math ncert supplementary exercise 10.4 :-2,3,4

no answer  sry

 
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x+2y-3 + k(4x-y+7) =0 x+2y-3+4xk-ky+7k=0 x(1+4k)+y(2-k)+7k-3=0 Slope of this line = -(1+4k)/(2-k) Also,this line is parallel to the line 5x+4y-20=0 whose slope is -(5/4) As parallel lines have same slope, (1+4k)/(2-k)=5/4 K=2/7 Substituting the value of k Thw reqd eqn is 15x+12y-7=0
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find the equation of the line through the intersection of the lines 2x+3y-4=0 and x-5y=7 that has its x-intercept equal to -4
 
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The equation of required line will be 2x + 3y-4 + k(x-5y-7) =0 then it has its X intercept -4 it means (-4,0) lies on this equation. by putting X = -4 and y = 0 we get the value of k that is -12/11. By putting the value of k in in above equation we get the required equation that is 10x + 93y +40= 0
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Find the equation of the line through the intersection of line 3x+4y=7 & x-y+2=0 and whose slope is 5
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Q2. Answer 15x+12y-7=0
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I hope this will help you🙏

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Question.4

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