup>-2pa
4pa=0
p=0<\/p>\n\n\n\n

\u0905\u0924\u0903 A y-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0938\u094d\u0925\u093f\u0924 \u0939\u0948 \u0964<\/p>\n\n\n\n

OA=$\\sqrt{AC^2-OC^2}$
$=\\sqrt{(2a)^2}-a^2$
$q=\\sqrt{3}a$<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e BE \u0924\u0925\u093e CF \u0915\u094d\u0930\u092e\u0936
B \u0924\u0925\u093e C \u0936\u093f\u0930\u094d\u0937 \u0938\u0947 \u0916\u093f\u091a\u0940 \u091c\u093e\u0928\u0947 \u0935\u093e\u0932\u0940 \u092e\u093e\u0927\u094d\u092e\u093f\u0915 \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2234E \u0924\u0925\u093e F \u0915\u094d\u0930\u092e\u0936\u0903 AC \u0924\u0925\u093e AC \u092e\u0927\u094d\u092f \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0964<\/p>\n\n\n\n

E \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915$=\\left(\\frac{a+0}{2},\\frac{\\sqrt{3}a+0}{2}\\right)$
$=\\left(\\frac{a}{2},\\frac{\\sqrt{3}a}{2}\\right)$<\/p>\n\n\n\n

F \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915$=\\left(\\frac{-a+0}{2},\\frac{0+\\sqrt{3}a}{2}\\right)$
$=\\left(\\frac{-a}{2},\\frac{\\sqrt{3}a}{2}\\right)$<\/p>\n\n\n\n

m(BE)$=\\left(\\dfrac{\\frac{\\sqrt{3}a}{2}-0}{\\frac{a}{2}-(-a)}\\right)$
$=\\dfrac{\\frac{\\sqrt{3}a}{2}}{\\frac{a+2a}{2}}$
$=\\frac{\\sqrt{3}a}{3a}=\\frac{\\sqrt{3}}{3}$<\/p>\n\n\n\n

m(AC)$=\\frac{\\sqrt{3}a-0}{0-a}=\\frac{\\sqrt{3}a}{-a}$
=-\u221a3<\/p>\n\n\n\n

m(BE)\u00d7m(AC)$=\\frac{\\sqrt{3}}{3}\\times (-\\sqrt{3})$
$=\\frac{-3}{3}=-1$<\/p>\n\n\n\n

\u2234BE\u27c2AC<\/p>\n\n\n\n

\u0905\u0924\u0903 \u092e\u093e\u0927\u094d\u092f\u093f\u0915\u093e BE \u0394ABC \u0915\u0940 \u092d\u0941\u091c\u093e AC \u092a\u0930 \u0932\u0902\u092c \u0939\u0948 \u0964<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Question 18<\/h4>\n\n\n\n

\u0938\u093e\u092c\u093f\u0924 \u0915\u0930\u0947 \u0915\u093f \u0915\u093f\u0938\u0940 \u0935\u093f\u0937\u092e\u0915\u094b\u0923 \u0938\u092e\u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u0935\u093f\u0915\u0930\u094d\u0923 \u092a\u0930\u0938\u094d\u092a\u0930 \u0932\u092e\u094d\u092c \u0939\u0948 \u0964
[Prove that the diagonals of a rhombus are at right angels]
<\/strong> Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

\u092e\u093e\u0928\u093e OABC \u090f\u0915 \u0935\u093f\u0937\u092e\u0915\u094b\u0923 \u0938\u092e\u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0939\u0948 , \u091c\u093f\u0938\u0915\u0940 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092d\u0941\u091c\u093e a \u0939\u0948 \u0964<\/p>\n\n\n\n

O \u092e\u0942\u0932 \u092c\u093f\u0902\u0926\u0941 \u0939\u0948 \u0924\u0925\u093e OA x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0939\u0948 \u0964 \u0938\u092e\u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0940 \u090a\u0901\u091a\u093e\u0908 b \u0939\u0948 \u0964<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e CE \u0924\u0925\u093e BF \u0932\u0902\u092c \u0939\u0948 x-\u0905\u0915\u094d\u0937 \u092a\u0930 ,<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e OE=C<\/p>\n\n\n\n

\u0394COE\u2a6d\u0394BAF (R.H.S)
OE=AF=C<\/p>\n\n\n\n

O \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915\u0943(0,0) ,<\/p>\n\n\n\n

A \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915\u0943=(a,0)<\/p>\n\n\n\n

B \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915=(a+c,b)<\/p>\n\n\n\n

C \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915=(c,b)<\/p>\n\n\n\n

\u0394COE \u092e\u0947 ,<\/p>\n\n\n\n

OC2<\/sup>=OE2<\/sup>+CE2<\/sup>
a2<\/sup>=c2<\/sup>+b2<\/sup><\/p>\n\n\n\n

m(OB)$=\\frac{b-0}{a+c-0}$
$=\\frac{b}{a+c}$<\/p>\n\n\n\n

m(AC)$=\\frac{b-0}{c-a}$
$=\\frac{b}{c-a}$<\/p>\n\n\n\n

m(OB)\u00d7m(AC)$=\\frac{b}{c+a}\\times \\frac{b}{c-a}$
$=\\frac{b^2}{c^2-a^2}=\\frac{b^2}{-b^2}$<\/p>\n\n\n\n

\u2234\u0935\u093f\u0937\u092e\u0915\u094b\u0923 \u0938\u092e\u091a\u0924\u0941\u0930\u094d\u092d\u0941\u091c \u0915\u0947 \u0935\u093f\u0915\u0930\u094d\u0923 \u092a\u0930\u0938\u094d\u092a\u0930 \u0932\u0902\u092c \u0939\u094b\u0924\u0947 \u0939\u0948 \u0964<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 11 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1  \u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u091d\u0941\u0915\u093e\u0935 \u0939\u0948 :[Find the slope of the line whose inclination is] (i) 0\u00b0Sol :\u03b8=0\u00b0 \u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932 ,m=tan0\u00b0=0 (ii) 60\u00b0Sol : (iii) 150\u00b0Sol :\u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932=tan150\u00b0=tan(180\u00b0-30\u00b0)=-tan30\u00b0$=-\\frac{1}{\\sqrt{3}}$ (iv) 45\u00b0Sol : Question 2 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092c\u093f\u0928\u094d\u0926\u0941\u0913 \u0938\u0947 \u091c\u093e\u0924\u0940  \u0939\u0941\u0908 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093c\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964[Find the slope of the line through […]<\/p>\n","protected":false},"author":302,"featured_media":626111,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[919],"tags":[],"boards":[],"class_list":["post-626095","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 18.1 - Mathematics Solution Class 11 Chapter 18 \u0938\u0930\u0932 \u0930\u0947\u0916\u093e - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1  \u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u091d\u0941\u0915\u093e\u0935 \u0939\u0948 : (i) 0\u00b0Sol :\u03b8=0\u00b0 \u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932 ,m=tan0\u00b0=0 (ii) 60\u00b0Sol : (iii) 150\u00b0Sol :\u0930\u0947\u0916\u093e \u0915\u093e\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 18.1 - Mathematics Solution Class 11 Chapter 18 \u0938\u0930\u0932 \u0930\u0947\u0916\u093e\" \/>\n<meta property=\"og:description\" content=\"Question 1  \u0909\u0938 \u0930\u0947\u0916\u093e \u0915\u0940 \u0922\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u093e \u091d\u0941\u0915\u093e\u0935 \u0939\u0948 : (i) 0\u00b0Sol :\u03b8=0\u00b0 \u0930\u0947\u0916\u093e \u0915\u093e \u0922\u093c\u093e\u0932 ,m=tan0\u00b0=0 (ii) 60\u00b0Sol : (iii) 150\u00b0Sol :\u0930\u0947\u0916\u093e \u0915\u093e\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-09T08:27:23+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-09T08:27:34+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-19-1-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"9 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 18.1 – Mathematics Solution Class 11 Chapter 18 \u0938\u0930\u0932 \u0930\u0947\u0916\u093e\",\"datePublished\":\"2023-09-09T08:27:23+00:00\",\"dateModified\":\"2023-09-09T08:27:34+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\"},\"wordCount\":1361,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-19-1-scaled.jpg\",\"articleSection\":[\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-18-1-mathematics-solution-class-11-chapter-18-saral-rekha\/\",\"name\":\"KC Sinha: Exercise 18.1 - 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