SOLUTION: Find the equations of the lines given the following information. passes through the point (7,4) and is perpindicular the line y=-3/4x+2. Any help would be appreciated...

Question 21512: Find the equations of the lines given the following information.
passes through the point (7,4) and is perpindicular the line y=-3/4x+2.
Any help would be appreciated...
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING TO KNOW THE METHOD OF OLVING THESE PROBLEMS
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SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO GIVEN:� There is a line (L1) that passes through the points(8,-3) and (3,3/4).� There is another line (L2) with slope m=2/3 thatintersects L1 at the point (-4,6).� A third line (L3) is parallel to L2 that passesthrough the (7,-13 1/2).� Yet another line (L4) is perpendicular to L3, andpasses through the point (1/2,5 2/3).� The fifth line (L5) has the equation2/5y-6/10x=24/5. Using whatever method, find the following: 2. The point of intersection of L1 and L33. The point of intersection of L1 and L44. The point of intersection of L1 and L55. The point of intersection of L2 and L36. The point of intersection of L2 and L47. The point of intersection of L2 and L58. The point of intersection of L3 and L4 9. The point of intersection of L3 and L510. The point of intersection of L4 and L5 PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF ASTRAIGHT LINE:slope(m)and intercept(c) form...y=mx+cpoint (x1,y1) and slope (m) form...y-y1=m(x-x1)two point (x1,y1)and(x2,y2)form..................... y-y1=((y2-y1)/(x2-x1))*(x-x1)standard linear form..ax+by+c=0..here by transformingwe get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing withslope intercept form we get ...slope = -a/b andintercept = -c/b*****************************************************line (L1) that passes through the points (8,-3) and(3,3/4).eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8)y+3=((15/4(-5)))(x-8)=(-3/4)(x-8)or multiplying with 4 throughout4y+12=-3x+243x+4y+12-24=03x+4y-12=0.........L1There is another line (L2) with slope m=2/3 thatintersects L1 at the point (-4,6).This means (-4,6)lies on both L1 and L2.(you can checkthe eqn.of L1 we got by substituting this point inequation of L1 and see whether it is satisfied).Soeqn.of L2...y-6=(2/3)(x+4)..multiplying with 3 throughout..3y-18=2x+8-2x+3y-26=0.........L2A third line (L3) is parallel to L2 that passesthrough the (7,-13 1/2).lines are parallel mean their slopes are same . so wekeep coefficients of x and y same for both parallellines and change the constant term only..eqn.of L2 from above is ...-2x+3y-26=0.........L2hence L3,its parallel will be ...-2x+3y+k=0..now itpasses through (7,-13 1/2)=(7,-13.5)......substitutingin L3..we get k-2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.ofL3 is........-2x+3y+54.5=0......................L3Yet another line (L4) is perpendicular to L3, andpasses through the point (1/2,5 2/3).lines are perpendicular when the product of theirslopes is equal to -1..so ,we interchange coefficientsof x and y from the first line and insert a negativesign to one of them and then change the constant term.L3 is........-2x+3y+54.5=0......................L3hence L4,its perpendicular will be ..3x+2y+p=0...L4this passes through (1/2,5 2/3)=(1/2,17/3).hence..3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is3x+2y-77/6=0..............L4The fifth line (L5) has the equation 2/5y-6/10x=24/5. now to find a point of intersection means to find apoint say P(x,y) which lies on both the lines ..thatis, it satisfies both the equations..so we have tosimply solve the 2 equations of the 2 lines for x andy to get their point of intersection.For example tofind the point of intersection of L1 and L3 we have tosolve for x and y the 2 equations....3x+4y-12=0.........L1....(1) and-2x+3y+40.5=0......L3.....(2)I TRUST YOU CAN CONTINUE FROM HERE TO GET THEANSWERS.If you have any doubts or get into anydifficulty ,please ask me.venugopal
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