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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Find the equations of the lines which cut off intercepts...
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Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.

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Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b...
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Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b = - 6 b = - 6/a a + b = 1 a - 6/a = 1 a^2 - 6 = a a^2 - a - 6 = 0 a^2 - 3a + 2a - 6 = 0 a(a - 3) + 2( a - 3) = 0 (a-3) ( a + 2) = 0 a = 3 or a = - 2 If a = 3, b would be -6/a = -2 If a = -2, then b would be -6/a = 3 Hence equation of line would be either of : x/ -2 + y/3 = 1 x/ 3 + y / -2 = 1 read less
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Let , X-intercept=a and Y-intercept=b ? Equation the lines: x/a +y/b =1 (intercept form)----(1) So we need to know values of a and b. Now according to the question, a+b=1 -------(2) and a.b= -6 -------(3) Solving for a and b, we would get two solutions: (a=3, b=-2) and (a=-2, b=3) Hence from...
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Let , X-intercept=a and Y-intercept=b ? Equation the lines: x/a +y/b =1 (intercept form)----(1) So we need to know values of a and b. Now according to the question, a+b=1 -------(2) and a.b= -6 -------(3) Solving for a and b, we would get two solutions: (a=3, b=-2) and (a=-2, b=3) Hence from (1), the required Equations of the lines: x/3 +y/(-2) =1 => 2x – 3y=6 x/(-2) + y/3 = 1=> 2y – 3x = 6 read less
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Intercept means the point where a line crosses the axis. x intercept, is denoted by 'a' and y intercept by 'b'. The intercept form of a line is x/a + y/b = 1 here a+b =1 and ab=-6. Solving these 2 equations a*(1-a)= -6 a -a^2 = -6 multiplying both sides by -1 a^2 - a = 6 a^2 - a - 6...
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Intercept means the point where a line crosses the axis. x intercept, is denoted by 'a' and y intercept by 'b'. The intercept form of a line is x/a + y/b = 1 here a+b =1 and ab=-6. Solving these 2 equations a*(1-a)= -6 a -a^2 = -6 multiplying both sides by -1 a^2 - a = 6 a^2 - a - 6 = 0 a^2 + (3-2)a + (-2*3) = 0 (a - 3) * (a + 2)= 0 a = 3; a = -2 Case 1: if a=3 then b= -2, with this the equation is x/3 + y/(-2) = 1 -2x+3y=-6. Simplifying, by multiplying with -1 on both sides we get 2x-3y=6.... this is one equation. Case 2. if a=-2 then b=3 which gives the equation of line as x/(-2) + y/3=1 Taking LCM and cross-multiplying we get 3x-2y=-6....this is the second equation read less
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Solution: Let equation of the required line be x/a + y/b = 1 a+b = 1 ---- (1), ab=- 6 a - b=sqrt(a-b)^2 - 4ab=sqrt(1+24)= plus or minus 5 ---- (2) Adding equations (1) and (2) 2a=6 => a=3 then b=-2 another set== -2 , b=3 Equations x/3 + y/(-2) = 1=> 2x -3y +6=0 3x - 2y...
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Headstart to assured 95+ score in math

2x - 3y = 6 or -3x + 2y = 6. both lines satisfy the criteria. :)
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a+b=1,ab=-6,(1-b)b=-6,square of b -b= 6,square of b -b - 6=0,square of b - 3b+2b-6=0,(b-3)(b+2)=0,b=3and b=-2, if b= -2,then a=the equation is x=3, the equation is x/3+y/-2 =1 and if b= 3,x=-2,the equation is( x/-2)+(y/3)=1.
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Tutor

X/(-2)+y/(3)=1 and x/(3)+y/(-2)=1
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http://www.mathisfunforum.com/viewtopic.php?pid=350267
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Math Tutor

x/3 - y/2 = 1
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2x-3y-6=0 and 3x-2y+6=0.
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