{"id":623856,"date":"2023-08-31T11:40:07","date_gmt":"2023-08-31T11:40:07","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623856"},"modified":"2023-08-31T11:40:47","modified_gmt":"2023-08-31T11:40:47","slug":"kc-sinha-exercise-8-1-mathematics-solution-class-9-chapter-8-lines-and-angles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-8-1-mathematics-solution-class-9-chapter-8-lines-and-angles\/","title":{"rendered":"KC Sinha: Exercise 8.1 – Mathematics Solution Class 9 Chapter 8 Lines and Angles"},"content":{"rendered":"\n\n\n\n\n

Exercise 8.1<\/h3>\n\n\n\n

Question 1 <\/h4>\n\n\n\n

Fill in the blanks in each of the following to make the statement true:<\/strong>
(i) Two distinct points in a plane determine __ line .<\/strong>
Sol : unique
Page 8.13
(ii) A line separates a plane into __ parts namely the __ and the itself.<\/strong>
Sol : three , two half planes , line
(iii) Two distinct __ in  a plane cannot have more than one point in common.<\/strong>
Sol :  lines
(iv) If any ray stands on a line , sum of the two adjacent angles are __<\/strong>
Sol :  180\u00b0
(v) If two lines intersect each other, vertically opposite angles are ___<\/strong>
Sol :  equal<\/p>\n\n\n\n

\n\n\n\n

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

Which of the following statements are true (T) and which are false (F) . Give reasons.<\/strong>
(i) Angles forming a linear pair can both be acute angles<\/strong>
Sol : F
(ii) Angles forming a linear pair are supplementary<\/strong>
Sol : T
(iii) Two distinct lines in a plane can have two points in common<\/strong>
Sol : F
(iv) If two lines intersect and one of the angles so formed is of measure 90\u00b0 <\/strong>then each of the other three angles is of measure 90\u00b0<\/strong>
Sol : T
(v) If angles forming a linear pair are equal, then each of these angles is of measure 90\u00b0 ?<\/strong>
Sol : T<\/p>\n\n\n\n

\n\n\n\n

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

Give answer to the following questions :<\/strong>
(i) If a ray stand on a line , what will be the sum of the two adjacent angles ?<\/strong>
Sol : 180\u00b0
(ii) If sum of two adjacent angles is two right angles , what type of angles will these be ?<\/strong>
Sol : supplementary
(iii) If two lines intersect , what is a the relation between vertically opposite angles ?<\/strong>
Sol : They are equal<\/p>\n\n\n\n

\n\n\n\n

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

Write the sum of all angles (in right angles) formed at any point in a plane.<\/strong>
Sol :
Four right angles or 360\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

Lines AB and CD intersect each other at a point O such that \u2220AOC=\u2220COB. What is the relation between these lines ?<\/strong>
Sol :
Diagram
\u2220AOC=\u2220COB
\u2220AOC+\u2220AOC=180\u00b0
2\u2220AOC=180\u00b0
\u2220AOC=90\u00b0
\"\"
\u2234AB\u22a5CD<\/p>\n\n\n\n

\n\n\n\n

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

If lines AB and CD intersect each other at a point O and \u2220AOC=135\u00b0 , then<\/strong>
\"\"
(a)<\/strong> \u2220AOD = __
Sol :
\u2220AOD+\u2220AOC=180\u00b0
\u2220AOD+135\u00b0=180\u00b0
\u2220AOD=180\u00b0-135\u00b0
\u2220AOD=45\u00b0<\/p>\n\n\n\n

(b)<\/strong> \u2220BOD = __
Sol :
\u2220BOD =\u2220AOC (V.O.A)
\u2220BOD =135\u00b0<\/p>\n\n\n\n

(c)<\/strong> \u2220COB = __
Sol :
\u2220COB =\u2220AOD (V.O.A)
\u2220COB =45\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

If a ray stands on a line, then angles formed between the bisectors of adjacent angles is __<\/strong>
Sol :
OP and OQ bisector of \u2220AOC and \u2220BOC.<\/p>\n\n\n\n

\u2220POC=$\\frac{1}{2}$\u2220AOC
2\u2220POC=\u2220AOC<\/p>\n\n\n\n

\u2220QOC=$\\frac{1}{2}$\u2220BOC
2\u2220QOC=\u2220BOC<\/p>\n\n\n\n

\u2220AOC+\u2220BOC=180\u00b0
2\u2220POC+\u2220QOC=180\u00b0
2\u2220POQ=180\u00b0
\u2220POQ=90\u00b0<\/p>\n\n\n\n

Right angle<\/p>\n\n\n\n

\n\n\n\n

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

Lines AB and CD intersect at point O, Write in degree the measure of the angles between bisectors of \u2220AOC and \u2220BOC .<\/strong>
Sol :
90\u00b0<\/p>\n\n\n\n

\n\n\n\n

Type 1<\/p>\n\n\n\n

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

In the given f\u200cigure, find the value of x<\/strong>
\"In
Sol :
\u2220ACD+\u2220BCD=180\u00b0
5x+4x=180\u00b0
9x=180\u00b0
$x=\\frac{180^{\\circ}}{9}$
x=20\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

In the following f\u200cigure, f\u200cind the value of y.<\/strong>
(i)<\/strong>
\"In
Sol :
\u2220AOC+\u2220COD+\u2220BOD=180\u00b0
5y+3y+2y=180\u00b0
10y=180\u00b0
$y=\\frac{180^{\\circ}}{10}$
y=18\u00b0<\/p>\n\n\n\n

(ii)<\/strong>
\"\"
Sol :
5y+2y+3y+5y+2y+3y=360\u00b0
20y=360\u00b0
$y=\\frac{360^{\\circ}}{20}$
y=18\u00b0<\/p>\n\n\n\n

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

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

In the given f\u200cigure, f\u200cind the value of y<\/strong>
<fig to be added>
Sol :<\/p>\n\n\n\n

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

30\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

In the given f\u200cigure, f\u200cind \u2220AOB in degree<\/strong>
<fig to be added>
Sol :
\u2220DOC+\u2220BOC+\u2220AOC=180\u00b0
x+108\u00b0+2x=180\u00b0
x+2x=72\u00b0
3x=72\u00b0
x=24\u00b0<\/p>\n\n\n\n

\u2220AOB=2x
=2\u00d724\u00b0
=48\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

In the given f\u200cigure, a is greater than b by one third of a right angles. Find the values of a and b .<\/strong>
<fig to be added>
[Hint: Given, $a-b=\\dfrac{1}{3}\\times 90^{\\circ}$
\u21d2a-b=30\u00b0..(i)
Also \u21d2a+b=180\u00b0..(ii)
On solving (i) and (ii) , we get
\u21d22a=210\u00b0
\u21d2a=105\u00b0 , b=180\u00b0-105\u00b0 = 75\u00b0 ]<\/p>\n\n\n\n

\n\n\n\n

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

If a ray stands on a line such that difference of adjacent angles so formed is 30\u00b0 , then find the measure of each adjacent angle in degree.<\/strong>
[Hint: Let a and b be the two adjacent angle. Then , a-b=30\u00b0 and a+b=180\u00b0
Solving , we get a=105\u00b0 , b=75\u00b0]
Sol :
Diagram<\/p>\n\n\n\n

$a-b=\\frac{1}{3}\\times 90^{\\circ}$
a-b=30\u00b0
a=30\u00b0+b..(i)<\/p>\n\n\n\n

\u2220AOC+\u2220BOC=180\u00b0
a+b=180\u00b0
30\u00b0+b+b=180\u00b0
30\u00b0+2b=180\u00b0
2b=180\u00b0-30\u00b0
2b=150\u00b0
b=75\u00b0<\/p>\n\n\n\n

a=30\u00b0+b
a=30\u00b0+75\u00b0
a=105\u00b0<\/p>\n\n\n\n

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

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

In the given figure , what value of x will make POQ a straight line ?<\/strong>
<fig to be added>
[Hint : For POQ to be a line , we must have
2x+3x+10\u00b0=180\u00b0
Hence $x=\\dfrac{170^{\\circ}}{5}=34^{\\circ}$ ]
Sol :
\u2220POR+\u2220QOR=180\u00b0
2x+3x+10\u00b0=180\u00b0<\/p>\n\n\n\n

5x=180\u00b0-10\u00b0<\/p>\n\n\n\n

5x=170\u00b0<\/p>\n\n\n\n

$x=\\frac{170}{5}$<\/p>\n\n\n\n

x=34\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

In the given figures (i) and (ii) , find the values of x in each case
(i)
<fig to be added>
Sol :
\u2220AOD=\u2220BOC
x+45\u00b0=150\u00b0
x=150\u00b0-45\u00b0
x=105\u00b0<\/p>\n\n\n\n

(ii)
<fig to be added>
Sol :
\u2220AOC+\u2220BOC=180\u00b0
6x+30\u00b0+4x=180\u00b0
10x+30\u00b0=180\u00b0
10x=180\u00b0-30\u00b0
10x=150\u00b0
$x=\\frac{150}{10}$
x=15\u00b0<\/p>\n\n\n\n

\n\n\n\n

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

What is the measure of the angle (in degree) which is twice of its supplementary angles ?<\/strong>
Sol :
One supplement angle=x
Twice of supplement angle=2x<\/p>\n\n\n\n

So , we know two supplement angle=180\u00b0<\/p>\n\n\n\n

Then , x+2x=180\u00b0
3x=180\u00b0
$x=\\frac{180^{\\circ}}{3}$
x=60\u00b0<\/p>\n\n\n\n

2x=2\u00d760\u00b0=120\u00b0<\/p>\n\n\n\n

\n\n\n\n

Type 2<\/p>\n\n\n\n

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

Ray OE bisects \u2220AOB and ray OF is opposite to ray OE. Show that \u2220FOB=\u2220FOA<\/strong>
Sol :
Diagram
Given That $\\overrightarrow{O E}$ bisects \u2220AOB
\u2234\u2220AOE=\u2220EOB<\/p>\n\n\n\n

\u2220AOE=\u2220EOB<\/p>\n\n\n\n

Here $\\overrightarrow{O F}$ and $\\overrightarrow{O E}$ form a straight line.<\/p>\n\n\n\n

\u2220EOA+\u2220AOE=180\u00b0
\u2220FOB+\u2220BOE=180\u00b0<\/p>\n\n\n\n

\u2234\u2220FOA+\u2220AOE=\u2220FOB+\u2220BOE
\u2220FOA=FOB<\/p>\n\n\n\n

\n\n\n\n

Question 19<\/h4>\n\n\n\n

If from any point O on a line PQ, two lines OR and OS are drawn in the opposite sides of PQ, such that \u2220POR=\u2220QOS , then prove that OR and OS lie in a line.<\/strong>
Sol :
Diagram
Given: PQ is a line
OR and OS are drawn in the opposite sides of PQ
\u2220POR=\u2220QOS<\/p>\n\n\n\n

To prove: SOR is a line<\/p>\n\n\n\n

Prove: \u2220POR+\u2220ROS=180\u00b0 (Linear pair)<\/p>\n\n\n\n

\u2220QOS+\u2220ROQ=180\u00b0<\/p>\n\n\n\n

\u2234SOR is a line
or
OR and OS lie on a line<\/p>\n\n\n\n

\n\n\n\n

Question 20<\/h4>\n\n\n\n

From any point O , four lines AO, OB , OC and OD are drawn respectively such that \u2220AOB=\u2220COD and \u2220BOC=\u2220DOA , prove that AOC and BOD are straight lines.<\/strong>
Sol :
Diagram
Given: \u2220AOB=\u2220COD..(i)
\u2220BOC=\u2220DOA..(ii)<\/p>\n\n\n\n

from figure-
\u2220AOB+\u2220BOC+\u2220COD+\u2220DOA=360\u00b0
\u21d22\u2220COD+2\u2220DOA=360\u00b0
\u21d2\u2220COD+\u2220DOA$=\\frac{360^{\\circ}}{2}$
\u2220AOC=180\u00b0<\/p>\n\n\n\n

Similarly \u2220BOD=180\u00b0<\/p>\n\n\n\n

Sum of all angles at a point on a straight line is 180\u00b0<\/p>\n\n\n\n

\n\n\n\n

Question 21<\/h4>\n\n\n\n

O is a point on line AB , OC and OD are perpendiculars drawn on AB in opposite directions. Prove that OC and OD lie in a straight line .<\/strong>
Sol :
Diagram<\/p>\n\n\n\n

Given: AB is a straight line<\/p>\n\n\n\n

OC\u27c2AB , OD\u27c2AB<\/p>\n\n\n\n

To Prove: OC and OD lie in a straight line<\/p>\n\n\n\n

Prove :
OC\u27c2AB,OD\u27c2AB<\/p>\n\n\n\n

\u2220COB=90\u00b0 , \u2220BOD=90\u00b0
\u2220COB+\u2220BOD=90\u00b0+90\u00b0=180\u00b0<\/p>\n\n\n\n

\u2234COD is a straight line<\/p>\n\n\n\n

\n\n\n\n

Question 22<\/h4>\n\n\n\n

Two lines AB and CD intersect each other at point O. If line OP bisects \u2220BOD , prove that if OP is produced backwards , then it bisects \u2220AOC. If OP and OQ are respectively bisectors of \u2220BOD and \u2220AOC . Show that the rays OP and OQ are in the same line<\/strong>
\"\"
[Hint : Since OP is the bisector of \u2220BOD
\u2234 \u22201=\u22206;
If OP is produced,
then \u22201=\u22204
and \u22206=\u22203 [vertically opposite angles]
\u2234 \u22203=\u22204
Thus, OQ is the bisector of \u2220AOC
Also \u22202=\u22205 [vertically opposite angles]<\/p>\n\n\n\n

Second part :<\/strong>
Since sum of the angles formed at a point is 360\u00b0
\u2234 \u22201+\u22202++\u22203+\u22204+\u22205+\u22206=360\u00b0
\u21d2 (\u22201+\u22206)+(\u22203+\u22204)+(\u22202+\u22205)=360\u00b0
\u21d2 2\u22201+2\u22203+2\u22202=360\u00b0 [Using above equations]
\u21d2 \u22201+\u22203+\u22202=180\u00b0 \u2234\u2220POQ=180\u00b0
Hence, OP and OQ are in the same line ]
Sol :<\/p>\n\n\n\n

Given: AB and CD intersect each other on P at O<\/p>\n\n\n\n

To prove: OQ bisects \u2220AOC<\/p>\n\n\n\n

Prove: OP bisect \u2220BOD<\/p>\n\n\n\n

\u2220DOP=\u2220BOP..(i)
\u2220BOP=\u2220AOQ..(ii) (V.O.A)
\u2220DOP=\u2220COQ..(iii) (V.O.A)<\/p>\n\n\n\n

From Equation (i), (ii) and (iii):
\u2234\u2220AOQ=\u2220COQ<\/p>\n\n\n\n

then OQ, bisects \u2220AOC<\/p>\n\n\n\n

\n\n\n\n

Question 23<\/h4>\n\n\n\n

(i) In the given figure, lines PQ and RS intersect at Point O. If \u2220POR:\u2220ROQ= 5:7 , find all the angles.<\/strong>
\"\"
[Hint: \u2235\u2220POR+\u2220ROQ=180\u00b0 [By linear pair axiom]
Given , $\\dfrac{\\angle POR}{\\angle ROQ}=\\dfrac{5}{7}$ or $\\dfrac{\\angle POR}{5}=\\dfrac{\\angle ROQ}{7}$ $\\dfrac{(\\angle POR + \\angle ROQ)}{12}=\\dfrac{180^{\\circ}}{12}$
$\\angle POR=5\\times \\dfrac{180^{\\circ}}{12}=75^{\\circ}$ and $\\angle ROQ=7\\times \\dfrac{180^{\\circ}}{12}=105^{\\circ}$
Now , \u2220POS=\u2220ROQ=105\u00b0 and \u2220SOQ=\u2220POR=75\u00b0]<\/p>\n\n\n\n

(ii) Three coplanar lines AB ,CD and EF intersect at a point O, forming angles as shown in the figure. Find the values of x, y , z and v.<\/strong>
\"\"
[Hint: Clearly , \u2220y=50\u00b0 [Vertically opposite angles]
\u2220z=90\u00b0 [Vertically opposite angles]
\u2220v=\u2220x [Vertically opposite angles]
Now, \u2220x=40\u00b0 ,\u2220y=50\u00b0 , \u2220z=90\u00b0 and \u2220v=40\u00b0]<\/p>\n\n\n\n

(iii) In the given figure , find the value of x and then find \u2220BOC, \u2220 FOC , \u2220COA
\"\"
[Hint: \u2235 \u2220DOE=\u2220FOC=2x
Now ray OF stands in line AOB]
\u2234\u2220BOF+\u2220FOC+\u2220COA=180\u00b0
\u21d25x+2x+3x=180\u00b0
\u21d210x=180\u00b0
\u21d2x=18\u00b0
\u2234\u2220BOF=5x=5\u00d718\u00b0=90\u00b0
\u2220FOC=2x=2\u00d718\u00b0=36\u00b0
\u2220COA=3x=3\u00d718\u00b0=54\u00b0<\/p>\n\n\n\n

(iv) In the given f\u200cigure. two straight lines PQ and RS intersect eaach other at O.
If \u2220POT=70\u00b0, find the value of a,b and c
<fig to be added>
[Hint : Since ray OT stands on line RS
\u2234 \u2220ROP+\u2220POT+\u2220TOS=180\u00b0
or 4b+70\u00b0+b=180\u00b0
5b=180\u00b0-70\u00b0=110\u00b0
\u21d2b=22\u00b0; since PQ and RS intersected at O, so
\u2220QOS=\u2220POR or a=4b
a=4\u00d722\u00b0=88\u00b0
Now \u2220ROQ=\u2220POS  [Vertically opposite angles]
\u2234 2c=70\u00b0+b=70\u00b0+22\u00b0=92\u00b0
or $c=\\dfrac{92}{2}=46^{\\circ}$]<\/p>\n\n\n\n

Alternatively,<\/strong>
2c+a=180\u00b0 [\u2235Ray OQ stands on line RS]
\u21d2 2c+88\u00b0=180\u00b0
\u21d2 2c=180\u00b0-88\u00b0
\u21d2 c=46\u00b0<\/p>\n\n\n\n

\n\n\n\n

Question 24<\/h4>\n\n\n\n

If a ray OC stands on AB such that \u2220AOC=\u2220COB , then show that \u2220AOC=90\u00b0<\/strong>
Sol :
Diagram<\/p>\n\n\n\n

Given: A ray OC stands on AB<\/p>\n\n\n\n

Such that \u2220AOC=\u2220COB<\/p>\n\n\n\n

To prove: \u2220AOC=90\u00b0<\/p>\n\n\n\n

Prove
\u2220ADC+\u2220COB=180\u00b0
2\u2220AOC=180\u00b0
$\\angle AOC =\\frac{180^{\\circ}}{2}$
\u2220AOC=90\u00b0<\/p>\n\n\n\n

\n\n\n\n

Question 25<\/h4>\n\n\n\n

Point O is the common end point of the rays OA , OB , OC , OD and OE. Show that \u2220AOB+\u2220BOC+\u2220COD+\u2220DOE+\u2220EOA=360\u00b0<\/strong><\/p>\n\n\n\n

[Hint: Draw a ray OP opposite to ray OA]
Sol :
Diagram
Given: O is the common end point of the ray OA,OB,OC,OD and OE<\/p>\n\n\n\n

To prove: \u2220AOB+\u2220BOC+\u2220COD+\u2220DOE+\u2220EOA=360\u00b0<\/p>\n\n\n\n

Prove: AOD is a straight line or OA and OD are two opposite ray<\/p>\n\n\n\n

\u2220EOA+\u2220DOE=180\u00b0(Linear pair) ..(i)
\u2220AOB+\u2220BOC+\u2220COD=180\u00b0..(i)<\/p>\n\n\n\n

On adding equation (i) and (ii)
\u2220AOB+\u2220BOC+\u2220COD+\u2220DOE+\u2220EOA=360\u00b0<\/p>\n\n\n\n

\n\n\n\n

Question 26<\/h4>\n\n\n\n

In the given f\u200cigure, if each of \u2220AOC and \u2220AOB is 90\u00b0 , show BOC is a line .<\/strong>
\"\"
Sol :
Given: Each of \u2220AOC and \u2220AOB is 90\u00b0<\/p>\n\n\n\n

To prove: BOC is a line<\/p>\n\n\n\n

Prove: \u2220AOC+\u2220AOB=90\u00b0+90\u00b0=180\u00b0<\/p>\n\n\n\n

\u2234BOC is a line<\/p>\n\n\n\n

\n\n\n\n

Question 27<\/h4>\n\n\n\n

In the given f\u200cigure, OE and OF bisect \u2220AOC and \u2220COB respectively and OE\u22a5OF . Show that points A,O,B are collinear.<\/strong>
\"\"
Sol :
Given: OE and OF bisect \u2220AOC and \u2220COB.OE\u27c2OF<\/p>\n\n\n\n

To Prove: AOB is a line<\/p>\n\n\n\n

Prove: OE and OF bisect \u2220AOC and \u2220COB<\/p>\n\n\n\n

$\\angle EOC=\\frac{1}{2}\\angle AOC$<\/p>\n\n\n\n

$\\angle COF=\\frac{1}{2}\\angle COB$<\/p>\n\n\n\n

\u2235OE\u27c2OF\u21d2\u2220EOC=90\u00b0<\/p>\n\n\n\n

\u2220EOC+\u2220COF=90\u00b0<\/p>\n\n\n\n

\u2220AOC+\u2220COB=180\u00b0<\/p>\n\n\n\n

\u2234AOB is a line<\/p>\n\n\n\n

\n\n\n\n

Question 28<\/h4>\n\n\n\n

In the given f\u200cigure, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of \u2220POS , and \u2220SOQ , if \u2220POS=x , then find \u2220ROT.<\/strong>
\"\"
Sol :
Given : Ray OS stands on a line POQ. Ray OR and ray OT are bisectors of \u2220POS and \u2220SOQ.<\/p>\n\n\n\n

To Prove: \u2220ROT=?<\/p>\n\n\n\n

Prove: OR and OT are bisectors of \u2220POS and \u2220SOQ.<\/p>\n\n\n\n

$\\angle ROS=\\frac{1}{2}\\angle POS$
\u21d22\u2220ROS=\u2220POS
\u21d22\u2220SOT=\u2220SOQ<\/p>\n\n\n\n

\u2220POS+\u2220SOQ=180\u00b0(from Linear Pair)
2\u2220ROS+2\u2220SOT=180\u00b0
2(\u2220ROS+\u2220SOT)=180\u00b0
\u2220ROS+\u2220SOT$=\\frac{180^{\\circ}}{90^{\\circ}}$
\u2220ROS+\u2220SOT=90\u00b0<\/p>\n\n\n\n

\u2220ROT=90\u00b0<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Exercise 8.1 Question 1  Fill in the blanks in each of the following to make the statement true:(i) Two distinct points in a plane determine __ line .Sol : uniquePage 8.13(ii) A line separates a plane into __ parts namely the __ and the itself.Sol : three , two half planes , line(iii) Two distinct […]<\/p>\n","protected":false},"author":302,"featured_media":623859,"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":[921],"tags":[],"boards":[],"class_list":["post-623856","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 8.1 - Mathematics Solution Class 9 Chapter 8 Lines and Angles - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Exercise 8.1 Question 1  Fill in the blanks in each of the following to make the statement true:(i) Two distinct points in a plane determine __ line\" \/>\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-8-1-mathematics-solution-class-9-chapter-8-lines-and-angles\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 8.1 - 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