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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: jmosh002 on February 06, 2012, 09:28:00 pm

Title: Help with gradient perpendicular question?
Post by: jmosh002 on February 06, 2012, 09:28:00 pm: Hi guys,
This is a question from methods 1/2.

Given that the lines (4x - 3y = 10) and (4x - Ly = M) are perpendicular and intersect at the point (4,2), find the values of L and M.

Could someone please explain this to me as i think i have forgotten?
Title: Re: Help with gradient perpendicular question?
Post by: Phy124 on February 06, 2012, 10:07:36 pm: Firstly, if 4x - Ly = M has an intersect at (4,2) then:

4(4) - L(2) = M

16 - 2L = M

Secondly if 4x - Ly = M is perpendicular to 4x - 3y = 10, then the gradient of 4x - Ly = M will be the negative reciprocal of 4x - 3y = 10

Rearrange 4x - 3y = 10 to find the gradient:

3y = 4x - 10

$y = \frac{4}{3}x-\frac{10}{3}$

The gradient of 4x - 3y = 10 is $\frac{4}{3}$

Therefore the gradient of 4x - Ly = M will be $- \frac{3}{4}$

Rearrange 4x - Ly = M:

Ly = 4x - M

Now we need to work out what L value will make the gradient of the equation (coefficient of x) = $- \frac{3}{4}$

$-\frac{3}{4} = \frac{4}{L}$

$-\frac{3}{4}L = 4$

$L = \frac{4}{-\frac{3}{4}}$

$L = -\frac{16}{3}$

Sub L back into 16 - 2L = M to find M:

$M = 16- 2\left (-\frac{16}{3} \right )$

$M=16+ \frac{32}{3}$

$M=\frac{48}{3}+ \frac{32}{3}$

$M=\frac{80}{3}$

I hope that's right, I'll have a read of it just to make sure ;)
Title: Re: Help with gradient perpendicular question?
Post by: jmosh002 on February 07, 2012, 07:25:21 am: Quote from: Phy124 on February 06, 2012, 10:07:36 pm
Firstly, if 4x - Ly = M has an intersect at (4,2) then:

4(4) - L(2) = M

16 - 2L = M

Secondly if 4x - Ly = M is perpendicular to 4x - 3y = 10, then the gradient of 4x - Ly = M will be the negative reciprocal of 4x - 3y = 10

Rearrange 4x - 3y = 10 to find the gradient:

3y = 4x - 10

$y = \frac{4}{3}x-\frac{10}{3}$

The gradient of 4x - 3y = 10 is $\frac{4}{3}$

Therefore the gradient of 4x - Ly = M will be $- \frac{3}{4}$

Rearrange 4x - Ly = M:

Ly = 4x - M

Now we need to work out what L value will make the gradient of the equation (coefficient of x) = $- \frac{3}{4}$

$-\frac{3}{4} = \frac{4}{L}$

$-\frac{3}{4}L = 4$

$L = \frac{4}{-\frac{3}{4}}$

$L = -\frac{16}{3}$

Sub L back into 16 - 2L = M to find M:

$M = 16- 2\left (-\frac{16}{3} \right )$

$M=16+ \frac{32}{3}$

$M=\frac{48}{3}+ \frac{32}{3}$

$M=\frac{80}{3}$

I hope that's right, I'll have a read of it just to make sure ;)

Thankyou so much!
Title: Re: Help with gradient perpendicular question?
Post by: butene on February 07, 2012, 08:16:01 pm: n/m