find the distance of the point [3,2] from the straight line whose slope is 5 and is passing through the point of intersection of lines x+2y=5and x-3y+5=o

To find the equation of the required line we havr to find out intersection of given line equations 1sti.e intersection of x+2y=5     1 & x-3y=-5         2Subtracting 2 from 1, we get5y=10y=2put y=2 in eq 1x=5-2y=5-2×2x=1 Now equation of a line having the slope m and passing through x1 ,y1, is     y-y1=mx-x1        where mslope & x1 ,y1coordinates satisfying the eq.y-2=5x-15x-y-3=0Now, distance of  3,2 from the  line 5x-y-3=0 is given by d=ax0+by0+ca2+b2   Where x0 ,y0 are coordinates of desired point, here 3,2=5×3-2-352+-12  =1026  = 102626 = 51326  units                                                                                        

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