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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Find the equation of the line which is equidistant...
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Find the equation of the line which is equidistant from parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0.

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Re-writing the equation of lines in the form y=mx+c where m is the slope and c is the y-intercept. LINE (1) 9x + 6y = 7 taking 9x term to RHS, dividing by 6 on both sides and then after simplifying we get, y=(-3/2)x + (7/6) The first line has a y-intercept of (7/6). Similarly, LINE (2) 3x+2y...
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Re-writing the equation of lines in the form y=mx+c where m is the slope and c is the y-intercept. LINE (1) 9x + 6y = 7 taking 9x term to RHS, dividing by 6 on both sides and then after simplifying we get, y=(-3/2)x + (7/6) The first line has a y-intercept of (7/6). Similarly, LINE (2) 3x+2y = -6 taking 3x term to RHS, dividing by 2 on both sides and then after simplifying we get, y = (-3/2)x - 3 The second line has a y-intercept of (-3) For both lines slope equal to (-3/2). The slope will be same for the new line as well. From this it is clear that the required line should have a y-intercept, which is exactly half way from these two y-intercepts. The distance between these two intercepts = [(7/6) - (-3)] = [(7/6)+3] taking LCM = [(7 + 18)/6] =25/6. Half way distance between them is thus = (25/6)*(1/2) = 25/12. To find the y-intercept of the new line, which let us call c' we have 2 ways. Either Subtract 25/12 from 7/6 or Add 25/12 to -3 NOTE:- Both will give the same answer for finding c'. Let us check both ways c' = (7/6) - (25/12) taking LCM c' = (14-25)/12 c' = (-11/12) case 2 c' = 25/12 + (-3) taking LCM c' = (25-36)/12 c'= -11/12. Thus equation of line parallel to the above 2 lines will have slope = (-3/2) and y-intercept = (-11/12). This gives us y= (-3/2)x - (11/12) multiplying by 12 on both sides, 12y = (-18)x -11. Bringing all terms to LHS we get 18x+12y +11 = 0. Ans read less
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1st line=9x+6y-7,=>6y=-9y+7=> 6y=- 9x + 7=> y=- (3/2)x + (7/6) ..... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y = - (3/2)x - 3 ........ (2) Line (2) has a slope = m = - 3/2 and y-intercept = c2 = - 3 A line equidistant...
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1st line=9x+6y-7,=>6y=-9y+7=> 6y=- 9x + 7=> y=- (3/2)x + (7/6) ..... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y=- (3/2)x - 3 ........ (2) Line (2) has a slope=m = - 3/2 and y-intercept = c2=- 3 A line equidistant from the two parallel lines is a line parallel to both and mid way between the two Slope of the 3rd line=m = - 3/2 y-intercept of the 3rd line=c = (c1 + c2)/2 = {(7/6) - 3}/2 = - 11/12 Hence the eqn. of this 3rd line is : y=mx + c=> y=- (3/2)x - 11/12 => 12y=- 18x - 11=> 18x + 12y + 11 = 0 read less
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Answers Home All Categories Arts & Humanities Beauty & Style Business & Finance Cars & Transportation Computers & Internet Consumer Electronics Dining Out Education & Reference Entertainment & Music Environment Family & Relationships ...
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Answers Home All Categories Arts & Humanities Beauty & Style Business & Finance Cars & Transportation Computers & Internet Consumer Electronics Dining Out Education & Reference Entertainment & Music Environment Family & Relationships Food & Drink Games & Recreation Health Home & Garden Local Businesses News & Events Pets Politics & Government Pregnancy & Parenting Science & Mathematics Social Science Society & Culture Sports Travel Yahoo Products International About Science & Mathematics > Mathematics Next Equation of line equidistant from parallel lines 9x+6y-7=0 and 3x+2y+6=0? Best AnswerAsker's Choice Dambarudhar answered 3 years ago 1st line : 9x + 6y - 7=0 => 6y=- 9x + 7=> y=- (3/2)x + (7/6) ....... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y=- (3/2)x - 3 ........ (2) Line (2) has a slope=m = - 3/2 and y-intercept = c2=- 3 A line equidistant from the two parallel lines is a line parallel to both and mid way between the two Slope of the 3rd line=m = - 3/2 y-intercept of the 3rd line=c = (c1 + c2)/2 = {(7/6) - 3}/2 = - 11/12 Hence the eqn. of this 3rd line is : y=mx + c=> y=- (3/2)x - 11/12 => 12y=- 18x - 11=> 18x + 12y + 11 = 0 read less
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Maths tuitions for CBSE Class XI - XII , IITJEE MAIN and Advanced

Convert them in one form by 3 X (eq. 2) i.e. 9x + 6y + 18 = 0 , now the value of c for the line equidistant from both shall be (18 - 7 ) / 2 = 11/2
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1st line : 9x + 6y - 7=0 => 6y=- 9x + 7=> y=- (3/2)x + (7/6) ....... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y = - (3/2)x - 3 ........ (2) Line (2) has a slope = m = - 3/2 and y-intercept = c2 = - 3 A line equidistant...
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1st line : 9x + 6y - 7=0 => 6y=- 9x + 7=> y=- (3/2)x + (7/6) ....... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y=- (3/2)x - 3 ........ (2) Line (2) has a slope=m = - 3/2 and y-intercept = c2=- 3 A line equidistant from the two parallel lines is a line parallel to both and mid way between the two Slope of the 3rd line=m = - 3/2 y-intercept of the 3rd line=c = (c1 + c2)/2 = {(7/6) - 3}/2 = - 11/12 Hence the eqn. of this 3rd line is : y=mx + c=> y=- (3/2)x - 11/12 => 12y=- 18x - 11=> 18x + 12y + 11 = 0 read less
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Solution: 3x +2y +6 = 0 can be rewritten as 9x +6y+18=0 Equation of any line equii distance from both the given lines is 9x + 6y +( 18-7)/2=0 9x +6y +11/2=0 (or) 18x +12y+11 = 0 is the required equation.
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The parallel lines are 9x + 6y - 7 = 0 and 9x + 6y + 18 = 0. Let say the line parallel to both lines and equidistant from them be 9x + 6y + d = 0. Hence, d + 7 = 18 - d (applying the distance formula between two parallel lines) d = 11/2 So the required line is 9x + 6y + 11/2 = 0 or, 18x +...
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The parallel lines are 9x + 6y - 7 = 0 and 9x + 6y + 18 = 0. Let say the line parallel to both lines and equidistant from them be 9x + 6y + d = 0. Hence, d + 7 = 18 - d (applying the distance formula between two parallel lines) d = 11/2 So the required line is 9x + 6y + 11/2 = 0 or, 18x + 12y + 11 = 0 read less
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54x + 36y +11=0
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18x + 12y + 11 = 0
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1st line : 9x + 6y - 7=0 => 6y=- 9x + 7=> y=- (3/2)x + (7/6) ....... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y = - (3/2)x - 3 ........ (2) Line (2) has a slope = m = - 3/2 and y-intercept = c2 = - 3 A line...
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1st line : 9x + 6y - 7=0 => 6y=- 9x + 7=> y=- (3/2)x + (7/6) ....... (1) Line (1) has a slope=m = - 3/2 and y-intercept = c1=7/6 2nd line : 3x + 2y + 6=0 => 2y=- 3x - 6=> y=- (3/2)x - 3 ........ (2) Line (2) has a slope=m = - 3/2 and y-intercept = c2=- 3 A line equidistant from the two parallel lines is a line parallel to both and mid way between the two Slope of the 3rd line=m = - 3/2 y-intercept of the 3rd line=c = (c1 + c2)/2 = {(7/6) - 3}/2 = - 11/12 Hence the eqn. of this 3rd line is : y=mx + c=> y=- (3/2)x - 11/12 => 12y=- 18x - 11=> 18x + 12y + 11 = 0 read less
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