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Mathematics 7 Online
OpenStudy (anonymous):

What is the equation of the line perpendicular to the line y =4/9 x – 2 that passes through the point (4,3)?

OpenStudy (anonymous):

Perpendicular lines have opposite slope. So since you know the first lines slope you can get the second

OpenStudy (anonymous):

explain please dnt understand

OpenStudy (anonymous):

is the slop 2

OpenStudy (anonymous):

To find the slope of the line perpendicular to this, you want to find the negative inverse. Then from there you can plug in your values to get the "b" value from the equation mx+b

OpenStudy (anonymous):

??

OpenStudy (anonymous):

Okay, so you have a slope of 1 in your original equation. Therefore the slope of the inverse is equal to -1/2. So now you have y=(-1/2)x+b. Plug in 4 for x and 3 for y and solve for b!

OpenStudy (anonymous):

let m1 be the slope of first line and m2 be the slope of second line for perpendicular lines m1 x m2= -1 your m1 here from the first equation is m1=1 since y=mx+c now m1=1 1x m2=-1 m2=-1 therefore equation of the second line use this equation y-y1=m(x-x1) x1= 3 and y1= 4 y-4= -1(x-3) y-4=-x-3 y= -x-3+4 y= -x+1 y= 1-x is your equation for the second line

OpenStudy (anonymous):

That's not quite right I don't think. You need to have the negative INVERSE for the perpendicular line

OpenStudy (anonymous):

@adutton for parallel line slopes are equal and for perpendicular line the product of slopes is -1

OpenStudy (anonymous):

@adutton that is what i did

OpenStudy (anonymous):

@adutton y=mx+c that you agree to tight now look that equation of first line it can be written as y= 1x-2 so m1=1 and since m2 is perpendicular to m1 so 1.m2=-1 m2=-1

OpenStudy (anonymous):

hey guys i edited the question i had left out the 4/9 part sorry

OpenStudy (anonymous):

Slope of the second line would be -9/4 x1=4 y1=3 y-y1=m(x-x1) y-3=-9/4(x-4) y-3=-9/4x+9 y=-9/4x+12

OpenStudy (anonymous):

let m1 be the slope of first line and m2 be the slope of second line for perpendicular lines m1 x m2= -1 your m1 here from the first equation is m1=4/9 since y=mx+c now m1=4/9 4/9 x m2=-1 m2=--9/4 therefore equation of the second line use this equation y-y1=m(x-x1) x1= 3 and y1= 4 y-4= (-9/4)(x-3) 4(y-4)=-9(x-3) 4y-16=-9x+27 4y=-9x+16+27 4y=-9x+43 y=x(-9/4)+(43/4) is your required equation

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